# Pascal's triangle

Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.

Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.

Its first few rows look like this:

```    1
1 1
1 2 1
1 3 3 1
```

where each element of each row is either 1 or the sum of the two elements right above it.

For example, the next row of the triangle would be:

1   (since the first element of each row doesn't have two elements above it)
4   (1 + 3)
6   (3 + 3)
4   (3 + 1)
1   (since the last element of each row doesn't have two elements above it)

So the triangle now looks like this:

```    1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```

Each row   n   (starting with row   0   at the top) shows the coefficients of the binomial expansion of   (x + y)n.

Write a function that prints out the first   n   rows of the triangle   (with   f(1)   yielding the row consisting of only the element 1).

This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.

Behavior for   n ≤ 0   does not need to be uniform, but should be noted.

## 11l

Translation of: Python
```F pascal(n)
V row = [1]
V k = [0]
L 0 .< max(n, 0)
print(row.join(‘ ’).center(16))
row = zip(row [+] k, k [+] row).map((l, r) -> l + r)

pascal(7)```
Output:
```       1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
```

## 360 Assembly

Translation of: PL/I
```*        Pascal's triangle         25/10/2015
PASCAL   CSECT
USING  PASCAL,R15         set base register
LA     R7,1               n=1
LOOPN    C      R7,=A(M)           do n=1 to m
BH     ELOOPN             if n>m then goto
MVC    U,=F'1'            u(1)=1
LA     R8,PG              pgi=@pg
LA     R6,1               i=1
LOOPI    CR     R6,R7              do i=1 to n
BH     ELOOPI             if i>n then goto
LR     R1,R6              i
SLA    R1,2               i*4
L      R3,T-4(R1)         t(i)
L      R4,T(R1)           t(i+1)
AR     R3,R4              t(i)+t(i+1)
ST     R3,U(R1)           u(i+1)=t(i)+t(i+1)
LR     R1,R6              i
SLA    R1,2               i*4
L      R2,U-4(R1)         u(i)
XDECO  R2,XD              edit u(i)
MVC    0(4,R8),XD+8       output u(i):4
LA     R8,4(R8)           pgi=pgi+4
LA     R6,1(R6)           i=i+1
B      LOOPI              end i
ELOOPI   MVC    T((M+1)*(L'T)),U   t=u
XPRNT  PG,80              print
LA     R7,1(R7)           n=n+1
B      LOOPN              end n
ELOOPN   XR     R15,R15            set return code
M        EQU    11                 <== input
T        DC     (M+1)F'0'          t(m+1) init 0
U        DC     (M+1)F'0'          u(m+1) init 0
PG       DC     CL80' '            pg     init ' '
XD       DS     CL12               temp
YREGS
END    PASCAL```
Output:
```   1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1
```

## 8th

One way, using array operations:

```\ print the array
: .arr \ a -- a
( . space ) a:each ;

: pasc \ a --
\ print the row
.arr cr
dup
\ create two rows from the first, one with a leading the other with a trailing 0
[0] 0 a:insert swap 0 a:push
\ add the arrays together to make the new one
' n:+ a:op ;

\ print the first 16 rows:
[1] ' pasc 16 times
```

Another way, using the relation between element 'n' and element 'n-1' in a row:

```: ratio \ m n -- num denom
tuck n:- n:1+ swap ;

\ one item in the row: n m
: pascitem \ n m -- n
r@ swap
ratio
n:*/ n:round int
dup . space ;

\ One row of Pascal's triangle
: pascline \ n --
>r 1 int dup . space
' pascitem
1 r@ loop rdrop drop cr ;

\ Calculate the first 'n' rows of Pascal's triangle:
: pasc \ n
' pascline 0 rot loop cr ;

15 pasc
```

## Action!

```PROC Main()
BYTE count=[10],row,item
CHAR ARRAY s(5)
INT v

FOR row=0 TO count-1
DO
v=1
FOR item=0 TO row
DO
StrI(v,s)
Position(2*(count-row)+4*item-s(0),row+1)
Print(s)
v=v*(row-item)/(item+1)
OD
PutE()
OD
RETURN```
Output:
```                  1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
```

The specification of auxiliary package "Pascal". "First_Row" outputs a row with a single "1", "Next_Row" computes the next row from a given row, and "Length" gives the number of entries in a row. The package is also used for the Catalan numbers solution [[1]]

```package Pascal is

type Row is array (Natural range <>) of Natural;

function Length(R: Row) return Positive;

function First_Row(Max_Length: Positive) return Row;

function Next_Row(R: Row) return Row;

end Pascal;
```

The implementation of that auxiliary package "Pascal":

```package body Pascal is

function First_Row(Max_Length: Positive) return Row is
R: Row(0 .. Max_Length) := (0 | 1 => 1, others => 0);
begin
return R;
end First_Row;

function Next_Row(R: Row) return Row is
S: Row(R'Range);
begin
S(0) := Length(R)+1;
S(Length(S)) := 1;
for J in reverse 2 .. Length(R) loop
S(J) := R(J)+R(J-1);
end loop;
S(1) := 1;
return S;
end Next_Row;

function Length(R: Row) return Positive is
begin
return R(0);
end Length;

end Pascal;
```

The main program, using "Pascal". It prints the desired number of rows. The number is read from the command line.

```with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line, Pascal; use Pascal;

procedure Triangle is

Row: Pascal.Row := First_Row(Number_Of_Rows);

begin
loop
-- print one row
for J in 1 .. Length(Row) loop
end loop;
exit when Length(Row) >= Number_Of_Rows;
Row := Next_Row(Row);
end loop;
end Triangle;
```
Output:
```>./triangle 12
1
1    1
1    2    1
1    3    3    1
1    4    6    4    1
1    5   10   10    5    1
1    6   15   20   15    6    1
1    7   21   35   35   21    7    1
1    8   28   56   70   56   28    8    1
1    9   36   84  126  126   84   36    9    1
1   10   45  120  210  252  210  120   45   10    1
1   11   55  165  330  462  462  330  165   55   11    1```

## ALGOL 68

```PRIO MINLWB = 8, MAXUPB = 8;
OP MINLWB = ([]INT a,b)INT: (LWB a<LWB b|LWB a|LWB b),
MAXUPB = ([]INT a,b)INT: (UPB a>UPB b|UPB a|UPB b);

OP + = ([]INT a,b)[]INT:(
[a MINLWB b:a MAXUPB b]INT out; FOR i FROM LWB out TO UPB out DO out[i]:= 0 OD;
out[LWB a:UPB a] := a; FOR i FROM LWB b TO UPB b DO out[i]+:= b[i] OD;
out
);

INT width = 4, stop = 9;
FORMAT centre = \$n((stop-UPB row+1)*width OVER 2)(q)\$;

FLEX[1]INT row := 1; # example of rowing #
FOR i WHILE
printf((centre, \$g(-width)\$, row, \$l\$));
# WHILE # i < stop DO
row := row[AT 1] + row[AT 2]
OD```
Output:
```                     1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
```

## ALGOL W

```begin
% prints the first n lines of Pascal's triangle lines %
% if n is <= 0, no output is produced                 %
procedure printPascalTriangle( integer value n ) ;
if n > 0 then begin
integer array pascalLine ( 1 :: n );
pascalLine( 1 ) := 1;
for line := 1 until n do begin
for i := line - 1 step - 1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
pascalLine( line ) := 1;
write( s_w := 0, " " );
for i := line until n do writeon( s_w := 0, "   " );
for i := 1 until line do writeon( i_w := 6, s_w := 0, pascalLine( i ) )
end for_line ;
end printPascalTriangle ;

printPascalTriangle( 8 )

end.```
Output:
```                              1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
```

## Amazing Hopper

```#include <jambo.h>
#define Mulbyandmoveto(_X_)   Mul by '_X_', Move to '_X_'

Main
filas=0, Get arg numeric '2', Move to 'filas'
i=0, r=""
Loop if( var 'i' Is less than 'filas' )
c=1, j=0
Set 'c' To str, Move to 'r'
Loop if ( var 'j' Is less than 'i' )
Set 'i' Minus 'j', Plus one 'j', Div it; Mul by and move to 'c'
Multi cat ' r, "\t", Str(c) '; Move to 'r'
++j
Back
Printnl 'r'
++i
Back
End```
Output:
```\$ hopper jm/pascal.jambo 14
1
1	1
1	2	1
1	3	3	1
1	4	6	4	1
1	5	10	10	5	1
1	6	15	20	15	6	1
1	7	21	35	35	21	7	1
1	8	28	56	70	56	28	8	1
1	9	36	84	126	126	84	36	9	1
1	10	45	120	210	252	210	120	45	10	1
1	11	55	165	330	462	462	330	165	55	11	1
1	12	66	220	495	792	924	792	495	220	66	12	1
1	13	78	286	715	1287	1716	1716	1287	715	286	78	13	1

```

## APL

Pascal' s triangle of order ⍵

### Dyalog APL

```{⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵}
```

example

```{⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵}5
```
```                   1
1      1
1      2      1
1     3      3      1
1     4      6      4     1
1     5     10     10     5     1
```

### GNU APL

GNU APL doesn't allow multiple statements within lambdas so the solution is phrased differently:

```{{⍉⍵∘.!⍵} 0,⍳⍵}
```

example

```{{⍉⍵∘.!⍵} 0,⍳⍵} 3
```
```1 0 0 0
1 1 0 0
1 2 1 0
1 3 3 1
```

## AppleScript

Drawing n rows from a generator:

```-------------------- PASCAL'S TRIANGLE -------------------

-- pascal :: Generator [[Int]]
on pascal()
script nextRow
on |λ|(row)
zipWith(my plus, {0} & row, row & {0})
end |λ|
end script
iterate(nextRow, {1})
end pascal

--------------------------- TEST -------------------------
on run
showPascal(take(7, pascal()))
end run

------------------------ FORMATTING ----------------------

-- showPascal :: [[Int]] -> String
on showPascal(xs)
set w to length of intercalate("   ", item -1 of xs)
script align
on |λ|(x)
|center|(w, space, intercalate("   ", x))
end |λ|
end script
unlines(map(align, xs))
end showPascal

------------------------- GENERIC ------------------------

-- center :: Int -> Char -> String -> String
on |center|(n, cFiller, strText)
set lngFill to n - (length of strText)
if lngFill > 0 then
set strPad to replicate(lngFill div 2, cFiller) as text
if lngFill mod 2 > 0 then
cFiller & strCenter
else
strCenter
end if
else
strText
end if
end |center|

-- intercalate :: String -> [String] -> String
on intercalate(sep, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, sep}
set s to xs as text
set my text item delimiters to dlm
return s
end intercalate

-- iterate :: (a -> a) -> a -> Generator [a]
on iterate(f, x)
script
property v : missing value
property g : mReturn(f)'s |λ|
on |λ|()
if missing value is v then
set v to x
else
set v to g(v)
end if
return v
end |λ|
end script
end iterate

-- length :: [a] -> Int
on |length|(xs)
set c to class of xs
if list is c or string is c then
length of xs
else
2 ^ 30 -- (simple proxy for non-finite)
end if
end |length|

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min

-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- plus :: Num -> Num -> Num
on plus(a, b)
a + b
end plus

-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
set out to {}
if n < 1 then return out
set dbl to {a}

repeat while (n > 1)
if (n mod 2) > 0 then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate

-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set end of ys to xs's |λ|()
end repeat
return ys
else
missing value
end if
end take

-- unlines :: [String] -> String
on unlines(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set str to xs as text
set my text item delimiters to dlm
str
end unlines

-- unwords :: [String] -> String
on unwords(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, space}
set s to xs as text
set my text item delimiters to dlm
return s
end unwords

-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
set lng to min(|length|(xs), |length|(ys))
if 1 > lng then return {}
set xs_ to take(lng, xs) -- Allow for non-finite
set ys_ to take(lng, ys) -- generators like cycle etc
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs_, item i of ys_)
end repeat
return lst
end tell
end zipWith
```
Output:
```              1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
1   6   15   20   15   6   1```

## Arturo

Translation of: Nim
```pascalTriangle: function [n][
triangle: new [[1]]

loop 1..dec n 'x [
'triangle ++ @[map couple (last triangle)++[0] [0]++(last triangle) 'x -> x\[0] + x\[1]]
]

return triangle
]

loop pascalTriangle 10 'row [
print pad.center join.with: " " map to [:string] row 'x -> pad.center x 5 60
]
```
Output:
```                             1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84    126   126   84    36     9     1```

## AutoHotkey

ahk forum: discussion

```n := 8, p0 := "1"        ; 1+n rows of Pascal's triangle
Loop %n% {
p := "p" A_Index, %p% := v := 1, q := "p" A_Index-1
Loop Parse, %q%, %A_Space%
If (A_Index > 1)
%p% .= " " v+A_LoopField, v := A_LoopField
%p% .= " 1"
}
; Triangular Formatted output
VarSetCapacity(tabs,n,Asc("`t"))
t .= tabs "`t1"
Loop %n% {
t .= "`n" SubStr(tabs,A_Index)
Loop Parse, p%A_Index%, %A_Space%
t .= A_LoopField "`t`t"
}
Gui Add, Text,, %t%      ; Show result in a GUI
Gui Show
Return

GuiClose:
ExitApp
```

Alternate

Works with: AutoHotkey L
```Msgbox % format(pascalstriangle())
Return

format(o) ; converts object to string
{
For k, v in o
s .= IsObject(v) ? format(v) "`n" : v " "
Return s
}
pascalstriangle(n=7) ; n rows of Pascal's triangle
{
p := Object(), z:=Object()
Loop, % n
Loop, % row := A_Index
col := A_Index
, p[row, col] := row = 1 and col = 1
? 1
: (p[row-1, col-1] = "" ; math operations on blanks return blanks; I want to assume zero
? 0
: p[row-1, col-1])
+ (p[row-1, col] = ""
? 0
: p[row-1, col])
Return p
}
```

n <= 0 returns empty

## AWK

```\$ awk 'BEGIN{for(i=0;i<6;i++){c=1;r=c;for(j=0;j<i;j++){c*=(i-j)/(j+1);r=r" "c};print r}}'
```
Output:
```
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

```

## Bait

```// Create a Pascal's triangle with a given number of rows.
// Returns an empty array for row_nr <= 0.
fun pascals_triangle(row_nr i32) [][]i32 {
mut rows := [][]i32

// Iterate over all rows
for r := 0; r < row_nr; r += 1 {
// Store the row above the current one
mut above := rows[r - 1]

// Fill the current row. It contains r + 1 numbers
for i := 0; i <= r; i += 1 {
// First number is always 1
if i == 0 {
rows.push([1]) // Push new row
}
// Last number is always 1
else if i == r {
rows[r].push(1)
}
// Other numbers are the sum of the two numbers above them
else {
rows[r].push(above[i - 1] + above[i])
}
}
}

return rows
}

// Helper function to pretty print triangles.
// It still get's ugly once numbers have >= 2 digits.
fun print_triangle(triangle [][]i32) {
for i, row in triangle {
// Create string with leading spaces
mut s := ' '.repeat(triangle.length - i - 1)

// Add each number to the string
for n in row {
s += n.str() + ' '
}

// Print and trim the extra trailing space
println(s.trim_right(' '))
}
}

fun main() {
print_triangle(pascals_triangle(7))
}```
Output:
```      1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
```

## BASIC

### Summing from Previous Rows

Works with: FreeBASIC
Works with: QBasic
Works with: QB64

This implementation uses an array to store one row of the triangle. DIM initializes the array values to zero. For first row, "1" is then stored in the array. To calculate values for next row, the value in cell (i-1) is added to each cell (i). This summing is done from right to left so that it can be done on-place, without using a tmp buffer. Because of symmetry, the values can be displayed from left to right.

Space for max 5 digit numbers is reserved when formatting the display. The maximum size of triangle is 100 rows, but in practice it is limited by screen space. If the user enters value less than 1, the first row is still always displayed.

```DIM i             AS Integer
DIM row           AS Integer
DIM nrows         AS Integer
DIM values(100)   AS Integer

INPUT "Number of rows: "; nrows
values(1) = 1
PRINT TAB((nrows)*3);"  1"
FOR row = 2 TO nrows
PRINT TAB((nrows-row)*3+1);
FOR i = row TO 1 STEP -1
values(i) = values(i) + values(i-1)
PRINT USING "##### "; values(i);
NEXT i
PRINT
NEXT row
```

### Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4
```10 let row = 12
20 for i = 0 to row
30   let c = 1
40   print tab (37-i*3);
50   for k = 0 to i
60     print using " #### ";c;
70     let c = c*(i-k)/(k+1)
80   next
90   print
100 next
```

### MSX Basic

Works with: MSX BASIC version any
Works with: GW-BASIC
```10 INPUT "Number of rows? "; R
20 FOR I = 0 TO R-1
30  LET C = 1
40  FOR K = 0 TO I
50   PRINT USING "####"; C;
60   LET C = C * (I-K) / (K+1)
70  NEXT K
80  PRINT
90 NEXT I
```

### QBasic

Same code as BASIC

### QB64

Same code as BASIC

### Yabasic

```input "Number of rows? " r
for i = 0 to r-1
c = 1
for k = 0 to i
print c using "####";
c = c*(i-k)/(k+1)
next
print
next
```

## Batch File

Based from the Fortran Code.

```@echo off
setlocal enabledelayedexpansion

::The Main Thing...
cls
echo.
set row=15
call :pascal
echo.
pause
exit /b 0
::/The Main Thing.

::The Functions...
:pascal
set /a prev=%row%-1
for /l %%I in (0,1,%prev%) do (
set c=1&set r=
for /l %%K in (0,1,%row%) do (
if not !c!==0 (
call :numstr !c!
set r=!r!!space!!c!
)
set /a c=!c!*^(%%I-%%K^)/^(%%K+1^)
)
echo !r!
)
goto :EOF

:numstr
::This function returns the number of whitespaces to be applied on each numbers.
set cnt=0&set proc=%1&set space=
:loop
set currchar=!proc:~%cnt%,1!
if not "!currchar!"=="" set /a cnt+=1&goto loop
set /a numspaces=5-!cnt!
for /l %%A in (1,1,%numspaces%) do set "space=!space! "
goto :EOF
::/The Functions.
```
Output:
```    1
1    1
1    2    1
1    3    3    1
1    4    6    4    1
1    5   10   10    5    1
1    6   15   20   15    6    1
1    7   21   35   35   21    7    1
1    8   28   56   70   56   28    8    1
1    9   36   84  126  126   84   36    9    1
1   10   45  120  210  252  210  120   45   10    1
1   11   55  165  330  462  462  330  165   55   11    1
1   12   66  220  495  792  924  792  495  220   66   12    1
1   13   78  286  715 1287 1716 1716 1287  715  286   78   13    1
1   14   91  364 1001 2002 3003 3432 3003 2002 1001  364   91   14    1

Press any key to continue . . .```

## BBC BASIC

```      nrows% = 10

colwidth% = 4
@% = colwidth% : REM Set column width
FOR row% = 1 TO nrows%
PRINT SPC(colwidth%*(nrows% - row%)/2);
acc% = 1
FOR element% = 1 TO row%
PRINT acc%;
acc% = acc% * (row% - element%) / element% + 0.5
NEXT
PRINT
NEXT row%
```
Output:
```                     1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1```

## BCPL

```get "libhdr"

let pascal(n) be
for i=0 to n-1
\$(  let c = 1
for j=1 to 2*(n-1-i) do wrch(' ')
for k=0 to i
\$(  writef("%I3 ",c)
c := c*(i-k)/(k+1)
\$)
wrch('*N')
\$)

let start() be pascal(8)```
Output:
```                1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1```

## Befunge

```0" :swor fo rebmuN">:#,_&> 55+, v
v01*p00-1:g00.:<1p011p00:\-1_v#:<
>g:1+10p/48*,:#^_\$ 55+,1+\: ^>\$\$@
```
Output:
```Number of rows: 10

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5  10  10  5  1
1  6  15  20  15  6  1
1  7  21  35  35  21  7  1
1  8  28  56  70  56  28  8  1
1  9  36  84  126  126  84  36  9  1  ```

## BQN

Displays n rows.

```Pascal ← {(0⊸∾+∾⟜0)⍟(↕𝕩)⋈1}

•Show¨Pascal 6
```
```⟨ 1 ⟩
⟨ 1 1 ⟩
⟨ 1 2 1 ⟩
⟨ 1 3 3 1 ⟩
⟨ 1 4 6 4 1 ⟩
⟨ 1 5 10 10 5 1 ⟩
```

## Bracmat

```( out\$"Number of rows? "
& get':?R
& -1:?I
&   whl
' ( 1+!I:<!R:?I
& 1:?C
& -1:?K
& !R+-1*!I:?tabs
& whl'(!tabs+-1:>0:?tabs&put\$\t)
&   whl
' ( 1+!K:~>!I:?K
& put\$(!C \t\t)
& !C*(!I+-1*!K)*(!K+1)^-1:?C
)
& put\$\n
)
&
)```
Output:
```Number of rows?
7
1
1               1
1               2               1
1               3               3               1
1               4               6               4               1
1               5               10              10              5               1
1               6               15              20              15              6               1```

## Burlesque

```blsq ) {1}{1 1}{^^2CO{p^?+}m[1+]1[+}15E!#s<-spbx#S
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1```

## C

Translation of: Fortran
```#include <stdio.h>

void pascaltriangle(unsigned int n)
{
unsigned int c, i, j, k;

for(i=0; i < n; i++) {
c = 1;
for(j=1; j <= 2*(n-1-i); j++) printf(" ");
for(k=0; k <= i; k++) {
printf("%3d ", c);
c = c * (i-k)/(k+1);
}
printf("\n");
}
}

int main()
{
pascaltriangle(8);
return 0;
}
```

### Recursive

```#include <stdio.h>

#define D 32
int pascals(int *x, int *y, int d)
{
int i;
for (i = 1; i < d; i++)
printf("%d%c", y[i] = x[i - 1] + x[i],
i < d - 1 ? ' ' : '\n');

return D > d ? pascals(y, x, d + 1) : 0;
}

int main()
{
int x[D] = {0, 1, 0}, y[D] = {0};
return pascals(x, y, 0);
}
```

```void triangleC(int nRows) {
if (nRows <= 0) return;
int *prevRow = NULL;
for (int r = 1; r <= nRows; r++) {
int *currRow = malloc(r * sizeof(int));
for (int i = 0; i < r; i++) {
int val = i==0 || i==r-1 ? 1 : prevRow[i-1] + prevRow[i];
currRow[i] = val;
printf(" %4d", val);
}
printf("\n");
free(prevRow);
prevRow = currRow;
}
free(prevRow);
}
```

## C#

Translation of: Fortran

Produces no output when n is less than or equal to zero.

```using System;

namespace RosettaCode {

class PascalsTriangle {

public static void CreateTriangle(int n) {
if (n > 0) {
for (int i = 0; i < n; i++) {
int c = 1;
Console.Write(" ".PadLeft(2 * (n - 1 - i)));
for (int k = 0; k <= i; k++) {
c = c * (i - k) / (k + 1);
}
Console.WriteLine();
}
}
}

public static void Main() {
CreateTriangle(8);
}
}
}
```

### Arbitrarily large numbers (BigInteger), arbitrary row selection

```using System;
using System.Linq;
using System.Numerics;
using System.Collections.Generic;

namespace RosettaCode
{
public static class PascalsTriangle
{
public static IEnumerable<BigInteger[]> GetTriangle(int quantityOfRows)
{
IEnumerable<BigInteger> range = Enumerable.Range(0, quantityOfRows).Select(num => new BigInteger(num));
return range.Select(num => GetRow(num).ToArray());
}

public static IEnumerable<BigInteger> GetRow(BigInteger rowNumber)
{
BigInteger denominator = 1;
BigInteger numerator = rowNumber;

BigInteger currentValue = 1;
for (BigInteger counter = 0; counter <= rowNumber; counter++)
{
yield return currentValue;
currentValue = BigInteger.Multiply(currentValue, numerator--);
currentValue = BigInteger.Divide(currentValue, denominator++);
}
yield break;
}

public static string FormatTriangleString(IEnumerable<BigInteger[]> triangle)
{
int maxDigitWidth = triangle.Last().Max().ToString().Length;
IEnumerable<string> rows = triangle.Select(arr =>
string.Join(" ", arr.Select(array => CenterString(array.ToString(), maxDigitWidth)) )
);
int maxRowWidth = rows.Last().Length;
return string.Join(Environment.NewLine, rows.Select(row => CenterString(row, maxRowWidth)));
}

private static string CenterString(string text, int width)
{
int spaces = width - text.Length;
int padLeft = (spaces / 2) + text.Length;
}
}
}
```

Example:

```static void Main()
{
IEnumerable<BigInteger[]> triangle = PascalsTriangle.GetTriangle(20);
string output = PascalsTriangle.FormatTriangleString(triangle)
Console.WriteLine(output);
}
```
Output:
```                                                           1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84    126   126   84    36     9     1
1    10    45    120   210   252   210   120   45    10     1
1    11    55    165   330   462   462   330   165   55    11     1
1    12    66    220   495   792   924   792   495   220   66    12     1
1    13    78    286   715  1287  1716  1716  1287   715   286   78    13     1
1    14    91    364  1001  2002  3003  3432  3003  2002  1001   364   91    14     1
1    15    105   455  1365  3003  5005  6435  6435  5005  3003  1365   455   105   15     1
1    16    120   560  1820  4368  8008  11440 12870 11440 8008  4368  1820   560   120   16     1
1    17    136   680  2380  6188  12376 19448 24310 24310 19448 12376 6188  2380   680   136   17     1
1    18    153   816  3060  8568  18564 31824 43758 48620 43758 31824 18564 8568  3060   816   153   18     1
1    19    171   969  3876  11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876   969   171   19     1
```

## C++

```#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
void Pascal_Triangle(int size) {

int a[100][100];
int i, j;

//first row and first coloumn has the same value=1
for (i = 1; i <= size; i++) {
a[i][1] = a[1][i] = 1;
}

//Generate the full Triangle
for (i = 2; i <= size; i++) {
for (j = 2; j <= size - i; j++) {
if (a[i - 1][j] == 0 || a[i][j - 1] == 0) {
break;
}
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}

/*
1 1 1 1
1 2 3
1 3
1

first print as above format-->

for (i = 1; i < size; i++) {
for (j = 1; j < size; j++) {
if (a[i][j] == 0) {
break;
}
printf("%8d",a[i][j]);
}
cout<<"\n\n";
}*/

// standard Pascal Triangle Format

int row,space;
for (i = 1; i < size; i++) {
space=row=i;
j=1;

while(space<=size+(size-i)+1){
cout<<" ";
space++;
}

while(j<=i){
if (a[row][j] == 0){
break;
}

if(j==1){
printf("%d",a[row--][j++]);
}
else
printf("%6d",a[row--][j++]);
}
cout<<"\n\n";
}

}

int main()
{
//freopen("out.txt","w",stdout);

int size;
cin>>size;
Pascal_Triangle(size);
}

}
```

### C++11 (with dynamic and semi-static vectors)

Constructs the whole triangle in memory before printing it. Uses vector of vectors as a 2D array with variable column size. Theoretically, semi-static version should work a little faster.

```// Compile with -std=c++11
#include<iostream>
#include<vector>
using namespace std;
void print_vector(vector<int> dummy){
for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
void print_vector_of_vectors(vector<vector<int>> dummy){
for (vector<vector<int>>::iterator i = dummy.begin(); i != dummy.end(); ++i)
print_vector(*i);
cout<<endl;
}
vector<vector<int>> dynamic_triangle(int dummy){
vector<vector<int>> result;
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
// The first row
row.push_back(1);
result.push_back(row);
// The second row
if (dummy > 1){
row.clear();
row.push_back(1); row.push_back(1);
result.push_back(row);
}
// The other rows
if (dummy > 2){
for (int i = 2; i < dummy; i++){
row.clear();
row.push_back(1);
for (int j = 1; j < i; j++)
row.push_back(result.back().at(j - 1) + result.back().at(j));
row.push_back(1);
result.push_back(row);
}
}
}
return result;
}
vector<vector<int>> static_triangle(int dummy){
vector<vector<int>> result;
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
result.resize(dummy); // This should work faster than consecutive push_back()s
// The first row
row.resize(1);
row.at(0) = 1;
result.at(0) = row;
// The second row
if (result.size() > 1){
row.resize(2);
row.at(0) = 1; row.at(1) = 1;
result.at(1) = row;
}
// The other rows
if (result.size() > 2){
for (int i = 2; i < result.size(); i++){
row.resize(i + 1); // This should work faster than consecutive push_back()s
row.front() = 1;
for (int j = 1; j < row.size() - 1; j++)
row.at(j) = result.at(i - 1).at(j - 1) + result.at(i - 1).at(j);
row.back() = 1;
result.at(i) = row;
}
}
}
return result;
}
int main(){
vector<vector<int>> triangle;
int n;
cout<<endl<<"The Pascal's Triangle"<<endl<<"Enter the number of rows: ";
cin>>n;
// Call the dynamic function
triangle = dynamic_triangle(n);
cout<<endl<<"Calculated using dynamic vectors:"<<endl<<endl;
print_vector_of_vectors(triangle);
// Call the static function
triangle = static_triangle(n);
cout<<endl<<"Calculated using static vectors:"<<endl<<endl;
print_vector_of_vectors(triangle);
return 0;
}
```

### C++11 (with a class)

A full fledged example with a class definition and methods to retrieve data, worthy of the title object-oriented.

```// Compile with -std=c++11
#include<iostream>
#include<vector>
using namespace std;
class pascal_triangle{
vector<vector<int>> data; // This is the actual data
void print_row(vector<int> dummy){
for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
public:
pascal_triangle(int dummy){ // Everything is done on the construction phase
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
data.resize(dummy); // Theoretically this should work faster than consecutive push_back()s
// The first row
row.resize(1);
row.at(0) = 1;
data.at(0) = row;
// The second row
if (data.size() > 1){
row.resize(2);
row.at(0) = 1; row.at(1) = 1;
data.at(1) = row;
}
// The other rows
if (data.size() > 2){
for (int i = 2; i < data.size(); i++){
row.resize(i + 1); // Theoretically this should work faster than consecutive push_back()s
row.front() = 1;
for (int j = 1; j < row.size() - 1; j++)
row.at(j) = data.at(i - 1).at(j - 1) + data.at(i - 1).at(j);
row.back() = 1;
data.at(i) = row;
}
}
}
}
~pascal_triangle(){
for (vector<vector<int>>::iterator i = data.begin(); i != data.end(); ++i)
i->clear(); // I'm not sure about the necessity of this loop!
data.clear();
}
void print_row(int dummy){
if (dummy < data.size())
for (vector<int>::iterator i = data.at(dummy).begin(); i != data.at(dummy).end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
void print(){
for (int i = 0; i < data.size(); i++)
print_row(i);
}
int get_coeff(int dummy1, int dummy2){
int result = 0;
if ((dummy1 < data.size()) && (dummy2 < data.at(dummy1).size()))
result = data.at(dummy1).at(dummy2);
return result;
}
vector<int> get_row(int dummy){
vector<int> result;
if (dummy < data.size())
result = data.at(dummy);
return result;
}
};
int main(){
int n;
cout<<endl<<"The Pascal's Triangle with a class!"<<endl<<endl<<"Enter the number of rows: ";
cin>>n;
pascal_triangle myptri(n);
cout<<endl<<"The whole triangle:"<<endl;
myptri.print();
cout<<endl<<"Just one row:"<<endl;
myptri.print_row(n/2);
cout<<endl<<"Just one coefficient:"<<endl;
cout<<myptri.get_coeff(n/2, n/4)<<endl<<endl;
return 0;
}
```

## Clojure

For n < 1, prints nothing, always returns nil. Copied from the Common Lisp implementation below, but with local functions and explicit tail-call-optimized recursion (recur).

```(defn pascal [n]
(let [newrow (fn newrow [lst ret]
(if lst
(recur (rest lst)
(conj ret (+ (first lst) (or (second lst) 0))))
ret))
genrow (fn genrow [n lst]
(when (< 0 n)
(do (println lst)
(recur (dec n) (conj (newrow lst []) 1)))))]
(genrow n [1])))
(pascal 4)
```

And here's another version, using the partition function to produce the sequence of pairs in a row, which are summed and placed between two ones to produce the next row:

```(defn nextrow [row]
(vec (concat [1] (map #(apply + %) (partition 2 1 row)) [1] )))

(defn pascal [n]
(assert (and (integer? n) (pos? n)))
(let [triangle (take n (iterate nextrow [1]))]
(doseq [row triangle]
(println row))))
```

The assert form causes the pascal function to throw an exception unless the argument is (integral and) positive.

Here's a third version using the iterate function

```(def pascal
(iterate
(fn [prev-row]
(->>
(concat [[(first prev-row)]] (partition 2 1 prev-row) [[(last prev-row)]])
(map (partial apply +) ,,,)))
[1]))
```

Another short version which returns an infinite pascal triangle as a list, using the iterate function.

```(def pascal
(iterate #(concat [1]
(map + % (rest %))
[1])
[1]))
```

One can then get the first n rows using the take function

```(take 10 pascal) ; returns a list of the first 10 pascal rows
```

Also, one can retrieve the nth row using the nth function

```(nth pascal 10) ;returns the nth row
```

## CoffeeScript

This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients.

```pascal = (n) ->
width = 6
for r in [1..n]
s = ws (width/2) * (n-r) # center row
output = (n) -> s += pad width, n
cell = 1
output cell
# Compute binomial coefficients as you go
# across the row.
for c in [1...r]
cell *= (r-c) / c
output cell
console.log s

ws = (n) ->
s = ''
s += ' ' for i in [0...n]
s

s = n.toString()
# There is probably a better way to do this.
cnt -= s.length
right = Math.floor(cnt / 2)
left = cnt - right
ws(left) + s + ws(right)

pascal(7)
```
Output:
```> coffee pascal.coffee
1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
```

## Commodore BASIC

```10 INPUT "HOW MANY";N
20 IF N<1 THEN END
30 DIM C(N)
40 DIM D(N)
50 LET C(1)=1
60 LET D(1)=1
70 FOR J=1 TO N
80 FOR I=1 TO N-J+1
90 PRINT "  ";
100 NEXT I
110 FOR I=1 TO J
120 PRINT C(I)" ";
130 NEXT I
140 PRINT
150 IF J=N THEN END
160 C(J+1)=1
170 D(J+1)=1
180 FOR I=1 TO J-1
190 D(I+1)=C(I)+C(I+1)
200 NEXT I
210 FOR I=1 TO J
220 C(I)=D(I)
230 NEXT I
240 NEXT J
```

Output:

```RUN
HOW MANY? 8
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
1   6   15   20   15   6   1
1   7   21   35   35   21   7   1
1   8   28   56   70   56   28   8    1
```

## Common Lisp

To evaluate, call (pascal n). For n < 1, it simply returns nil.

```(defun pascal (n)
(genrow n '(1)))

(defun genrow (n l)
(when (plusp n)
(print l)
(genrow (1- n) (cons 1 (newrow l)))))

(defun newrow (l)
(if (null (rest l))
'(1)
(cons (+ (first l) (second l))
(newrow (rest l)))))
```

An iterative solution with loop, using nconc instead of collect to keep track of the last cons. Otherwise, it would be necessary to traverse the list to do a (rplacd (last a) (list 1)).

```(defun pascal-next-row (a)
(loop :for q :in a
:and p = 0 :then q
:as s = (list (+ p q))
:nconc s :into a
:finally (rplacd s (list 1))
(return a)))

(defun pascal (n)
(loop :for a = (list 1) :then (pascal-next-row a)
:repeat n
:collect a))
```

Another iterative solution, this time using pretty-printing to automatically print the triangle in the shape of a triangle in the terminal. The print-pascal-triangle function computes and uses the length of the printed last row to decide how wide the triangle should be.

```(defun next-pascal-triangle-row (list)
`(1
,.(mapcar #'+ list (rest list))
1))

(defun pascal-triangle (number-of-rows)
(loop repeat number-of-rows
for row = '(1) then (next-pascal-triangle-row row)
collect row))

(defun print-pascal-triangle (number-of-rows)
(let* ((triangle (pascal-triangle number-of-rows))
(max-row-length (length (write-to-string (first (last triangle))))))
(format t
(format nil "~~{~~~D:@<~~{~~A ~~}~~>~~%~~}" max-row-length)
triangle)))
```

For example:

```(print-pascal-triangle 4)
```
```   1
1 1
1 2 1
1 3 3 1
```
```(print-pascal-triangle 8)
```
```         1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
```

## Component Pascal

```MODULE PascalTriangle;
IMPORT StdLog, DevCommanders, TextMappers;

TYPE
Expansion* = POINTER TO ARRAY OF LONGINT;

PROCEDURE Show*(e: Expansion);
VAR
i: INTEGER;
BEGIN
i := 0;
WHILE (i < LEN(e)) & (e[i] # 0) DO
StdLog.Int(e[i]);
INC(i)
END;
StdLog.Ln
END Show;

PROCEDURE GenFor*(p: LONGINT): Expansion;
VAR
expA,expB: Expansion;
i,j: LONGINT;

PROCEDURE Swap(VAR x,y: Expansion);
VAR
swap: Expansion;
BEGIN
swap := x; x := y; y := swap
END Swap;

BEGIN
ASSERT(p >= 0);
NEW(expA,p + 2);NEW(expB,p + 2);
FOR i := 0 TO p DO
IF i = 0 THEN expA[0] := 1
ELSE
FOR j := 0 TO i DO
IF j = 0 THEN
expB[j] := expA[j]
ELSE
expB[j] := expA[j - 1] + expA[j]
END
END;
Swap(expA,expB)
END;
END;
expB := NIL; (* for the GC *)
RETURN expA
END GenFor;

PROCEDURE Do*;
VAR
s: TextMappers.Scanner;
exp: Expansion;
BEGIN
s.ConnectTo(DevCommanders.par.text);
s.SetPos(DevCommanders.par.beg);
s.Scan;
WHILE (~s.rider.eot) DO
IF (s.type = TextMappers.char) & (s.char = '~') THEN
RETURN
ELSIF (s.type = TextMappers.int) THEN
exp := GenFor(s.int);
Show(exp)
END;
s.Scan
END
END Do;

END PascalTriangle.
```
`Execute: ^Q PascalTriangle.Do 0 1 2 3 4 5 6 7 8 9 10 11 12~`
Output:
``` 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
```

## D

### Less functional Version

```int[][] pascalsTriangle(in int rows) pure nothrow {
auto tri = new int[][rows];
foreach (r; 0 .. rows) {
int v = 1;
foreach (c; 0 .. r+1) {
tri[r] ~= v;
v = (v * (r - c)) / (c + 1);
}
}
return tri;
}

void main() {
immutable t = pascalsTriangle(10);
assert(t == [[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]);
}
```

### More functional Version

```import std.stdio, std.algorithm, std.range;

auto pascal() pure nothrow {
return [1].recurrence!q{ zip(a[n - 1] ~ 0, 0 ~ a[n - 1])
.map!q{ a[0] + a[1] }
.array };
}

void main() {
pascal.take(5).writeln;
}
```
Output:
`[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]`

### Alternative Version

There is similarity between Pascal's triangle and Sierpinski triangle. Their difference are the initial line and the operation that act on the line element to produce next line. The following is a generic pascal's triangle implementation for positive number of lines output (n).

```import std.stdio, std.string, std.array, std.format;

string Pascal(alias dg, T, T initValue)(int n) {
string output;

void append(in T[] l) {
output ~= " ".replicate((n - l.length + 1) * 2);
foreach (e; l)
output ~= format("%4s", format("%4s", e));
output ~= "\n";
}

if (n > 0) {
T[][] lines = [[initValue]];
append(lines[0]);
foreach (i; 1 .. n) {
lines ~= lines[i - 1] ~ initValue; // length + 1
foreach (int j; 1 .. lines[i-1].length)
lines[i][j] = dg(lines[i-1][j], lines[i-1][j-1]);
append(lines[i]);
}
}
return output;
}

string delegate(int n) genericPascal(alias dg, T, T initValue)() {
mixin Pascal!(dg, T, initValue);
return &Pascal;
}

void main() {
auto pascal = genericPascal!((int a, int b) => a + b, int, 1)();
static char xor(char a, char b) { return a == b ? '_' : '*'; }
auto sierpinski = genericPascal!(xor, char, '*')();

foreach (i; [1, 5, 9])
writef(pascal(i));
// an order 4 sierpinski triangle is a 2^4 lines generic
// Pascal triangle with xor operation
foreach (i; [16])
writef(sierpinski(i));
}
```
Output:
```     1
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
*
*   *
*   _   *
*   *   *   *
*   _   _   _   *
*   *   _   _   *   *
*   _   *   _   *   _   *
*   *   *   *   *   *   *   *
*   _   _   _   _   _   _   _   *
*   *   _   _   _   _   _   _   *   *
*   _   *   _   _   _   _   _   *   _   *
*   *   *   *   _   _   _   _   *   *   *   *
*   _   _   _   *   _   _   _   *   _   _   _   *
*   *   _   _   *   *   _   _   *   *   _   _   *   *
*   _   *   _   *   _   *   _   *   _   *   _   *   _   *
*   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *```

## Dart

```import 'dart:io';

pascal(n) {
if(n<=0) print("Not defined");

else if(n==1) print(1);

else {
List<List<int>> matrix = new List<List<int>>();
for (var i = 2; i < n; i++) {
List<int> list = new List<int>();
for (var j = 1; j<i; j++) {
}
}
for(var i=0; i<n; i++) {
for(var j=0; j<=i; j++) {
stdout.write(matrix[i][j]);
stdout.write(' ');
}
stdout.write('\n');
}
}
}

void main() {
pascal(0);
pascal(1);
pascal(3);
pascal(6);
}
```

## Delphi

```program PascalsTriangle;

procedure Pascal(r:Integer);
var
i, c, k:Integer;
begin
for i := 0 to r - 1 do
begin
c := 1;
for k := 0 to i do
begin
Write(c:3);
c := c * (i - k) div (k + 1);
end;
Writeln;
end;
end;

begin
Pascal(9);
end.
```

## DWScript

Doesn't print anything for negative or null values.

```procedure Pascal(r : Integer);
var
i, c, k : Integer;
begin
for i:=0 to r-1 do begin
c:=1;
for k:=0 to i do begin
Print(Format('%4d', [c]));
c:=(c*(i-k)) div (k+1);
end;
PrintLn('');
end;
end;

Pascal(9);
```
Output:
```   1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1```

## E

So as not to bother with text layout, this implementation generates a HTML fragment. It uses a single mutable array, appending one 1 and adding to each value the preceding value.

```def pascalsTriangle(n, out) {
def row := [].diverge(int)
out.print("<table style='text-align: center; border: 0; border-collapse: collapse;'>")
for y in 1..n {
out.print("<tr>")
row.push(1)
def skip := n - y
if (skip > 0) {
out.print(`<td colspan="\$skip"></td>`)
}
for x => v in row {
out.print(`<td>\$v</td><td></td>`)
}
for i in (1..!y).descending() {
row[i] += row[i - 1]
}
out.println("</tr>")
}
out.print("</table>")
}```
```def out := <file:triangle.html>.textWriter()
try {
pascalsTriangle(15, out)
} finally {
out.close()
}
makeCommand("yourFavoriteWebBrowser")("triangle.html")```

## EasyLang

```numfmt 0 4
proc pascal n . .
r[] = [ 1 ]
for i to n
rn[] = [ ]
l = 0
for j to n - len r[]
write "  "
.
for r in r[]
write r
rn[] &= l + r
l = r
.
print ""
rn[] &= l
swap r[] rn[]
.
.
pascal 13
```

## Eiffel

```note
description    : "Prints pascal's triangle"
output         : "[
Per requirements of the RosettaCode example, execution will print the first n rows of pascal's triangle
]"
date           : "19 December 2013"
authors        : "Sandro Meier", "Roman Brunner"
revision       : "1.0"
libraries      : "Relies on HASH_TABLE from EIFFEL_BASE library"
implementation : "[
Recursive implementation to calculate the n'th row.
]"
warning        : "[
Will not work for large n's (INTEGER_32)
]"

class
APPLICATION

inherit
ARGUMENTS

create
make

feature {NONE} -- Initialization

make
local
n:INTEGER
do
create {HASH_TABLE[ARRAY[INTEGER],INTEGER]}pascal_lines.make (n) --create the hash_table object
io.new_line
n:=25
draw(n)
end
feature
line(n:INTEGER):ARRAY[INTEGER]
--Calculates the n'th line
local
upper_line:ARRAY[INTEGER]
i:INTEGER
do
if	n=1 then	--trivial case first line
create Result.make_filled (0, 1, n+2)
Result.put (0, 1)
Result.put (1, 2)
Result.put (0, 3)
elseif pascal_lines.has (n) then	--checks if the result was already calculated
Result := pascal_lines.at (n)
else	--calculates the n'th line recursively
create Result.make_filled(0,1,n+2) --for caluclation purposes add a 0 at the beginning of each line
Result.put (0, 1)
upper_line:=line(n-1)
from
i:=1
until
i>upper_line.count-1
loop
Result.put(upper_line[i]+upper_line[i+1],i+1)
i:=i+1
end
Result.put (0, n+2)	--for caluclation purposes add a 0 at the end of each line
pascal_lines.put (Result, n)
end
end

draw(n:INTEGER)
--draw n lines of pascal's triangle
local
space_string:STRING
width, i:INTEGER

do
space_string:=" "		--question of design: add space_string at the beginning of each line
width:=line(n).count
space_string.multiply (width)
from
i:=1
until
i>n
loop
space_string.remove_tail (1)
io.put_string (space_string)
across line(i) as c
loop
if
c.item/=0
then
io.put_string (c.item.out+" ")
end
end
io.new_line
i:=i+1
end
end

feature --Access
pascal_lines:HASH_TABLE[ARRAY[INTEGER],INTEGER]
end
```

## Elixir

```defmodule Pascal do
def triangle(n), do: triangle(n,[1])

def triangle(0,list), do: list
def triangle(n,list) do
IO.inspect list
new_list = Enum.zip([0]++list, list++[0]) |> Enum.map(fn {a,b} -> a+b end)
triangle(n-1,new_list)
end
end

Pascal.triangle(8)
```
Output:
```[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
```

## Emacs Lisp

### Using mapcar and append, returing a list of rows

```(require 'cl-lib)

(defun next-row (row)
(cl-mapcar #'+ (cons 0 row)
(append row '(0))))

(defun triangle (row rows)
(if (= rows 0)
'()
(cons row (triangle (next-row row) (- rows 1)))))
```
Output:

Call the function from the REPL, IELM:

```ELISP> (triangle (list 1) 6)
((1)
(1 1)
(1 2 1)
(1 3 3 1)
(1 4 6 4 1)
(1 5 10 10 5 1))
```

### Translation from Pascal

```(defun pascal (r)
(dotimes (i r)
(let ((c 1))
(dotimes (k (+ i 1))
(princ (format "%d " c))
(setq c (/ (* c (- i k))
(+ k 1))))
(terpri))))
```
Output:

From the REPL:

```ELISP> (princ (pascal 6))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
```

### Returning a string

Same as the translation from Pascal, but now returning a string.

```(defun pascal (r)
(let ((out ""))
(dotimes (i r)
(let ((c 1))
(dotimes (k (+ i 1))
(setq out (concat out  (format "%d " c)))
(setq c (/ (* c (- i k))
(+ k 1))))
(setq out (concat out "\n"))))
out))
```
Output:

Now, since this one returns a string, it is possible to insert the result in the current buffer:

```(insert (pascal 6))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
```

## Erlang

```-import(lists).
-export([pascal/1]).

pascal(1)-> [[1]];
pascal(N) ->
L = pascal(N-1),
[H|_] = L,
[lists:zipwith(fun(X,Y)->X+Y end,[0]++H,H++[0])|L].
```
Output:
```  Eshell V5.5.5  (abort with ^G)
1> pascal:pascal(5).
[[1,4,6,4,1],[1,3,3,1],[1,2,1],[1,1],[1]]
```

## ERRE

```PROGRAM PASCAL_TRIANGLE

PROCEDURE PASCAL(R%)
LOCAL I%,C%,K%
FOR I%=0 TO R%-1 DO
C%=1
FOR K%=0 TO I% DO
WRITE("###";C%;)
C%=(C%*(I%-K%)) DIV (K%+1)
END FOR
PRINT
END FOR
END PROCEDURE

BEGIN
PASCAL(9)
END PROGRAM```

Output:

```  1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1
```

## Euphoria

### Summing from Previous Rows

```sequence row
row = {}
for m = 1 to 10 do
row = row & 1
for n = length(row)-1 to 2 by -1 do
row[n] += row[n-1]
end for
print(1,row)
puts(1,'\n')
end for```
Output:
``` {1}
{1,1}
{1,2,1}
{1,3,3,1}
{1,4,6,4,1}
{1,5,10,10,5,1}
{1,6,15,20,15,6,1}
{1,7,21,35,35,21,7,1}
{1,8,28,56,70,56,28,8,1}
{1,9,36,84,126,126,84,36,9,1}
```

## Excel

### LAMBDA

Binding the names PASCAL and BINCOEFF to the following lambda expressions in the Name Manager of the Excel WorkBook, to define Pascal's triangle in terms of binomial coefficients:

```PASCAL
=LAMBDA(n,
BINCOEFF(n - 1)(
SEQUENCE(1, n, 0, 1)
)
)

BINCOEFF
=LAMBDA(n,
LAMBDA(k,
QUOTIENT(FACT(n), FACT(k) * FACT(n - k))
)
)
```
Output:
 =PASCAL(A2) fx A B C D E F G H I J K 1 Row number PASCAL's TRIANGLE 2 1 1 3 2 1 1 4 3 1 2 1 5 4 1 3 3 1 6 5 1 4 6 4 1 7 6 1 5 10 10 5 1 8 7 1 6 15 20 15 6 1 9 8 1 7 21 35 35 21 7 1 10 9 1 8 28 56 70 56 28 8 1 11 10 1 9 36 84 126 126 84 36 9 1

Or defining the whole triangle as a single grid, by binding the name TRIANGLE to an additional lambda:

```TRIANGLE
=LAMBDA(n,
LET(
ixs, SEQUENCE(n, n, 0, 1),
x, MOD(ixs, n),
y, QUOTIENT(ixs, n),
IF(x <= y,
BINCOEFF(y)(x),
""
)
)
)
```
Output:
 =TRIANGLE(10) fx A B C D E F G H I J K 1 PASCAL's TRIANGLE 2 1 3 1 1 4 1 2 1 5 1 3 3 1 6 1 4 6 4 1 7 1 5 10 10 5 1 8 1 6 15 20 15 6 1 9 1 7 21 35 35 21 7 1 10 1 8 28 56 70 56 28 8 1 11 1 9 36 84 126 126 84 36 9 1

## F#

```let rec nextrow l =
match l with
| []      -> []
| h :: [] -> [1]
| h :: t  -> h + t.Head :: nextrow t

let pascalTri n = List.scan(fun l i -> 1 :: nextrow l) [1] [1 .. n]

for row in pascalTri(10) do
for i in row do
printf "%s" (i.ToString() + ", ")
printfn "%s" "\n"
```

## Factor

This implementation works by summing the previous line content. Result for n < 1 is the same as for n == 1.

```USING: grouping kernel math sequences ;

: (pascal) ( seq -- newseq )
dup last 0 prefix 0 suffix 2 <clumps> [ sum ] map suffix ;

: pascal ( n -- seq )
1 - { { 1 } } swap [ (pascal) ] times ;
```

It works as:

```5 pascal .
{ { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }
```

## Fantom

```class Main
{
Int[] next_row (Int[] row)
{
new_row := [1]
(row.size-1).times |i|
{
}

return new_row
}

Void print_pascal (Int n)  // no output for n <= 0
{
current_row := [1]
n.times
{
echo (current_row.join(" "))
current_row = next_row (current_row)
}
}

Void main ()
{
print_pascal (10)
}
}```

## FOCAL

```1.1 S OLD(1)=1; T %4.0, 1, !
1.2 F N=1,10; D 2
1.3 Q

2.1 S NEW(1)=1
2.2 F X=1,N; S NEW(X+1)=OLD(X)+OLD(X+1)
2.3 F X=1,N+1; D 3
2.4 T !

3.1 S OLD(X)=NEW(X)
3.2 T %4.0, OLD(X)```
Output:
```=    1
=    1=    1
=    1=    2=    1
=    1=    3=    3=    1
=    1=    4=    6=    4=    1
=    1=    5=   10=   10=    5=    1
=    1=    6=   15=   20=   15=    6=    1
=    1=    7=   21=   35=   35=   21=    7=    1
=    1=    8=   28=   56=   70=   56=   28=    8=    1
=    1=    9=   36=   84=  126=  126=   84=   36=    9=    1
=    1=   10=   45=  120=  210=  252=  210=  120=   45=   10=    1
```

## Forth

```: init ( n -- )
here swap cells erase  1 here ! ;
: .line ( n -- )
cr here swap 0 do dup @ . cell+ loop drop ;
: next ( n -- )
here swap 1- cells here + do
i @ i cell+ +!
-1 cells +loop ;
: pascal ( n -- )
dup init   1  .line
1 ?do i next i 1+ .line loop ;
```

This is a bit more efficient.

Translation of: C
```: PascTriangle
cr dup 0
?do
1 over 1- i - 2* spaces i 1+ 0 ?do dup 4 .r j i - * i 1+ / loop cr drop
loop drop
;

13 PascTriangle
```

## Fortran

Works with: Fortran version 90 and later

Prints nothing for n<=0. Output formatting breaks down for n>20

```PROGRAM Pascals_Triangle

CALL Print_Triangle(8)

END PROGRAM Pascals_Triangle

SUBROUTINE Print_Triangle(n)

IMPLICIT NONE
INTEGER, INTENT(IN) :: n
INTEGER :: c, i, j, k, spaces

DO i = 0, n-1
c = 1
spaces = 3 * (n - 1 - i)
DO j = 1, spaces
END DO
DO k = 0, i
c = c * (i - k) / (k + 1)
END DO
WRITE(*,*)
END DO

END SUBROUTINE Print_Triangle
```

## FreeBASIC

```' FB 1.05.0 Win64

Sub pascalTriangle(n As UInteger)
If n = 0 Then Return
Dim prevRow(1 To n) As UInteger
Dim currRow(1 To n) As UInteger
Dim start(1 To n) As UInteger  ''stores starting column for each row
start(n) = 1
For i As Integer = n - 1 To 1 Step -1
start(i) = start(i + 1) + 3
Next
prevRow(1) = 1
Print Tab(start(1));
Print 1U
For i As UInteger = 2 To n
For j As UInteger = 1 To i
If j = 1 Then
Print Tab(start(i)); "1";
currRow(1) = 1
ElseIf j = i Then
Print "     1"
currRow(i) = 1
Else
currRow(j) = prevRow(j - 1) + prevRow(j)
Print Using "######"; currRow(j); "    ";
End If
Next j
For j As UInteger = 1 To i
prevRow(j) = currRow(j)
Next j
Next i
End Sub

pascalTriangle(14)
Print
Print "Press any key to quit"
Sleep
```
Output:
```                                       1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1
1    10    45   120   210   252   210   120    45    10     1
1    11    55   165   330   462   462   330   165    55    11     1
1    12    66   220   495   792   924   792   495   220    66    12     1
1    13    78   286   715  1287  1716  1716  1287   715   286    78    13     1
```

## Frink

This version takes a little effort to automatically format the tree based upon the width of the largest numbers in the bottom row. It automatically calculates this easily using Frink's builtin function for efficiently calculating (even large) binomial coefficients with cached factorials and binary splitting.

```pascal[rows] :=
{
widest = length[toString[binomial[rows-1, (rows-1) div 2]]]

for row = 0 to rows-1
{
line = repeat[" ", round[(rows-row)* (widest+1)/2]]
for col = 0 to row
line = line + padRight[binomial[row, col], widest+1, " "]

println[line]
}
}

pascal[10]```
Output:
```                    1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6   1
1   7   21  35  35  21  7   1
1   8   28  56  70  56  28  8   1
1   9   36  84  126 126 84  36  9   1
```

## FunL

### Summing from Previous Rows

Translation of: Scala
```import lists.zip

def
pascal( 1 ) = [1]
pascal( n ) = [1] + map( (a, b) -> a + b, zip(pascal(n-1), pascal(n-1).tail()) ) + [1]```

### Combinations

```import integers.choose

def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]```

### Pascal's Triangle

```def triangle( height ) =
width = max( map(a -> a.toString().length(), pascal(height)) )

if 2|width
width++

for n <- 1..height
print( ' '*((width + 1)\2)*(height - n) )
println( map(a -> format('%' + width + 'd ', a), pascal(n)).mkString() )

triangle( 10 )```
Output:
```                    1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

Test case

## GAP

```Pascal := function(n)
local i, v;
v := [1];
for i in [1 .. n] do
Display(v);
v := Concatenation([0], v) + Concatenation(v, [0]);
od;
end;

Pascal(9);
# [ 1 ]
# [ 1, 1 ]
# [ 1, 2, 1 ]
# [ 1, 3, 3, 1 ]
# [ 1, 4, 6, 4, 1 ]
# [ 1, 5, 10, 10, 5, 1 ]
# [ 1, 6, 15, 20, 15, 6, 1 ]
# [ 1, 7, 21, 35, 35, 21, 7, 1 ]
# [ 1, 8, 28, 56, 70, 56, 28, 8, 1 ]
```

## Go

No output for n < 1. Otherwise, output formatted left justified.

```package main

import "fmt"

func printTriangle(n int) {
// degenerate cases
if n <= 0 {
return
}
fmt.Println(1)
if n == 1 {
return
}
// iterate over rows, zero based
a := make([]int, (n+1)/2)
a[0] = 1
for row, middle := 1, 0; row < n; row++ {
// generate new row
even := row&1 == 0
if even {
a[middle+1] = a[middle] * 2
}
for i := middle; i > 0; i-- {
a[i] += a[i-1]
}
// print row
for i := 0; i <= middle; i++ {
fmt.Print(a[i], " ")
}
if even {
middle++
}
for i := middle; i >= 0; i-- {
fmt.Print(a[i], " ")
}
fmt.Println("")
}
}

func main() {
printTriangle(4)
}
```

Output:

```1
1 1
1 2 1
1 3 3 1
```

## Groovy

### Recursive

In the spirit of the Haskell "think in whole lists" solution here is a list-driven, minimalist solution:

```def pascal
pascal = { n -> (n <= 1) ? [1] : [[0] + pascal(n - 1), pascal(n - 1) + [0]].transpose().collect { it.sum() } }
```

However, this solution is horribly inefficient (O(n**2)). It slowly grinds to a halt on a reasonably powerful PC after about line 25 of the triangle.

Test program:

```def count = 15
(1..count).each { n ->
printf ("%2d:", n); (0..(count-n)).each { print "    " }; pascal(n).each{ printf("%6d  ", it) }; println ""
}
```
Output:
``` 1:                                                                 1
2:                                                             1       1
3:                                                         1       2       1
4:                                                     1       3       3       1
5:                                                 1       4       6       4       1
6:                                             1       5      10      10       5       1
7:                                         1       6      15      20      15       6       1
8:                                     1       7      21      35      35      21       7       1
9:                                 1       8      28      56      70      56      28       8       1
10:                             1       9      36      84     126     126      84      36       9       1
11:                         1      10      45     120     210     252     210     120      45      10       1
12:                     1      11      55     165     330     462     462     330     165      55      11       1
13:                 1      12      66     220     495     792     924     792     495     220      66      12       1
14:             1      13      78     286     715    1287    1716    1716    1287     715     286      78      13       1
15:         1      14      91     364    1001    2002    3003    3432    3003    2002    1001     364      91      14       1  ```

## GW-BASIC

```10 INPUT "Number of rows? ",R
20 FOR I=0 TO R-1
30 C=1
40 FOR K=0 TO I
50 PRINT USING "####";C;
60 C=C*(I-K)/(K+1)
70 NEXT
80 PRINT
90 NEXT
```

Output:

```Number of rows? 7
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
```

An approach using the "think in whole lists" principle: Each row in the triangle can be calculated from the previous row by adding a shifted version of itself to it, keeping the ones at the ends. The Prelude function zipWith can be used to add two lists, but it won't keep the old values when one list is shorter. So we need a similar function

```zapWith :: (a -> a -> a) -> [a] -> [a] -> [a]
zapWith f xs [] = xs
zapWith f [] ys = ys
zapWith f (x:xs) (y:ys) = f x y : zapWith f xs ys
```

Now we can shift a list and add it to itself, extending it by keeping the ends:

```extendWith f [] = []
extendWith f xs@(x:ys) = x : zapWith f xs ys
```

And for the whole (infinite) triangle, we just iterate this operation, starting with the first row:

```pascal = iterate (extendWith (+)) [1]
```

For the first n rows, we just take the first n elements from this list, as in

```*Main> take 6 pascal
[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]
```

A shorter approach, plagiarized from [2]

```-- generate next row from current row
nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])

-- returns the first n rows
pascal = iterate nextRow [1]
```

Alternatively, using list comprehensions:

```pascal :: [[Integer]]
pascal =
(1 : [ 0 | _ <- head pascal])
: [zipWith (+) (0:row) row | row <- pascal]
```
```*Pascal> take 5 <\$> (take 5 \$ triangle)
[[1,0,0,0,0],[1,1,0,0,0],[1,2,1,0,0],[1,3,3,1,0],[1,4,6,4,1]]
```

With binomial coefficients:

```fac = product . enumFromTo 1

binCoef n k = fac n `div` (fac k * fac (n - k))

pascal = ((fmap . binCoef) <*> enumFromTo 0) . pred
```

Example:

```*Main> putStr \$ unlines \$ map unwords \$ map (map show) \$ pascal 10
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
```

## HicEst

```   CALL Pascal(30)

SUBROUTINE Pascal(rows)
CHARACTER fmt*6
WRITE(Text=fmt, Format='"i", i5.5') 1+rows/4

DO row = 0, rows-1
n = 1
DO k = 0, row
col = rows*(rows-row+2*k)/4
WRITE(Row=row+1, Column=col, F=fmt) n
n = n * (row - k) / (k + 1)
ENDDO
ENDDO
END```

## Icon and Unicon

The code below is slightly modified from the library version of pascal which prints 0's to the full width of the carpet. It also presents the data as an isoceles triangle.

```link math

procedure main(A)
every n := !A do  {    # for each command line argument
n := integer(\n) | &null
pascal(n)
}
end

procedure pascal(n)		#: Pascal triangle
/n := 16
write("width=", n, " height=", n)	# carpet header
fw := *(2 ^ n)+1
every i := 0 to n - 1 do {
writes(repl(" ",fw*(n-i)/2))
every j := 0 to n - 1 do
writes(center(binocoef(i, j),fw) | break)
write()
}
end
```

Sample output:

```->pascal 1 4 8
width=1 height=1
1
width=4 height=4
1
1  1
1  2  1
1  3  3  1
width=8 height=8
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6   1
1   7   21  35  35  21  7   1
->```

## IDL

```Pro Pascal, n
;n is the number of lines of the triangle to be displayed
r=[1]
print, r
for i=0, (n-2) do begin
pascalrow,r
endfor
End

Pro PascalRow, r
for i=0,(n_elements(r)-2) do begin
r[i]=r[i]+r[i+1]
endfor
r= [1, r]
print, r

End
```

## IS-BASIC

```100 PROGRAM "PascalTr.bas"
110 TEXT 80
120 LET ROW=12
130 FOR I=0 TO ROW
140   LET C=1
150   PRINT TAB(37-I*3);
160   FOR K=0 TO I
170     PRINT USING " #### ":C;
180     LET C=C*(I-K)/(K+1)
190   NEXT
200   PRINT
210 NEXT```
Output:
```                                         1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1
1    10    45   120   210   252   210   120    45    10     1
1    11    55   165   330   462   462   330   165    55    11     1
1    12    66   220   495   792   924   792   495   220    66    12     1```

## ivy

```op pascal N = transp (0 , iota N) o.! -1 , iota N
pascal 5
1  0  0  0  0  0
1  1  0  0  0  0
1  2  1  0  0  0
1  3  3  1  0  0
1  4  6  4  1  0
1  5 10 10  5  1```

## J

```   !~/~ i.5
1 0 0 0 0
1 1 0 0 0
1 2 1 0 0
1 3 3 1 0
1 4 6 4 1
```
```   ([: ":@-.&0"1 !~/~)@i. 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```
```   (-@|. |."_1 [: ":@-.&0"1 !~/~)@i. 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```

However, multi-digit numbers take up additional space, which looks slightly odd. But we can work around that by adding additional padding and shifting the lines a bit more:

```   (|."_1~ 0-3*i.@-@#) ;@((<'%6d') sprintf each -.&0)"1 !~/~i.10
1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1
```

Also... when we mix positive and negative numbers it stops being a triangle:

```   i:5
_5 _4 _3 _2 _1 0 1 2 3 4 5
!~/~i:5
1  0  0  0 0 1 _5 15 _35 70 _126
_4  1  0  0 0 1 _4 10 _20 35  _56
6 _3  1  0 0 1 _3  6 _10 15  _21
_4  3 _2  1 0 1 _2  3  _4  5   _6
1 _1  1 _1 1 1 _1  1  _1  1   _1
0  0  0  0 0 1  0  0   0  0    0
0  0  0  0 0 1  1  0   0  0    0
0  0  0  0 0 1  2  1   0  0    0
0  0  0  0 0 1  3  3   1  0    0
0  0  0  0 0 1  4  6   4  1    0
0  0  0  0 0 1  5 10  10  5    1
!/~i:5
1  _4   6 _4  1 0 0 0 0 0  0
0   1  _3  3 _1 0 0 0 0 0  0
0   0   1 _2  1 0 0 0 0 0  0
0   0   0  1 _1 0 0 0 0 0  0
0   0   0  0  1 0 0 0 0 0  0
1   1   1  1  1 1 1 1 1 1  1
_5  _4  _3 _2 _1 0 1 2 3 4  5
15  10   6  3  1 0 0 1 3 6 10
_35 _20 _10 _4 _1 0 0 0 1 4 10
70  35  15  5  1 0 0 0 0 1  5
_126 _56 _21 _6 _1 0 0 0 0 0  1
```

See the talk page for explanation of earlier version

## Java

### Summing from Previous Rows

Works with: Java version 1.5+
```import java.util.ArrayList;
...//class definition, etc.
public static void genPyrN(int rows){
if(rows < 0) return;
//save the last row here
ArrayList<Integer> last = new ArrayList<Integer>();
System.out.println(last);
for(int i= 1;i <= rows;++i){
//work on the next row
ArrayList<Integer> thisRow= new ArrayList<Integer>();
for(int j= 1;j < i;++j){//loop the number of elements in this row
//sum from the last row
}
last= thisRow;//save this row
System.out.println(thisRow);
}
}
```

### Combinations

This method is limited to 21 rows because of the limits of long. Calling pas with an argument of 22 or above will cause intermediate math to wrap around and give false answers.

```public class Pas{
public static void main(String[] args){
//usage
pas(20);
}

public static void pas(int rows){
for(int i = 0; i < rows; i++){
for(int j = 0; j <= i; j++){
System.out.print(ncr(i, j) + " ");
}
System.out.println();
}
}

public static long ncr(int n, int r){
return fact(n) / (fact(r) * fact(n - r));
}

public static long fact(int n){
long ans = 1;
for(int i = 2; i <= n; i++){
ans *= i;
}
return ans;
}
}
```

### Using arithmetic calculation of each row element

This method is limited to 30 rows because of the limits of integer calculations (probably when calculating the multiplication). If m is declared as long then 62 rows can be printed.

```public class Pascal {
private static void printPascalLine (int n) {
if (n < 1)
return;
int m = 1;
System.out.print("1 ");
for (int j=1; j<n; j++) {
m = m * (n-j)/j;
System.out.print(m);
System.out.print(" ");
}
System.out.println();
}

public static void printPascal (int nRows) {
for(int i=1; i<=nRows; i++)
printPascalLine(i);
}
}
```

## JavaScript

### ES5

#### Imperative

Works with: SpiderMonkey
Works with: V8
```// Pascal's triangle object
function pascalTriangle (rows) {

// Number of rows the triangle contains
this.rows = rows;

// The 2D array holding the rows of the triangle
this.triangle = new Array();
for (var r = 0; r < rows; r++) {
this.triangle[r] = new Array();
for (var i = 0; i <= r; i++) {
if (i == 0 || i == r)
this.triangle[r][i] = 1;
else
this.triangle[r][i] = this.triangle[r-1][i-1]+this.triangle[r-1][i];
}
}

// Method to print the triangle
this.print = function(base) {
if (!base)
base = 10;

// Private method to calculate digits in number
var digits = function(n,b) {
var d = 0;
while (n >= 1) {
d++;
n /= b;
}
return d;
}

// Calculate max spaces needed
var spacing = digits(this.triangle[this.rows-1][Math.round(this.rows/2)],base);

// Private method to add spacing between numbers
var insertSpaces = function(s) {
var buf = "";
while (s > 0) {
s--;
buf += " ";
}
return buf;
}

// Print the triangle line by line
for (var r = 0; r < this.triangle.length; r++) {
var l = "";
for (var s = 0; s < Math.round(this.rows-1-r); s++) {
l += insertSpaces(spacing);
}
for (var i = 0; i < this.triangle[r].length; i++) {
if (i != 0)
l += insertSpaces(spacing-Math.ceil(digits(this.triangle[r][i],base)/2));
l += this.triangle[r][i].toString(base);
if (i < this.triangle[r].length-1)
l += insertSpaces(spacing-Math.floor(digits(this.triangle[r][i],base)/2));
}
print(l);
}
}

}

// Display 4 row triangle in base 10
var tri = new pascalTriangle(4);
tri.print();
// Display 8 row triangle in base 16
tri = new pascalTriangle(8);
tri.print(16);
```

Output:

```\$ d8 pascal.js
1
1 1
1 2 1
1 3 3 1
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   a   a   5   1
1   6   f   14  f   6   1
1   7   15  23  23  15  7   1```

#### Functional

```(function (n) {
'use strict';

// PASCAL TRIANGLE --------------------------------------------------------

// pascal :: Int -> [[Int]]
function pascal(n) {
return foldl(function (a) {
var xs = a.slice(-1)[0]; // Previous row
return append(a, [zipWith(
function (a, b) {
return a + b;
},
append([0], xs),
append(xs, [0])
)]);
}, [
[1] // Initial seed row
], enumFromTo(1, n - 1));
};

// GENERIC FUNCTIONS ------------------------------------------------------

// (++) :: [a] -> [a] -> [a]
function append(xs, ys) {
return xs.concat(ys);
};

// enumFromTo :: Int -> Int -> [Int]
function enumFromTo(m, n) {
return Array.from({
length: Math.floor(n - m) + 1
}, function (_, i) {
return m + i;
});
};

// foldl :: (b -> a -> b) -> b -> [a] -> b
function foldl(f, a, xs) {
return xs.reduce(f, a);
};

// foldr (a -> b -> b) -> b -> [a] -> b
function foldr(f, a, xs) {
return xs.reduceRight(f, a);
};

// map :: (a -> b) -> [a] -> [b]
function map(f, xs) {
return xs.map(f);
};

// min :: Ord a => a -> a -> a
function min(a, b) {
return b < a ? b : a;
};

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
function zipWith(f, xs, ys) {
return Array.from({
length: min(xs.length, ys.length)
}, function (_, i) {
return f(xs[i], ys[i]);
});
};

// TEST and FORMAT --------------------------------------------------------
var lstTriangle = pascal(n);

// [[a]] -> bool -> s -> s
return '{| class="wikitable" ' + (strStyle ? 'style="' + strStyle +
'"' : '') + lstRows.map(function (lstRow, iRow) {
var strDelim = blnHeaderRow && !iRow ? '!' : '|';
return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
return typeof v === 'undefined' ? ' ' : v;
})
.join(' ' + strDelim + strDelim + ' ');
})
.join('') + '\n|}';
}

var lstLastLine = lstTriangle.slice(-1)[0],
lngBase = lstLastLine.length * 2 - 1,
nWidth = lstLastLine.reduce(function (a, x) {
var d = x.toString()
.length;
return d > a ? d : a;
}, 1) * lngBase;

return [wikiTable(lstTriangle.map(function (lst) {
return lst.join(';;')
.split(';');
})
.map(function (line, i) {
var lstPad = Array((lngBase - line.length) / 2);
}), false, 'text-align:center;width:' + nWidth + 'em;height:' + nWidth +
'em;table-layout:fixed;'), JSON.stringify(lstTriangle)].join('\n\n');
})(7);
```
Output:
 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
```[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1],[1,6,15,20,15,6,1]]
```

### ES6

```(() => {
"use strict";

// ---------------- PASCAL'S TRIANGLE ----------------

// pascal :: Generator [[Int]]
const pascal = () =>
iterate(
xs => zipWith(
a => b => a + b
)(
[0, ...xs]
)(
[...xs, 0]
)
)([1]);

// ---------------------- TEST -----------------------
// main :: IO ()
const main = () =>
showPascal(
take(10)(
pascal()
)
);

// showPascal :: [[Int]] -> String
const showPascal = xs => {
const w = last(xs).join("   ").length;

return xs.map(
ys => center(w)(" ")(ys.join("   "))
)
.join("\n");
};

// ---------------- GENERIC FUNCTIONS ----------------

// center :: Int -> Char -> String -> String
const center = n =>
// Size of space -> filler Char ->
// String -> Centered String
c => s => {
const gap = n - s.length;

return 0 < gap ? (() => {
const
margin = c.repeat(Math.floor(gap / 2)),
dust = c.repeat(gap % 2);

return `\${margin}\${s}\${margin}\${dust}`;
})() : s;
};

// iterate :: (a -> a) -> a -> Gen [a]
const iterate = f =>
// An infinite list of repeated
// applications of f to x.
function* (x) {
let v = x;

while (true) {
yield v;
v = f(v);
}
};

// last :: [a] -> a
const last = xs =>
0 < xs.length ? xs.slice(-1)[0] : undefined;

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => "GeneratorFunction" !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : Array.from({
length: n
}, () => {
const x = xs.next();

return x.done ? [] : [x.value];
}).flat();

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// A list constructed by zipping with a
// custom function, rather than with the
// default tuple constructor.
xs => ys => xs.map(
(x, i) => f(x)(ys[i])
).slice(
0, Math.min(xs.length, ys.length)
);

// MAIN ---
return main();
})();
```
Output:
```                      1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
1   6   15   20   15   6   1
1   7   21   35   35   21   7   1
1   8   28   56   70   56   28   8   1
1   9   36   84   126   126   84   36   9   1```

#### Recursive

```const aux = n => {
if(n <= 1) return [1]
const prevLayer = aux(n - 1)
const shifted = [0, ...prevLayer]
return shifted.map((x, i) => (prevLayer[i] || 0) + x)
}
const pascal = n => {
for(let i = 1; i <= n; i++) {
console.log(aux(i).join(' '))
}
}
pascal(8)```
Output:
```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
```

#### Recursive - memoized

```const aux = (() => {
const layers = [[1], [1]]
return n => {
if(layers[n]) return layers[n]
const prevLayer = aux(n - 1)
const shifted = [0, ...prevLayer]
layers[n] = shifted.map((x, i) => (prevLayer[i] || 0) + x)
return layers[n]
}
})()
const pascal = n => {
for(let i = 1; i <= n; i++) {
console.log(aux(i).join(' '))
}
}
pascal(8)```

## jq

Works with: jq version 1.4

pascal(n) as defined here produces a stream of n arrays, each corresponding to a row of the Pascal triangle. The implementation avoids any arithmetic except addition.

```# pascal(n) for n>=0; pascal(0) emits an empty stream.
def pascal(n):
def _pascal:  # input: the previous row
. as \$in
| .,
if length >= n then empty
else
reduce range(0;length-1) as \$i
([1]; . + [ \$in[\$i] + \$in[\$i + 1] ]) + [1] | _pascal
end;
if n <= 0 then empty else [1] | _pascal end ;```

Example:

```pascal(5)
```
Output:
```\$ jq -c -n -f pascal_triangle.jq
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]```

Using recurse/1

Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function.

```def pascal(n):
if n <= 0 then empty
else [1]
| recurse( if length >= n then empty
else . as \$in
| reduce range(0;length-1) as \$i
([1]; . + [ \$in[\$i] + \$in[\$i + 1] ]) + [1]
end)
end;```

## Julia

```function pascal(n)
if n<=0
print("n has to have a positive value")
end
x=0
while x<=n
for a=0:x
print(binomial(x,a)," ")
end
println("")
x+=1
end
end
```
```pascal(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
```

Another solution using matrix exponentiation.

```iround(x) = round(Int64, x)

triangle(n) = iround.(exp(diagm(-1=> 1:n)))

function pascal(n)
t=triangle(n)
println.(join.([filter(!iszero, t[i,:]) for i in 1:(n+1)], " "))
end```
Output:
```pascal(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
```

Yet another solution using a static vector

```function pascal(n)
(n<=0) && error("Pascal trinalge can not have zero or negative rows")
r=Vector{Int}(undef,n)
pr=Vector{Int}(undef,n)
pr[1]=r[1]=1
println(@view pr[1])
for i=2:n
r[1]=r[i]=1
for j=2:i-1
r[j]=pr[j-1]+pr[j]
end
println(join(view(r,1:i), " "))
r,pr=pr,r
end
end```
Output:
```pascal(8)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
```

## K

```pascal:{(x-1){+':x,0}\1}
pascal 6
(1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1)```

## Kotlin

```fun pas(rows: Int) {
for (i in 0..rows - 1) {
for (j in 0..i)
print(ncr(i, j).toString() + " ")
println()
}
}

fun ncr(n: Int, r: Int) = fact(n) / (fact(r) * fact(n - r))

fun fact(n: Int) : Long {
var ans = 1.toLong()
for (i in 2..n)
ans *= i
return ans
}

fun main(args: Array<String>) = pas(args[0].toInt())```

## Lambdatalk

```1) Based on this expression of pascalian binomial:

Cnp = [n*(n-1)...(n-p+1)]/[p*(p-1)...2*1]

2) we define the following function:

{def C
{lambda {:n :p}
{/ {* {S.serie :n {- :n :p -1} -1}}
{* {S.serie :p 1 -1}}}}}

{C 16 8}
-> 12870

3) Writing

1{S.map {lambda {:n} {br}1
{S.map {C :n} {S.serie 1 {- :n 1}}} 1}
{S.serie 2 16}}
displays:

1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1```

## Liberty BASIC

```input "How much rows would you like? "; n
dim a\$(n)

for i=  0 to n
c = 1
o\$ =""
for k =0 to i
o\$ =o\$ ; c; " "
c =c *(i-k)/(k+1)
next k
a\$(i)=o\$
next i

maxLen = len(a\$(n))
for i=  0 to n
print space\$((maxLen-len(a\$(i)))/2);a\$(i)
next i

end```

## Locomotive Basic

```10 CLS
20 INPUT "Number of rows? ", rows:GOSUB 40
30 END
40 FOR i=0 TO rows-1
50 c=1
60 FOR k=0 TO i
70 PRINT USING "####";c;
80 c=c*(i-k)/(k+1)
90 NEXT
100 PRINT
110 NEXT
120 RETURN```

Output:

```Number of rows? 7
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
```

## Logo

```to pascal :n
if :n = 1 [output [1]]
localmake "a pascal :n-1
output (sentence first :a (map "sum butfirst :a butlast :a) last :a)
end

for [i 1 10] [print pascal :i]```

## Logtalk

Our implementation will have an object `pascals` with work done in the method `triangle/2`. We will be caching results for time efficiency at the cost of space efficiency,and the `reset/0` method will flush that cache should it grow to be a problem. The resulting object looks like this:

```:- object(pascals).

:- uses(integer, [plus/3, succ/2]).

:- public(reset/0).

reset :-
retractall(triangle_(_,_,_)).

:- private(triangle_/3).
:- dynamic(triangle_/3).

:- public(triangle/2).

triangle(N, Lines) :-
triangle(N, _, DiffLines),
difflist::as_list(DiffLines, Lines).

% Shortcut with cached value if it exists.
triangle(N, Line, DiffLines) :- triangle_(N, Line, DiffLines), !.

triangle(N, Line, DiffLines) :-
succ(N0, N),
triangle(N0, Line0, DiffLines0),
ZL = [0|Line0],
list::append(Line0, [0], ZR),
meta::map(plus, ZL, ZR, Line),
asserta(triangle_(N, Line, DiffLines)).

triangle(1, [1], [[1]|X]-X).

:- end_object.```
Output:

Using the SWI-Prolog back-end:

```?- logtalk_load([meta(loader), types(loader), pascals], [optimize(on)]).
% messages elided
true.

?- pascals::triangle(17, Ls), logtalk::print_message(information, user, Ls).
% - [1]
% - [1,1]
% - [1,2,1]
% - [1,3,3,1]
% - [1,4,6,4,1]
% - [1,5,10,10,5,1]
% - [1,6,15,20,15,6,1]
% - [1,7,21,35,35,21,7,1]
% - [1,8,28,56,70,56,28,8,1]
% - [1,9,36,84,126,126,84,36,9,1]
% - [1,10,45,120,210,252,210,120,45,10,1]
% - [1,11,55,165,330,462,462,330,165,55,11,1]
% - [1,12,66,220,495,792,924,792,495,220,66,12,1]
% - [1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1]
% - [1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1]
% - [1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1]
% - [1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1]
Ls = [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4|...], [1, 5, 10|...], [1, 6|...], [1|...], [...|...]|...].

?-
```

## Lua

```function nextrow(t)
local ret = {}
t[0], t[#t+1] = 0, 0
for i = 1, #t do ret[i] = t[i-1] + t[i] end
return ret
end

function triangle(n)
t = {1}
for i = 1, n do
print(unpack(t))
t = nextrow(t)
end
end```

## Maple

```f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);

f(3);```
```   1
1 1
1 2 1
```

## Mathematica /Wolfram Language

```n=7;
Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[
{Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]```

A more graphical output with arrows would involve the plotting functionality with Graph[]:

```nmax := 10;
pascal[nmax_] := Module[
{vals = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, n}],
ids = Table[{n, k}, {n, 0, nmax}, {k, 0, n}],
labels, left, right, leftright, edgeLabels
},
left = DeleteCases[Flatten[Flatten[ids, 1] /. {n_, k_} /; (n >= k + 1) :> {{n, k + 1} -> {n + 1, k + 1}}, 1], _?NumberQ];
right = DeleteCases[Flatten[Flatten[ids, 1] /. {n_, k_} /; (n > k) :> {{n, k} -> {n + 1, k + 1}}, 1], _?NumberQ];
leftright = DeleteCases[left \[Union] right, _ -> {b_, _} /; b > nmax];
edgeLabels = (# -> Style["+", Medium] & /@ leftright);
Graph[Flatten[ids, 1], leftright
, VertexLabels -> MapAt[Placed[#, Center] &, labels, {All, 2}]
, GraphLayout -> "SpringEmbedding"
, VertexSize -> 0.8, EdgeLabels -> edgeLabels
, PlotLabel -> "Pascal's Triangle"
]
];
pascal[nmax]```

## MATLAB / Octave

A matrix containing the pascal triangle can be obtained this way:

`pascal(n);`
```>> pascal(6)
ans =

1     1     1     1     1     1
1     2     3     4     5     6
1     3     6    10    15    21
1     4    10    20    35    56
1     5    15    35    70   126
1     6    21    56   126   252

```

The binomial coefficients can be extracted from the Pascal triangle in this way:

`  binomCoeff = diag(rot90(pascal(n)))',`
```>> for k=1:6,diag(rot90(pascal(k)))', end
ans =  1
ans =

1   1

ans =

1   2   1

ans =

1   3   3   1

ans =

1   4   6   4   1

ans =

1    5   10   10    5    1

```

Another way to get a formated pascals triangle is to use the convolution method:

```>>
x = [1  1] ;
y = 1;
for k=8:-1:1
fprintf(['%', num2str(k), 'c'], zeros(1,3)),
fprintf('%6d', y), fprintf('\n')
y = conv(y,x);

end
```

The result is:

```>>

1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
```

## Maxima

```sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))\$

display_pascal_triangle(n) := for i from 0 thru 6 do disp(sjoin(makelist(binomial(i, j), j, 0, i), " "));

display_pascal_triangle(6);
/* "1"
"1 1"
"1 2 1"
"1 3 3 1"
"1 4 6 4 1"
"1 5 10 10 5 1"
"1 6 15 20 15 6 1" */```

## Metafont

(The formatting starts to be less clear when numbers start to have more than two digits)

```vardef bincoeff(expr n, k) =
save ?;
? := (1 for i=(max(k,n-k)+1) upto n: * i endfor )
/ (1 for i=2 upto min(k, n-k): * i endfor); ?
enddef;

def pascaltr expr c =
string s_;
for i := 0 upto (c-1):
s_ := "" for k=0 upto (c-i): & "  " endfor;
s_ := s_ for k=0 upto i: & decimal(bincoeff(i,k))
& "  " if bincoeff(i,k)<9: & " " fi endfor;
message s_;
endfor
enddef;

pascaltr(4);
end```

## Microsoft Small Basic

Translation of: GW-BASIC
```TextWindow.Write("Number of rows? ")
For i = 0 To r - 1
c = 1
For k = 0 To i
TextWindow.CursorLeft = (k + 1) * 4 - Text.GetLength(c)
TextWindow.Write(c)
c = c * (i - k) / (k + 1)
EndFor
TextWindow.WriteLine("")
EndFor```

Output:

```Number of rows? 7
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
```

## Modula-2

```MODULE Pascal;
FROM FormatString IMPORT FormatString;

PROCEDURE PrintLine(n : INTEGER);
VAR
buf : ARRAY[0..63] OF CHAR;
m,j : INTEGER;
BEGIN
IF n<1 THEN RETURN END;
m := 1;
WriteString("1 ");
FOR j:=1 TO n-1 DO
m := m * (n - j) DIV j;
FormatString("%i ", buf, m);
WriteString(buf)
END;
WriteLn
END PrintLine;

PROCEDURE Print(n : INTEGER);
VAR i : INTEGER;
BEGIN
FOR i:=1 TO n DO
PrintLine(i)
END
END Print;

BEGIN
Print(10);

END Pascal.```

## NetRexx

```/* NetRexx */
options replace format comments java crossref symbols nobinary

numeric digits 1000 -- allow very large numbers
parse arg rows .
if rows = '' then rows = 11 -- default to 11 rows
printPascalTriangle(rows)
return

-- -----------------------------------------------------------------------------
method printPascalTriangle(rows = 11) public static
lines = ''
mx = (factorial(rows - 1) / factorial(rows % 2) / factorial(rows - 1 - rows % 2)).length() -- width of widest number

loop row = 1 to rows
n1 = 1.center(mx)
line = n1
loop col = 2 to row
n2 = col - 1
n1 = n1 * (row - n2) / n2
line = line n1.center(mx)
end col
lines[row] = line.strip()
end row

-- display triangle
ml = lines[rows].length() -- length of longest line
loop row = 1 to rows
say lines[row].centre(ml)
end row

return

-- -----------------------------------------------------------------------------
method factorial(n) public static
fac = 1
loop n_ = 2 to n
fac = fac * n_
end n_
return fac /*calc. factorial*/```
Output:
```                    1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84  126 126 84  36   9   1
1  10  45  120 210 252 210 120 45  10   1
```

## Nial

Like J

(pascal.nial)

```factorial is recur [ 0 =, 1 first, pass, product, -1 +]
combination is fork [ > [first, second], 0 first,
/ [factorial second, * [factorial - [second, first], factorial first] ]
]
pascal is transpose each combination cart [pass, pass] tell```

Using it

```|loaddefs 'pascal.nial'
|pascal 5```

## Nim

```import sequtils, strutils

proc printPascalTriangle(n: int) =
## Print a Pascal triangle.

# Build the triangle.
var triangle: seq[seq[int]]
for _ in 1..<n:
triangle.add zip(triangle[^1] & @[0], @[0] & triangle[^1]).mapIt(it[0] + it[1])

# Build the lines to display.
let length = len(\$max(triangle[^1]))  # Maximum length of number.
var lines: seq[string]
for row in triangle:

# Display the lines.
let lineLength = lines[^1].len    # Length of largest line (the last one).
for line in lines:
echo line.center(lineLength)

printPascalTriangle(10)```
Output:
```                   1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84  126 126 84  36   9   1 ```

A more optimized solution that doesn't require importing, but produces, naturally, uglier output, would look like this:

```const ROWS = 10
const TRILEN = toInt(ROWS * (ROWS + 1) / 2) # Sum of arth progression
var triangle = newSeqOfCap[Natural](TRILEN) # Avoid reallocations

proc printPascalTri(row: Natural, result: var seq[Natural]) =
for i in 2..row-1: add(result, result[^row] + result[^(row-1)])

echo result[^row..^1]
if row + 1 <= ROWS: printPascalTri(row + 1, result)

printPascalTri(1, triangle)```
Output:
```@[1]
@[1, 1]
@[1, 2, 1]
@[1, 3, 3, 1]
@[1, 4, 6, 4, 1]
@[1, 5, 10, 10, 5, 1]
@[1, 6, 15, 20, 15, 6, 1]
@[1, 7, 21, 35, 35, 21, 7, 1]
@[1, 8, 28, 56, 70, 56, 28, 8, 1]
@[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]```

## OCaml

```(* generate next row from current row *)
let next_row row =
List.map2 (+) ([0] @ row) (row @ [0])

(* returns the first n rows *)
let pascal n =
let rec loop i row =
if i = n then []
else row :: loop (i+1) (next_row row)
in loop 0 [1]```

## Octave

```function pascaltriangle(h)
for i = 0:h-1
for k = 0:h-i
printf("  ");
endfor
for j = 0:i
printf("%3d ", bincoeff(i, j));
endfor
printf("\n");
endfor
endfunction

pascaltriangle(4);```

## Oforth

No result if n <= 0

`: pascal(n)  [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop ;`
Output:
```10 pascal
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
```

## Oz

```declare
fun {NextLine Xs}
{List.zip 0|Xs {Append Xs [0]}
fun {\$ Left Right}
Left + Right
end}
end

fun {Triangle N}
{List.take {Iterate [1] NextLine} N}
end

fun lazy {Iterate I F}
I|{Iterate {F I} F}
end

%% Only works nicely for N =< 5.
proc {PrintTriangle T}
N = {Length T}
in
for
Line in T
Indent in N-1..0;~1
do
for _ in 1..Indent do {System.printInfo " "} end
for L in Line do {System.printInfo L#" "} end
{System.printInfo "\n"}
end
end
in
{PrintTriangle {Triangle 5}}```

For n = 0, prints nothing. For negative n, throws an exception.

## PARI/GP

```pascals_triangle(N)= {
my(row=[],prevrow=[]);
for(x=1,N,
if(x>5,break(1));
row=eval(Vec(Str(11^(x-1))));
print(row));
prevrow=row;
for(y=6,N,
for(p=2,#prevrow,
row[p]=prevrow[p-1]+prevrow[p]);
row=concat(row,1);
prevrow=row;
print(row);
);
}```

## Pascal

```Program PascalsTriangle(output);

procedure Pascal(r : Integer);
var
i, c, k : Integer;
begin
for i := 0 to r-1 do
begin
c := 1;
for k := 0 to i do
begin
write(c:3);
c := (c * (i-k)) div (k+1);
end;
writeln;
end;
end;

begin
Pascal(9)
end.```

Output:

```% ./PascalsTriangle
1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1
```

See Pascal.

## Perl

These functions perform as requested in the task: they print out the first n lines. If n <= 0, they print nothing. The output is simple (no fancy formatting).

```sub pascal {
my \$rows = shift;
my @next = (1);
for my \$n (1 .. \$rows) {
print "@next\n";
@next = (1, (map \$next[\$_]+\$next[\$_+1], 0 .. \$n-2), 1);
}
}```

If you want more than 68 rows, then use either "use bigint" or "use Math::GMP qw/:constant/" inside the function to enable bigints. We can also use a binomial function which will expand to bigints if many rows are requested:

Library: ntheory
```use ntheory qw/binomial/;
sub pascal {
my \$rows = shift;
for my \$n (0 .. \$rows-1) {
print join(" ", map { binomial(\$n,\$_) } 0 .. \$n), "\n";
}
}```

Here is a non-obvious version using bignum, which is limited to the first 23 rows because of the algorithm used:

```use bignum;
sub pascal_line { \$_[0] ? unpack "A(A6)*", 1000001**\$_[0] : 1 }
sub pascal { print "@{[map -+-\$_, pascal_line \$_]}\n" for 0..\$_[0]-1 }```

This triangle is build using the 'sock' or 'hockey stick' pattern property. Here I use the word tartaglia and not pascal because in my country it's called after the Niccolò Fontana, known also as Tartaglia. A full graphical implementation of 16 properties that can be found in the triangle can be found at mine Tartaglia's triangle

```#!/usr/bin/perl
use strict;
use warnings;

{
my @tartaglia ;
sub tartaglia {
my (\$x,\$y) = @_;
if (\$x == 0 or \$y == 0)  { \$tartaglia[\$x][\$y]=1 ; return 1};
my \$ret ;
foreach my \$yps (0..\$y){
\$ret += ( \$tartaglia[\$x-1][\$yps] || tartaglia(\$x-1,\$yps) );
}
\$tartaglia[\$x][\$y] = \$ret;
return \$ret;
}
}
sub tartaglia_row {
my \$y = shift;
my \$x = 0;
my @row;
\$row[0] = &tartaglia(\$x,\$y+1);
foreach my \$pos (0..\$y-1) {push @row, tartaglia(++\$x,--\$y)}
return @row;
}

for (0..5) {print join ' ', tartaglia_row(\$_),"\n"}
print "\n\n";

print tartaglia(3,3),"\n";
my @third = tartaglia_row(5);
print "@third\n";```

which output

```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

20
1 5 10 10 5 1
```

## Phix

```sequence row = {}
for m = 1 to 13 do
row = row & 1
for n=length(row)-1 to 2 by -1 do
row[n] += row[n-1]
end for
printf(1,repeat(' ',(13-m)*2))
for i=1 to length(row) do
printf(1," %3d",row[i])
end for
puts(1,'\n')
end for
```
Output:
```                           1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1
1  11  55 165 330 462 462 330 165  55  11   1
1  12  66 220 495 792 924 792 495 220  66  12   1
```

"Reflected" Pascal's triangle, it uses symmetry property to "mirror" second part. It determines even and odd strings. automatically.

## PHP

```<?php
//Author Ivan Gavryshin @dcc0
function tre(\$n) {
\$ck=1;
\$kn=\$n+1;

if(\$kn%2==0) {
\$kn=\$kn/2;
\$i=0;
}
else
{

\$kn+=1;
\$kn=\$kn/2;
\$i= 1;
}

for (\$k = 1; \$k <= \$kn-1; \$k++) {
\$ck = \$ck/\$k*(\$n-\$k+1);
\$arr[] = \$ck;
echo  "+" . \$ck ;

}

if (\$kn>1) {
echo \$arr[i];
\$arr=array_reverse(\$arr);
for (\$i; \$i<= \$kn-1; \$i++) {
echo  "+" . \$arr[\$i]  ;
}

}

}
//set amount of strings here
while (\$n<=20) {
++\$n;
echo tre(\$n);
echo "<br/>";
}

?>```

==PHP ==

```function pascalsTriangle(\$num){
\$c = 1;
\$triangle = Array();
for(\$i=0;\$i<=\$num;\$i++){
\$triangle[\$i] = Array();
if(!isset(\$triangle[\$i-1])){
\$triangle[\$i][] = \$c;
}else{
for(\$j=0;\$j<count(\$triangle[\$i-1])+1;\$j++){
\$triangle[\$i][] = (isset(\$triangle[\$i-1][\$j-1]) && isset(\$triangle[\$i-1][\$j])) ? \$triangle[\$i-1][\$j-1] + \$triangle[\$i-1][\$j] : \$c;
}
}
}
return \$triangle;
}

\$tria = pascalsTriangle(8);
foreach(\$tria as \$val){
foreach(\$val as \$value){
echo \$value . ' ';
}
echo '<br>';
}```
```                                       1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
```

## Picat

```%Author: Petar Kabashki
spatr([]) = [].
spatr([_|T]) = A, T = [] => A = [].
spatr([H|T]) = A, T = [TH|_] => A = [H+TH] ++ spatr(T).

table
patr(0) = [1].
patr(1) = [1, 1].
patr(N) = A, N > 1 => Apre = patr(N-1), A = [1] ++ spatr(Apre) ++ [1].

foreach(I in 0 .. 10) println(patr(I)) end.```
```
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]
[1,5,10,10,5,1]
[1,6,15,20,15,6,1]
[1,7,21,35,35,21,7,1]
[1,8,28,56,70,56,28,8,1]
[1,9,36,84,126,126,84,36,9,1]
[1,10,45,120,210,252,210,120,45,10,1]```

## PicoLisp

Translation of: C
```(de pascalTriangle (N)
(for I N
(space (* 2 (- N I)))
(let C 1
(for K I
(prin (align 3 C) " ")
(setq C (*/ C (- I K) K)) ) )
(prinl) ) )```

## PL/I

```declare (t, u)(40) fixed binary;
declare (i, n) fixed binary;

t,u = 0;
get (n);
if n <= 0 then return;

do n = 1 to n;
u(1) = 1;
do i = 1 to n;
u(i+1) = t(i) + t(i+1);
end;
put skip edit ((u(i) do i = 1 to n)) (col(40-2*n), (n+1) f(4));
t = u;
end;```
```                                        1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1```

## Potion

```printpascal = (n) :
if (n < 1) :
1 print
(1)
. else :
prev = printpascal(n - 1)
prev append(0)
curr = (1)
n times (i):
curr append(prev(i) + prev(i + 1))
.
"\n" print
curr join(", ") print
curr
.
.

## PowerShell

```\$Infinity = 1
\$NewNumbers = \$null
\$Numbers = \$null
\$Result = \$null
\$Number = \$null
\$Power = \$args[0]

Write-Host \$Power

For(
\$i=0;
\$i -lt \$Infinity;
\$i++
)
{
\$Numbers = New-Object Object[] 1
\$Numbers[0] = \$Power
For(
\$k=0;
\$k -lt \$NewNumbers.Length;
\$k++
)
{
\$Numbers = \$Numbers + \$NewNumbers[\$k]
}
If(
\$i -eq 0
)
{
\$Numbers = \$Numbers + \$Power
}
\$NewNumbers = New-Object Object[] 0
Try
{
For(
\$j=0;
\$j -lt \$Numbers.Length;
\$j++
)
{
\$Result = \$Numbers[\$j] + \$Numbers[\$j+1]
\$NewNumbers = \$NewNumbers + \$Result
}
}
Catch [System.Management.Automation.RuntimeException]
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Break;
}
Foreach(
\$Number in \$Numbers
)
{
If(
\$Number.ToString() -eq "+unendlich"
)
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Exit
}
}
Write-Host \$Numbers
\$Infinity++
}```

Save the above code to a .ps1 script file and start it by calling its name and providing N.

```PS C:\> & '.\Pascals Triangle.ps1' 1

----

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
```

## Prolog

Difference-lists are used to make quick append.

```pascal(N) :-
pascal(1, N, [1], [[1]|X]-X, L),
maplist(my_format, L).

pascal(Max, Max, L, LC, LF) :-
!,
make_new_line(L, NL),
append_dl(LC, [NL|X]-X, LF-[]).

pascal(N, Max, L, NC, LF) :-
build_new_line(L, NL),
append_dl(NC, [NL|X]-X, NC1),
N1 is N+1,
pascal(N1, Max, NL, NC1, LF).

build_new_line(L, R) :-
build(L, 0, X-X, R).

build([], V, RC, RF) :-
append_dl(RC, [V|Y]-Y, RF-[]).

build([H|T], V, RC, R) :-
V1 is V+H,
append_dl(RC, [V1|Y]-Y, RC1),
build(T, H, RC1, R).

append_dl(X1-X2, X2-X3, X1-X3).

% to have a correct output !
my_format([H|T]) :-
write(H),
maplist(my_writef, T),
nl.

my_writef(X) :-
writef(' %5r', [X]).```

Output :

``` ?- pascal(15).
1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1
1    10    45   120   210   252   210   120    45    10     1
1    11    55   165   330   462   462   330   165    55    11     1
1    12    66   220   495   792   924   792   495   220    66    12     1
1    13    78   286   715  1287  1716  1716  1287   715   286    78    13     1
1    14    91   364  1001  2002  3003  3432  3003  2002  1001   364    91    14     1
1    15   105   455  1365  3003  5005  6435  6435  5005  3003  1365   455   105    15     1
true.```

### An alternative

The above use of difference lists is a really innovative example of late binding. Here's an alternative source which, while possibly not as efficient (or as short) as the previous example, may be a little easier to read and understand.

```%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Produce a pascal's triangle of depth N
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%  Prolog is declarative.  The predicate pascal/3 below says that to produce
%  a row of depth N, we can do so by first producing the row at depth(N-1),
%  and then adding the paired values in that row.  The triangle is produced
%  by prepending the row at N-1 to the preceding rows as recursion unwinds.
%  The triangle produced by pascal/3 is upside down and lacks the last row,
%  so pascal/2 prepends the last row to the triangle and reverses it.
%  Finally, pascal/1 produces the triangle, iterates each row and prints it.
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
pascal_row([V], [V]).                           % No more value pairs to add
pascal_row([V0, V1|T], [V|Rest]) :-             % Add values from preceding row
V is V0 + V1, !, pascal_row([V1|T], Rest).  % Drops initial value (1).

pascal(1, [1], []).    % at depth 1, this row is [1] and no preceding rows.
pascal(N, [1|ThisRow], [Last|Preceding]) :- % Produce a row of depth N
succ(N0, N),                            % N is the successor to N0
pascal(N0, Last, Preceding),            % Get the previous row
!, pascal_row(Last, ThisRow).           % Calculate this row from the previous

pascal(N, Triangle) :-
pascal(N, Last, Rows),             % Retrieve row at depth N and preceding rows
!, reverse([Last|Rows], Triangle). % Add last row to triangle and reverse order

pascal(N) :-
pascal(N, Triangle), member(Row, Triangle), % Iterate and write each row
write(Row), nl, fail.
pascal(_).```
• Output*:
```?- pascal(5).
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]```

## PureBasic

```Procedure pascaltriangle( n.i)

For i=  0 To  n
c = 1
For k=0 To i
Print(Str( c)+" ")
c = c * (i-k)/(k+1);
Next ;k
PrintN(" "); nächste zeile
Next ;i

EndProcedure

OpenConsole()
Parameter.i = Val(ProgramParameter(0))
pascaltriangle(Parameter);
Input()```

## Python

### Procedural

```def pascal(n):
"""Prints out n rows of Pascal's triangle.
It returns False for failure and True for success."""
row = [1]
k = [0]
for x in range(max(n,0)):
print row
row=[l+r for l,r in zip(row+k,k+row)]
return n>=1```

or by creating a scan function:

```def scan(op, seq, it):
a = []
result = it
a.append(it)
for x in seq:
result = op(result, x)
a.append(result)
return a

def pascal(n):
def nextrow(row, x):
return [l+r for l,r in zip(row+[0,],[0,]+row)]

return scan(nextrow, range(n-1), [1,])

for row in pascal(4):
print(row)```

### Functional

Deriving finite and non-finite lists of pascal rows from a simple nextPascal step function:

Works with: Python version 3.7
```'''Pascal's triangle'''

from itertools import (accumulate, chain, islice)

# nextPascal :: [Int] -> [Int]
def nextPascal(xs):
'''A row of Pascal's triangle
derived from a preceding row.'''
return list(
map(add, [0] + xs, xs + [0])
)

# pascalTriangle :: Generator [[Int]]
def pascalTriangle():
'''A non-finite stream of
Pascal's triangle rows.'''
return iterate(nextPascal)([1])

# finitePascalRows :: Int -> [[Int]]
def finitePascalRows(n):
'''The first n rows of Pascal's triangle.'''
return accumulate(
chain(
[[1]], range(1, n)
),
lambda a, _: nextPascal(a)
)

# ------------------------ TESTS -------------------------
# main :: IO ()
def main():
'''Test of two different approaches:
- taking from a non-finite stream of rows,
- or constructing a finite list of rows.'''
print('\n'.join(map(
showPascal,
[
islice(
pascalTriangle(),       # Non finite,
7
),
finitePascalRows(7)         # finite.
]
)))

# showPascal :: [[Int]] -> String
def showPascal(xs):
'''Stringification of a list of
Pascal triangle rows.'''
ys = list(xs)

def align(w):
return lambda ns: center(w)(
' '
)('   '.join(map(str, ns)))

w = len('   '.join((map(str, ys[-1]))))
return '\n'.join(map(align(w), ys))

# ----------------------- GENERIC ------------------------

# center :: Int -> Char -> String -> String
def center(n):
'''String s padded with c to approximate centre,
fitting in but not truncated to width n.'''
def go(c, s):
qr = divmod(n - len(s), 2)
q = qr[0]
return (q * c) + s + ((q + qr[1]) * c)

return lambda c: lambda s: go(c, s)

# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated
applications of f to x.
'''
def go(x):
v = x
while True:
yield v
v = f(v)

return go

# MAIN ---
if __name__ == '__main__':
main()```
Output:
```             1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
1   6   15   20   15   6   1
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1
1   6   15   20   15   6   1```

## q

```pascal:{(x-1){0+':x,0}\1}
pascal 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1```

## Qi

```(define iterate
_ _ 0 -> []
F V N -> [V|(iterate F (F V) (1- N))])

(define next-row
R -> (MAPCAR + [0|R] (append R [0])))

(define pascal
N -> (iterate next-row [1] N))```

## Quackery

The behaviour of `pascal` for values less than 1 is the same as its behaviour for 1.

```  [ over size -
space swap of
swap join ]           is justify  ( \$ n -->   )

[ witheach
[ number\$
5 justify echo\$ ]
cr ]                  is echoline (   [ -->   )

[ [] 0 rot 0 join
witheach
[ tuck +
rot join swap ]
drop ]                is nextline (   [ --> [ )

[ ' [ 1 ] swap
1 - times
[ dup echoline
nextline ]
echoline ]            is pascal   (   n -->   )

16 pascal```
Output:
```    1
1    1
1    2    1
1    3    3    1
1    4    6    4    1
1    5   10   10    5    1
1    6   15   20   15    6    1
1    7   21   35   35   21    7    1
1    8   28   56   70   56   28    8    1
1    9   36   84  126  126   84   36    9    1
1   10   45  120  210  252  210  120   45   10    1
1   11   55  165  330  462  462  330  165   55   11    1
1   12   66  220  495  792  924  792  495  220   66   12    1
1   13   78  286  715 1287 1716 1716 1287  715  286   78   13    1
1   14   91  364 1001 2002 3003 3432 3003 2002 1001  364   91   14    1
1   15  105  455 1365 3003 5005 6435 6435 5005 3003 1365  455  105   15    1
```

## R

Translation of: Octave
```pascalTriangle <- function(h) {
for(i in 0:(h-1)) {
s <- ""
for(k in 0:(h-i)) s <- paste(s, "  ", sep="")
for(j in 0:i) {
s <- paste(s, sprintf("%3d ", choose(i, j)), sep="")
}
print(s)
}
}```

Here's an R version:

```pascalTriangle <- function(h) {
lapply(0:h, function(i) choose(i, 0:i))
}```

## Racket

Iterative version by summing rows up to ${\displaystyle n}$.

```#lang racket

(define (pascal n)
(define (next-row current-row)
(map + (cons 0 current-row)
(append current-row '(0))))
(reverse
(for/fold ([triangle '((1))])
([row (in-range 1 n)])
(cons (next-row (first triangle)) triangle))))```

## Raku

(formerly Perl 6)

Works with: rakudo version 2015-10-03

### using a lazy sequence generator

The following routine returns a lazy list of lines using the sequence operator (...). With a lazy result you need not tell the routine how many you want; you can just use a slice subscript to get the first N lines:

```sub pascal {
[1], { [0, |\$_ Z+ |\$_, 0] } ... *
}

.say for pascal[^10];```

One problem with the routine above is that it might recalculate the sequence each time you call it. Slightly more idiomatic would be to define the sequence as a lazy constant. Here we use the @ sigil to indicate that the sequence should cache its values for reuse, and use an explicit parameter \$prev for variety:

```constant @pascal = [1], -> \$prev { [0, |\$prev Z+ |\$prev, 0] } ... *;

.say for @pascal[^10];```

Since we use ordinary subscripting, non-positive inputs throw an index-out-of-bounds error.

### recursive

```multi sub pascal (1) { \$[1] }
multi sub pascal (Int \$n where 2..*) {
my @rows = pascal \$n - 1;
|@rows, [0, |@rows[*-1] Z+ |@rows[*-1], 0 ];
}

.say for pascal 10;```

Non-positive inputs throw a multiple-dispatch error.

### iterative

Translation of: Perl
```sub pascal (\$n where \$n >= 1) {
say my @last = 1;
for 1 .. \$n - 1 -> \$row {
@last = 1, |map({ @last[\$_] + @last[\$_ + 1] }, 0 .. \$row - 2), 1;
say @last;
}
}

pascal 10;```

Non-positive inputs throw a type check error.

Output:
```[1]
[1 1]
[1 2 1]
[1 3 3 1]
[1 4 6 4 1]
[1 5 10 10 5 1]
[1 6 15 20 15 6 1]
[1 7 21 35 35 21 7 1]
[1 8 28 56 70 56 28 8 1]
[1 9 36 84 126 126 84 36 9 1]```

## RapidQ

### Summing from Previous Rows

Translation of: BASIC

The main difference to BASIC implementation is the output formatting. RapidQ does not support PRINT USING. Instead, function FORMAT\$() is used. TAB() is not supported, so SPACE\$() was used instead.

Another difference is that in RapidQ, DIM does not clear array values to zero. But if dimensioning is done with DEF..., you can give the initial values in curly braces. If less values are given than there are elements in the array, the remaining positions are initialized to zero. RapidQ does not require simple variables to be declared before use.

```DEFINT values(100) = {0,1}

INPUT "Number of rows: "; nrows
PRINT SPACE\$((nrows)*3);"  1"
FOR row = 2 TO nrows
PRINT SPACE\$((nrows-row)*3+1);
FOR i = row TO 1 STEP -1
values(i) = values(i) + values(i-1)
PRINT FORMAT\$("%5d ", values(i));
NEXT i
PRINT
NEXT row```

### Using binary coefficients

Translation of: BASIC
```INPUT "Number of rows: "; nrows
FOR row = 0 TO nrows-1
c = 1
PRINT SPACE\$((nrows-row)*3);
FOR i = 0 TO row
PRINT FORMAT\$("%5d ", c);
c = c * (row - i) / (i+1)
NEXT i
PRINT
NEXT row```

## Red

```Red[]
pascal-triangle: function [
n [ integer! ] "number of rows"
][
row: make vector! [ 1 ]
loop n [
print row
left: copy row
right: copy row
insert left 0
append right 0
row: left + right
]
]```

Output:

```pascal-triangle 7
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
```

## Retro

```2 elements i j
: pascalTriangle
cr dup
[ dup !j 1 swap 1+ [ !i dup putn space @j @i - * @i 1+ / ] iter cr drop ] iter drop
;
13 pascalTriangle```

## REXX

There is no practical limit for this REXX version, triangles up to 46 rows have been generated (without wrapping) in a screen window with a width of 620 characters.

If the number (of rows) specified is negative,   the output is written to a (disk) file instead.   Triangles with over a   1,000   rows have been easily created.
The output file created (that is written to disk) is named     PASCALS.n     where   n   is the absolute value of the number entered.

Note:   Pascal's triangle is also known as:

•   Khayyam's triangle
•   Khayyam─Pascal's triangle
•   Tartaglia's triangle
•   Yang Hui's triangle
```/*REXX program displays (or writes to a file)   Pascal's triangle  (centered/formatted).*/
numeric digits 3000                              /*be able to handle gihugeic triangles.*/
parse arg nn .                                   /*obtain the optional argument from CL.*/
if nn=='' | nn==","  then nn= 10                 /*Not specified?  Then use the default.*/
n= abs(nn)                                       /*N  is the number of rows in triangle.*/
w= length( !(n-1)  %  !(n%2)  %  !(n - 1 - n%2)) /*W:  the width of biggest integer.    */
ww= (n-1) * (W + 1)   +   1                      /*WW:  "    "    " triangle's last row.*/
@.= 1;      \$.= @.;          unity= right(1, w)  /*defaults rows & lines; aligned unity.*/
/* [↓]  build rows of Pascals' triangle*/
do    r=1  for n;              rm= r-1   /*Note:  the first column is always  1.*/
do c=2  to rm;              cm= c-1   /*build the rest of the columns in row.*/
@.r.c= @.rm.cm  +  @.rm.c             /*assign value to a specific row & col.*/
\$.r  = \$.r   right(@.r.c, w)          /*and construct a line for output (row)*/
end   /*c*/                           /* [↑]    C  is the column being built.*/
if r\==1  then \$.r= \$.r  unity           /*for  rows≥2,  append a trailing  "1".*/
end      /*r*/                           /* [↑]    R  is the  row   being built.*/
/* [↑]  WIDTH: for nicely looking line.*/
do r=1  for n;     \$\$= center(\$.r, ww)      /*center this particular Pascals' row. */
if nn>0  then say                       \$\$  /*SAY    if   NN    is positive,  else */
else call lineout 'PASCALS.'n, \$\$  /*write this Pascal's row ───►  a file.*/
end   /*r*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; !=1;  do j=2  to arg(1); != !*j; end /*j*/;  return !  /*compute factorial*/```
output   when using the input of:     11
```                    1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1
```
output   when using the input of:     22

(Output shown at   4/5   size.)

```                                                                         1
1      1
1      2      1
1      3      3      1
1      4      6      4      1
1      5     10     10      5      1
1      6     15     20     15      6      1
1      7     21     35     35     21      7      1
1      8     28     56     70     56     28      8      1
1      9     36     84    126    126     84     36      9      1
1     10     45    120    210    252    210    120     45     10      1
1     11     55    165    330    462    462    330    165     55     11      1
1     12     66    220    495    792    924    792    495    220     66     12      1
1     13     78    286    715   1287   1716   1716   1287    715    286     78     13      1
1     14     91    364   1001   2002   3003   3432   3003   2002   1001    364     91     14      1
1     15    105    455   1365   3003   5005   6435   6435   5005   3003   1365    455    105     15      1
1     16    120    560   1820   4368   8008  11440  12870  11440   8008   4368   1820    560    120     16      1
1     17    136    680   2380   6188  12376  19448  24310  24310  19448  12376   6188   2380    680    136     17      1
1     18    153    816   3060   8568  18564  31824  43758  48620  43758  31824  18564   8568   3060    816    153     18      1
1     19    171    969   3876  11628  27132  50388  75582  92378  92378  75582  50388  27132  11628   3876    969    171     19      1
1     20    190   1140   4845  15504  38760  77520 125970 167960 184756 167960 125970  77520  38760  15504   4845   1140    190     20      1
1     21    210   1330   5985  20349  54264 116280 203490 293930 352716 352716 293930 203490 116280  54264  20349   5985   1330    210     21      1
```

## Ring

```row = 5
for i = 0 to row - 1
col = 1
see left("     ",row-i)
for k = 0 to i
see "" + col + " "
col = col*(i-k)/(k+1)
next
see nl
next```

Output:

```     1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```

## RPL

```« 0 SWAP FOR n
"" 0 n FOR p
n p COMB + " " +
NEXT
n 1 + DISP
NEXT
7 FREEZE
» 'PASCAL' STO
```
```8 PASCAL
```
Output:
```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 …
```

RPL screens are limited to 22 characters.

## Ruby

```def pascal(n)
raise ArgumentError, "must be positive." if n < 1
yield ar = [1]
(n-1).times do
ar.unshift(0).push(0) # tack a zero on both ends
yield ar = ar.each_cons(2).map(&:sum)
end
end

pascal(8){|row| puts row.join(" ").center(20)}```
Output:
```         1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
```

Or for more or less a translation of the two line Haskell version (with inject being abused a bit I know):

```def next_row(row) ([0] + row).zip(row + [0]).collect {|l,r| l + r } end

def pascal(n) n.times.inject([1]) {|x,_| next_row x } end

8.times{|i| p pascal(i)}```
Output:
```[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
```

## Run BASIC

```input "number of rows? ";r
for i = 0 to r - 1
c = 1
print left\$("                          ",(r*2)-(i*2));
for k = 0 to i
print using("####",c);
c = c*(i-k)/(k+1)
next
print
next```

Output:

```Number of rows? ?5
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1```

## Rust

Translation of: C
```fn pascal_triangle(n: u64)
{

for i in 0..n {
let mut c = 1;
for _j in 1..2*(n-1-i)+1 {
print!(" ");
}
for k in 0..i+1 {
print!("{:2} ", c);
c = c * (i-k)/(k+1);
}
println!();
}
}```

## Scala

### Functional solutions

#### Summing: Recursive row definition

```  def tri(row: Int): List[Int] =
row match {
case 1 => List(1)
case n: Int => 1 +: ((tri(n - 1) zip tri(n - 1).tail) map { case (a, b) => a + b }) :+ 1
}```

Function to pretty print n rows:

```def prettyTri(n:Int) = (1 to n) foreach {i => print(" "*(n-i)); tri(i) map (c => print(c + " ")); println}

prettyTri(5)```
Output:
```    1
1 1
1 2 1
1 3 3 1
1 4 6 4 1```

#### Summing: Scala Stream (Recursive & Memoization)

```object Blaise extends App {
def pascalTriangle(): Stream[Vector[Int]] =
Vector(1) #:: Stream.iterate(Vector(1, 1))(1 +: _.sliding(2).map(_.sum).toVector :+ 1)

val output = pascalTriangle().take(15).map(_.mkString(" "))
val longest = output.last.length

println("Pascal's Triangle")
output.foreach(line => println(s"\${" " * ((longest - line.length) / 2)}\$line"))
}```
Output:

See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).

## Scheme

Works with: Scheme version R${\displaystyle ^{5}}$RS
```(define (next-row row)
(map + (cons 0 row) (append row '(0))))

(define (triangle row rows)
(if (= rows 0)
'()
(cons row (triangle (next-row row) (- rows 1)))))

(triangle (list 1) 5)```

Output:

`((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))`

## Seed7

```\$ include "seed7_05.s7i";

const proc: main is func
local
var integer: numRows is 0;
var array integer: values is [] (0, 1);
var integer: row is 0;
var integer: index is 0;
begin
write("Number of rows: ");
for row range 2 to numRows do
write("" lpad (numRows - row) * 3);
values &:= [] 0;
for index range succ(row) downto 2 do
values[index] +:= values[pred(index)];
write(" " <& values[index] lpad 5);
end for;
writeln;
end for;
end func;```

## Sidef

```func pascal(rows) {
var row = [1]
{ | n|
say row.join(' ')
row = [1, {|i| row[i] + row[i+1] }.map(0 .. n-2)..., 1]
} << 1..rows
}

pascal(10)```

## Stata

First, a few ways to compute a "Pascal matrix". With the first, the upper triangle is made of missing values (zeros with the other two).

```function pascal1(n) {
return(comb(J(1,n,0::n-1),J(n,1,0..n-1)))
}

function pascal2(n) {
a = I(n)
a[.,1] = J(n,1,1)
for (i=3; i<=n; i++) {
a[i,2..i-1] = a[i-1,2..i-1]+a[i-1,1..i-2]
}
return(a)
}

function pascal3(n) {
a = J(n,n,0)
for (i=1; i<n; i++) {
a[i+1,i] = i
}
s = p = I(n)
k = 1
for (i=0; i<n; i++) {
p = p*a/k++
s = s+p
}
return(s)
}```

Now print the Pascal triangle.

```function print_pascal_triangle(n) {
a = pascal1(n)
for (i=1; i<=n; i++) {
for (j=1; j<=i; j++) {
printf("%10.0f",a[i,j])
}
printf("\n")
}
}

print_pascal_triangle(5)
1
1         1
1         2         1
1         3         3         1
1         4         6         4         1```

## Swift

```func pascal(n:Int)->[Int]{
if n==1{
let a=[1]
print(a)
return a
}
else{
var a=pascal(n:n-1)
var temp=a
for i in 0..<a.count{
if i+1==a.count{
temp.append(1)
break
}
temp[i+1] = a[i]+a[i+1]
}
a=temp
print(a)
return a
}
}
let waste = pascal(n:10)```

## Tcl

### Summing from Previous Rows

```proc pascal_iterative n {
if {\$n < 1} {error "undefined behaviour for n < 1"}
set row [list 1]
lappend rows \$row
set i 1
while {[incr i] <= \$n} {
set prev \$row
set row [list 1]
for {set j 1} {\$j < [llength \$prev]} {incr j} {
lappend row [expr {[lindex \$prev [expr {\$j - 1}]] + [lindex \$prev \$j]}]
}
lappend row 1
lappend rows \$row
}
return \$rows
}

puts [join [pascal_iterative 6] \n]```
```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1```

### Using binary coefficients

Translation of: BASIC
```proc pascal_coefficients n {
if {\$n < 1} {error "undefined behaviour for n < 1"}
for {set i 0} {\$i < \$n} {incr i} {
set c 1
set row [list \$c]
for {set j 0} {\$j < \$i} {incr j} {
set c [expr {\$c * (\$i - \$j) / (\$j + 1)}]
lappend row \$c
}
lappend rows \$row
}
return \$rows
}

puts [join [pascal_coefficients 6] \n]```

### Combinations

Translation of: Java

Thanks to Tcl 8.5's arbitrary precision integer arithmetic, this solution is not limited to a couple of dozen rows. Uses a caching factorial calculator to improve performance.

```package require Tcl 8.5

proc pascal_combinations n {
if {\$n < 1} {error "undefined behaviour for n < 1"}
for {set i 0} {\$i < \$n} {incr i} {
set row [list]
for {set j 0} {\$j <= \$i} {incr j} {
lappend row [C \$i \$j]
}
lappend rows \$row
}
return \$rows
}

proc C {n k} {
expr {[ifact \$n] / ([ifact \$k] * [ifact [expr {\$n - \$k}]])}
}

set fact_cache {1 1}
proc ifact n {
global fact_cache
if {\$n < [llength \$fact_cache]} {
return [lindex \$fact_cache \$n]
}
set i [expr {[llength \$fact_cache] - 1}]
set sum [lindex \$fact_cache \$i]
while {\$i < \$n} {
incr i
set sum [expr {\$sum * \$i}]
lappend fact_cache \$sum
}
return \$sum
}

puts [join [pascal_combinations 6] \n]```

### Comparing Performance

```set n 100
puts "calculate \$n rows:"
foreach proc {pascal_iterative pascal_coefficients pascal_combinations} {
puts "\$proc: [time [list \$proc \$n] 100]"
}```
Output:
```calculate 100 rows:
pascal_iterative: 2800.14 microseconds per iteration
pascal_coefficients: 8760.98 microseconds per iteration
pascal_combinations: 38176.66 microseconds per iteration```

## TI-83 BASIC

### Using Addition of Previous Rows

```PROGRAM:PASCALTR
:Lbl IN
:ClrHome
:Disp "NUMBER OF ROWS"
:Input N
:If N < 1:Goto IN
:{N,N}→dim([A])
:"CHEATING TO MAKE IT FASTER"
:For(I,1,N)
:1→[A](1,1)
:End
:For(I,2,N)
:For(J,2,I)
:[A](I-1,J-1)+[A](I-1,J)→[A](I,J)
:End
:End
:[A]```

### Using nCr Function

```PROGRAM:PASCALTR
:Lbl IN
:ClrHome
:Disp "NUMBER OF ROWS"
:Input N
:If N < 1:Goto IN
:{N,N}→dim([A])
:For(I,2,N)
:For(J,2,I)
:(I-1) nCr (J-1)→[A](I,J)
:End
:End
:[A]```

## Turing

```proc pascal (n : int)
for i : 0 .. n
var c := 1
for k : 0 .. i
put c : 4 ..
c := c * (i - k) div (k + 1)
end for
put ""
end for
end pascal

pascal(5)```

Output:

```  1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
```

## TypeScript

Translation of: XPL0
```// Pascal's triangle

function pascal(n: number): void {
// Display the first n rows of Pascal's triangle
// if n<=0 then nothing is displayed
var ld: number[] = new Array(40); // Old
var nw: number[] = new Array(40); // New
for (var row = 0; row < n; row++) {
nw[0] = 1;
for (var i = 1; i <= row; i++)
nw[i] = ld[i - 1] + ld[i];
process.stdout.write(" ".repeat((n - row - 1) * 2));
for (var i = 0; i <= row; i++) {
if (nw[i] < 100)
process.stdout.write(" ");
process.stdout.write(nw[i].toString());
if (nw[i] < 10)
process.stdout.write(" ");
process.stdout.write(" ");
}
nw[row + 1] = 0;
// We do not copy data from nw to ld
// but we work with references.
var tmp = ld;
ld = nw;
nw = tmp;
console.log();
}
}

pascal(13);```
Output:
```                         1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6   1
1   7   21  35  35  21  7   1
1   8   28  56  70  56  28  8   1
1   9   36  84 126 126  84  36  9   1
1   10  45 120 210 252 210 120  45  10  1
1   11  55 165 330 462 462 330 165  55  11  1
1   12  66 220 495 792 924 792 495 220  66  12  1
```

## uBasic/4tH

```Input "Number Of Rows: "; N
@(1) = 1
Print Tab((N+1)*3);"1"

For R = 2 To N
Print Tab((N-R)*3+1);
For I = R To 1 Step -1
@(I) = @(I) + @(I-1)
Print Using "______";@(i);
Next
Next

Print
End```

Output:

```Number Of Rows: 10
1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1

0 OK, 0:380
```

## UNIX Shell

Works with: Bourne Again SHell

Any n <= 1 will print the "1" row.

```#! /bin/bash
pascal() {
local -i n=\${1:-1}
if (( n <= 1 )); then
echo 1
else
local output=\$( \$FUNCNAME \$((n - 1)) )
set -- \$( tail -n 1 <<<"\$output" )   # previous row
echo "\$output"
printf "1 "
while [[ -n \$1 ]]; do
printf "%d " \$(( \$1 + \${2:-0} ))
shift
done
echo
fi
}
pascal "\$1"```

## Ursala

Zero maps to the empty list. Negatives are inexpressible. This solution uses a library function for binomial coefficients.

```#import std
#import nat

pascal = choose**ziDS+ iota*t+ iota+ successor```

This solution uses direct summation. The algorithm is to insert zero at the head of a list (initially the unit list <1>), zip it with its reversal, map the sum over the list of pairs, iterate n times, and return the trace.

```#import std
#import nat

pascal "n" = (next"n" sum*NiCixp) <1>```

test program:

```#cast %nLL

example = pascal 10```
Output:
```<
<1>,
<1,1>,
<1,2,1>,
<1,3,3,1>,
<1,4,6,4,1>,
<1,5,10,10,5,1>,
<1,6,15,20,15,6,1>,
<1,7,21,35,35,21,7,1>,
<1,8,28,56,70,56,28,8,1>,
<1,9,36,84,126,126,84,36,9,1>>```

## VBA

```Option Base 1
Private Sub pascal_triangle(n As Integer)
Dim odd() As String
Dim eve() As String
ReDim odd(1)
ReDim eve(2)
odd(1) = "  1"
For i = 1 To n
If i Mod 2 = 1 Then
Debug.Print String\$(2 * n - 2 * i, " ") & Join(odd, " ")
eve(1) = "  1"
ReDim Preserve eve(i + 1)
For j = 2 To i
eve(j) = Format(CStr(Val(odd(j - 1)) + Val(odd(j))), "@@@")
Next j
eve(i + 1) = "  1"
Else
Debug.Print String\$(2 * n - 2 * i, " ") & Join(eve, " ")
odd(1) = "  1"
ReDim Preserve odd(i + 1)
For j = 2 To i
odd(j) = Format(CStr(Val(eve(j - 1)) + Val(eve(j))), "@@@")
Next j
odd(i + 1) = "  1"
End If
Next i
End Sub
Public Sub main()
pascal_triangle 13
End Sub```
Output:
```                          1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1
1   9  36  84 126 126  84  36   9   1
1  10  45 120 210 252 210 120  45  10   1
1  11  55 165 330 462 462 330 165  55  11   1
1  12  66 220 495 792 924 792 495 220  66  12   1```

## VBScript

Derived from the BASIC version.

```Pascal_Triangle(WScript.Arguments(0))
Function Pascal_Triangle(n)
Dim values(100)
values(1) = 1
WScript.StdOut.Write values(1)
WScript.StdOut.WriteLine
For row = 2 To n
For i = row To 1 Step -1
values(i) = values(i) + values(i-1)
WScript.StdOut.Write values(i) & " "
Next
WScript.StdOut.WriteLine
Next
End Function```
Output:

Invoke from a command line.

```F:\VBScript>cscript /nologo rosettacode-pascals_triangle.vbs 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
```

## Vedit macro language

### Summing from Previous Rows

Translation of: BASIC

Vedit macro language does not have actual arrays (edit buffers are normally used for storing larger amounts of data). However, a numeric register can be used as index to access another numeric register. For example, if #99 contains value 2, then #@99 accesses contents of numeric register #2.

```#100 = Get_Num("Number of rows: ", STATLINE)
#0=0; #1=1
Ins_Char(' ', COUNT, #100*3-2) Num_Ins(1)
for (#99 = 2; #99 <= #100; #99++) {
Ins_Char(' ', COUNT, (#100-#99)*3)
#@99 = 0
for (#98 = #99; #98 > 0; #98--) {
#97 = #98-1
#@98 += #@97
Num_Ins(#@98, COUNT, 6)
}
Ins_Newline
}```

### Using binary coefficients

Translation of: BASIC
```#1 = Get_Num("Number of rows: ", STATLINE)
for (#2 = 0; #2 < #1; #2++) {
#3 = 1
Ins_Char(' ', COUNT, (#1-#2-1)*3)
for (#4 = 0; #4 <= #2; #4++) {
Num_Ins(#3, COUNT, 6)
#3 = #3 * (#2-#4) / (#4+1)
}
Ins_Newline
}```

## Visual Basic

Works with: Visual Basic version VB6 Standard
```Sub pascaltriangle()
'Pascal's triangle
Const m = 11
Dim t(40) As Integer, u(40) As Integer
Dim i As Integer, n As Integer, s As String, ss As String
ss = ""
For n = 1 To m
u(1) = 1
s = ""
For i = 1 To n
u(i + 1) = t(i) + t(i + 1)
s = s & u(i) & " "
t(i) = u(i)
Next i
ss = ss & s & vbCrLf
Next n
MsgBox ss, , "Pascal's triangle"
End Sub 'pascaltriangle```
Output:
```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
```

## Visual Basic .NET

Translation of: C#
```Imports System.Numerics

Module Module1
Iterator Function GetRow(rowNumber As BigInteger) As IEnumerable(Of BigInteger)
Dim denominator As BigInteger = 1
Dim numerator = rowNumber

Dim currentValue As BigInteger = 1
For counter = 0 To rowNumber
Yield currentValue
currentValue = currentValue * numerator
numerator = numerator - 1
currentValue = currentValue / denominator
denominator = denominator + 1
Next
End Function

Function GetTriangle(quantityOfRows As Integer) As IEnumerable(Of BigInteger())
Dim range = Enumerable.Range(0, quantityOfRows).Select(Function(num) New BigInteger(num))
Return range.Select(Function(num) GetRow(num).ToArray())
End Function

Function CenterString(text As String, width As Integer)
Dim spaces = width - text.Length
Dim padLeft = (spaces / 2) + text.Length
End Function

Function FormatTriangleString(triangle As IEnumerable(Of BigInteger())) As String
Dim maxDigitWidth = triangle.Last().Max().ToString().Length
Dim rows = triangle.Select(Function(arr) String.Join(" ", arr.Select(Function(array) CenterString(array.ToString(), maxDigitWidth))))
Dim maxRowWidth = rows.Last().Length
Return String.Join(Environment.NewLine, rows.Select(Function(row) CenterString(row, maxRowWidth)))
End Function

Sub Main()
Dim triangle = GetTriangle(20)
Dim output = FormatTriangleString(triangle)
Console.WriteLine(output)
End Sub

End Module```
Output:
```                                                           1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5     10    10    5     1
1     6     15    20    15    6     1
1     7     21    35    35    21    7     1
1     8     28    56    70    56    28    8     1
1     9     36    84   126   126    84    36    9     1
1     10    45   120   210   252   210   120    45    10    1
1     11    55   165   330   462   462   330   165    55    11    1
1     12    66   220   495   792   924   792   495   220    66    12    1
1     13    78   286   715  1287  1716  1716  1287   715   286    78    13    1
1     14    91   364  1001  2002  3003  3432  3003  2002  1001   364    91    14    1
1     15   105   455  1365  3003  5005  6435  6435  5005  3003  1365   455   105    15    1
1     16   120   560  1820  4368  8008  11440 12870 11440 8008  4368  1820   560   120    16    1
1     17   136   680  2380  6188  12376 19448 24310 24310 19448 12376 6188  2380   680   136    17    1
1     18   153   816  3060  8568  18564 31824 43758 48620 43758 31824 18564 8568  3060   816   153    18    1
1     19   171   969  3876  11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876   969   171    19    1```

## Wren

Library: Wren-fmt
Library: Wren-math
```import "./fmt" for Fmt
import "./math" for Int

var pascalTriangle = Fn.new { |n|
if (n <= 0) return
for (i in 0...n) {
System.write("   " * (n-i-1))
for (j in 0..i) {
Fmt.write("\$3d   ", Int.binomial(i, j))
}
System.print()
}
}

pascalTriangle.call(13)```
Output:
```                                      1
1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5    10    10     5     1
1     6    15    20    15     6     1
1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1
1     9    36    84   126   126    84    36     9     1
1    10    45   120   210   252   210   120    45    10     1
1    11    55   165   330   462   462   330   165    55    11     1
1    12    66   220   495   792   924   792   495   220    66    12     1
```

## X86 Assembly

Works with: NASM
Works with: Windows

uses: io.inc - Macro library from SASM

```%include "io.inc"

section .text
global CMAIN
CMAIN:
mov ebx, 7 ;size
call mloop
ret

mloop:
mov edx, 0 ;edx stands for the nth line
looping:
push ebx
push edx
call line
pop edx
pop ebx
inc edx
cmp edx, ebx
jl looping
xor eax, eax
ret

line:
mov ecx, 0 ;ecx stands for the nth character in each line
mlp:
push ecx
push edx
call nCk
pop edx
pop ecx
PRINT_DEC 4, eax ;print returned number
PRINT_STRING " "
inc ecx
cmp ecx, edx ;if
jle mlp
NEWLINE
ret

nCk:
;ecx : j
;edx : i
mov esi, edx
call fac ;i!
push eax ;save i!
mov esi, ecx
call fac ;j!
push eax ;save j!
mov ebx, edx
sub ebx, ecx ;(i-j)
mov esi, ebx
call fac ;(i-j)!
pop ebx ;(i-j)! is in eax
mul ebx ;(i-j)! * j!
mov ecx, eax
pop eax ; get i!
div ecx ; ; last step : i! divided by (i-j)! * j!
ret

fac:
push ecx
push edx
mov eax, 1
mov ecx, esi
cmp ecx, 0 ; 0! returns 1
je facz
lp:
mul ecx ;multiplies eax by ecx and then decrements ecx until ecx is 0
dec ecx
cmp ecx, 0
jg lp
jmp end
facz:
mov eax, 1
end:
pop edx
pop ecx
ret```
Output:
```1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
```

## XBasic

Translation of: GW-BASIC
Works with: Windows XBasic
```PROGRAM "pascal"
VERSION "0.0001"

DECLARE FUNCTION Entry()

FUNCTION Entry()
r@@ = UBYTE(INLINE\$("Number of rows? "))
FOR i@@ = 0 TO r@@ - 1
c%% = 1
FOR k@@ = 0 TO i@@
PRINT FORMAT\$("####", c%%);
c%% = c%% * (i@@ - k@@) / (k@@ + 1)
NEXT k@@
PRINT
NEXT i@@
END FUNCTION
END PROGRAM```
Output:
```Number of rows? 7
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
```

## XPL0

```include c:\cxpl\codes;

proc Pascal(N);         \Display the first N rows of Pascal's triangle
int  N;                 \if N<=0 then nothing is displayed
int  Row, I, Old(40), New(40);
[for Row:= 0 to N-1 do
[New(0):= 1;
for I:= 1 to Row do New(I):= Old(I-1) + Old(I);
for I:= 1 to (N-Row-1)*2 do ChOut(0, ^ );
for I:= 0 to Row do
[if New(I)<100 then ChOut(0, ^ );
IntOut(0, New(I));
if New(I)<10 then ChOut(0, ^ );
ChOut(0, ^ );
];
New(Row+1):= 0;
I:= Old;  Old:= New;  New:= I;
CrLf(0);
];
];

Pascal(13)```
Output:
```                         1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10  10  5   1
1   6   15  20  15  6   1
1   7   21  35  35  21  7   1
1   8   28  56  70  56  28  8   1
1   9   36  84 126 126  84  36  9   1
1   10  45 120 210 252 210 120  45  10  1
1   11  55 165 330 462 462 330 165  55  11  1
1   12  66 220 495 792 924 792 495 220  66  12  1
```

## zkl

Translation of: C
```fcn pascalTriangle(n){ // n<=0-->""
foreach i in (n){
c := 1;
print(" "*(2*(n-1-i)));
foreach k in (i+1){
print("%3d ".fmt(c));
c = c * (i-k)/(k+1);
}
println();
}
}

pascalTriangle(8);```
Output:
```                1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5  10  10   5   1
1   6  15  20  15   6   1
1   7  21  35  35  21   7   1
```

## ZX Spectrum Basic

In edit mode insert:

``` 10 INPUT "How many rows? ";n
15 IF n<1 THEN GO TO 210
20 DIM c(n)
25 DIM d(n)
30 LET c(1)=1
35 LET d(1)=1
40 FOR r=1 TO n
50 FOR i=1 TO (n-r)
60 PRINT " ";
70 NEXT i
80 FOR i=1 TO r
90 PRINT c(i);" ";
100 NEXT i
110 PRINT
120 IF r>=n THEN GO TO 140
130 LET d(r+1)=1
140 FOR i=2 TO r
150 LET d(i)=c(i-1)+c(i)
160 NEXT i
165 IF r>=n THEN GO TO 200
170 FOR i=1 TO r+1
180 LET c(i)=d(i)
190 NEXT i
200 NEXT r```

Then in command mode (basically don't put a number in front):

`RUN`
Output:
```        1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
```