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Catalan numbers (view source)

Revision as of 18:56, 1 May 2024

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Line 21:
*[[Evaluate binomial coefficients]]
<br><br>
 
=={{header|11l}}==
<langsyntaxhighlight lang="11l">V c = 1
L(n) 1..15
print(c)
c = 2 * (2 * n - 1) * c I/ (n + 1)</langsyntaxhighlight>
{{out}}
<pre>
Line 45 ⟶ 44:
2674440
</pre>
 
=={{header|360 Assembly}}==
Very compact version.
<langsyntaxhighlight lang="360asm">CATALAN CSECT 08/09/2015
USING CATALAN,R15
LA R7,1 c=1
Line 69 ⟶ 67:
PG DS CL24
YREGS
END CATALAN</langsyntaxhighlight>
{{out}}
<pre>
Line 88 ⟶ 86:
15 9694845
</pre>
 
=={{header|ABAP}}==
This works for ABAP Version 7.40 and above
 
<syntaxhighlight lang="abap">
<lang ABAP>
report z_catalan_numbers.
 
Line 125 ⟶ 122:
write / |C({ sy-index - 1 }) = { catalan_numbers=>get_nth_number( sy-index - 1 ) }|.
enddo.
</syntaxhighlight>
</lang>
 
{{out}}
Line 146 ⟶ 143:
C(14) = 2674440
</pre>
 
=={{header|Action!}}==
{{libheader|Action! Tool Kit}}
<langsyntaxhighlight Actionlang="action!">INCLUDE "D2:REAL.ACT" ;from the Action! Tool Ki
 
PROC Main()
Line 168 ⟶ 164:
PrintF("C(%B)=",n) PrintRE(c)
OD
RETURN</langsyntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Catalan_numbers.png Screenshot from Atari 8-bit computer]
Line 188 ⟶ 184:
C(15)=9694845
</pre>
 
=={{header|Ada}}==
<langsyntaxhighlight Adalang="ada">with Ada.Text_IO; use Ada.Text_IO;
 
procedure Test_Catalan is
Line 205 ⟶ 200:
Put_Line (Integer'Image (N) & " =" & Integer'Image (Catalan (N)));
end loop;
end Test_Catalan;</langsyntaxhighlight>
{{out|Sample output}}
<pre>
Line 225 ⟶ 220:
15 = 9694845
</pre>
 
=={{header|ALGOL 68}}==
<langsyntaxhighlight lang="algol68"># calculate the first few catalan numbers, using LONG INT values #
# (64-bit quantities in Algol 68G which can handle up to C23) #
 
Line 258 ⟶ 252:
FOR i FROM 0 TO 15 DO
print( ( whole( i, -2 ), ": ", whole( catalan( i ), 0 ), newline ) )
OD</langsyntaxhighlight>
{{out}}
<pre>
Line 278 ⟶ 272:
15: 9694845
</pre>
 
=={{header|ALGOL W}}==
<langsyntaxhighlight lang="algolw">begin
% print the catalan numbers up to C15 %
integer Cprev;
Line 289 ⟶ 282:
write( s_w := 0, i_w := 3, n, ": ", i_w := 9, Cprev );
end for_n
end.</langsyntaxhighlight>
{{out}}
<pre>
Line 309 ⟶ 302:
15: 9694845
</pre>
 
=={{header|APL}}==
<langsyntaxhighlight lang="apl"> {(!2×⍵)÷(!⍵+1)×!⍵}(⍳15)-1</langsyntaxhighlight>
{{out}}
<pre>1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440</pre>
 
=={{header|Arturo}}==
<langsyntaxhighlight lang="rebol">catalan: function [n][
if? n=0 -> 1
else -> div (catalan n-1) * (4*n)-2 n+1
Line 326 ⟶ 317:
pad.left to :string catalan i 20
]
]</langsyntaxhighlight>
{{out}}
Line 346 ⟶ 337:
14 2674440
15 9694845</pre>
 
=={{header|AutoHotkey}}==
As AutoHotkey has no BigInt, the formula had to be tweaked to prevent overflow. It still fails after n=22
<langsyntaxhighlight AHKlang="ahk">Loop 15
out .= "`n" Catalan(A_Index)
Msgbox % clipboard := SubStr(out, 2)
Line 368 ⟶ 358:
 
Return i
}</langsyntaxhighlight>
{{out}}
<pre>1
Line 385 ⟶ 375:
2674440
9694845</pre>
 
=={{header|AWK}}==
<langsyntaxhighlight AWKlang="awk"># syntax: GAWK -f CATALAN_NUMBERS.AWK
BEGIN {
for (i=0; i<=15; i++) {
Line 402 ⟶ 391:
}
return(ans)
}</langsyntaxhighlight>
{{out}}
<pre>
Line 429 ⟶ 418:
Use of <code>REDIM PRESERVE</code> means this will not work in QBasic (although that could be worked around if desired).
 
<langsyntaxhighlight lang="qbasic">DECLARE FUNCTION catalan (n as INTEGER) AS SINGLE
 
REDIM SHARED results(0) AS SINGLE
Line 446 ⟶ 435:
END IF
catalan = results(n)
END FUNCTION</langsyntaxhighlight>
{{out}}
1 1
Line 463 ⟶ 452:
14 2674440
15 9694845
 
==={{header|Applesoft BASIC}}===
{{works with|Chipmunk Basic}}
{{works with|QBasic}}
<syntaxhighlight lang="qbasic">10 HOME : REM 10 CLS for Chipmunk Basic/QBasic
20 DIM c(15)
30 c(0) = 1
40 PRINT 0, c(0)
50 FOR n = 0 TO 14
60 c(n + 1) = 0
70 FOR i = 0 TO n
80 c(n + 1) = c(n + 1) + c(i) * c(n - i)
90 NEXT i
100 PRINT n + 1, c(n + 1)
110 NEXT n
120 END</syntaxhighlight>
 
==={{header|BASIC256}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="freebasic">function factorial(n)
if n = 0 then return 1
return n * factorial(n - 1)
end function
 
function catalan1(n)
prod = 1
for i = n + 2 to 2 * n
prod *= i
next i
return int(prod / factorial(n))
end function
 
function catalan2(n)
if n = 0 then return 1
sum = 0
for i = 0 to n - 1
sum += catalan2(i) * catalan2(n - 1 - i)
next i
return sum
end function
 
function catalan3(n)
if n = 0 then return 1
return catalan3(n - 1) * 2 * (2 * n - 1) \ (n + 1)
end function
 
print "n", "First", "Second", "Third"
print "-", "-----", "------", "-----"
print
for i = 0 to 15
print i, catalan1(i), catalan2(i), catalan3(i)
next i</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
 
==={{header|BBC BASIC}}===
<syntaxhighlight lang="bbcbasic"> 10 FOR i% = 1 TO 15
20 PRINT FNcatalan(i%)
30 NEXT
40 END
50 DEF FNcatalan(n%)
60 IF n% = 0 THEN = 1
70 = 2 * (2 * n% - 1) * FNcatalan(n% - 1) / (n% + 1)</syntaxhighlight>
{{out}}
<pre> 1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440</pre>
 
==={{header|Chipmunk Basic}}===
{{works with|Chipmunk Basic|3.6.4}}
{{trans|Run BASIC}}
<syntaxhighlight lang="qbasic">10 FOR i = 1 TO 15
20 PRINT i;" ";catalan(i)
30 NEXT
40 END
50 SUB catalan(n)
60 catalan = 1
70 IF n <> 0 THEN catalan = ((2*((2*n)-1))/(n+1))*catalan(n-1)
80 END SUB</syntaxhighlight>
 
==={{header|Craft Basic}}===
<syntaxhighlight lang="basic">dim c[16]
 
let c[0] = 1
 
for n = 0 to 15
 
let p = n + 1
let c[p] = 0
 
for i = 0 to n
 
let q = n - i
let c[p] = c[p] + c[i] * c[q]
 
next i
 
print n, " ", c[n]
 
next n</syntaxhighlight>
{{out| Output}}<pre>
0 1
1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845
</pre>
 
==={{header|FreeBASIC}}===
<syntaxhighlight lang="freebasic">' FB 1.05.0 Win64
 
Function factorial(n As UInteger) As UInteger
If n = 0 Then Return 1
Return n * factorial(n - 1)
End Function
 
Function catalan1(n As UInteger) As UInteger
Dim prod As UInteger = 1
For i As UInteger = n + 2 To 2 * n
prod *= i
Next
Return prod / factorial(n)
End Function
 
Function catalan2(n As UInteger) As UInteger
If n = 0 Then Return 1
Dim sum As UInteger = 0
For i As UInteger = 0 To n - 1
sum += catalan2(i) * catalan2(n - 1 - i)
Next
Return sum
End Function
 
Function catalan3(n As UInteger) As UInteger
If n = 0 Then Return 1
Return catalan3(n - 1) * 2 * (2 * n - 1) \ (n + 1)
End Function
 
Print "n", "First", "Second", "Third"
Print "-", "-----", "------", "-----"
Print
For i As UInteger = 0 To 15
Print i, catalan1(i), catalan2(i), catalan3(i)
Next
Print
Print "Press any key to quit"
Sleep</syntaxhighlight>
 
{{out}}
<pre>
n First Second Third
- ----- ------ -----
 
0 1 1 1
1 1 1 1
2 2 2 2
3 5 5 5
4 14 14 14
5 42 42 42
6 132 132 132
7 429 429 429
8 1430 1430 1430
9 4862 4862 4862
10 16796 16796 16796
11 58786 58786 58786
12 208012 208012 208012
13 742900 742900 742900
14 2674440 2674440 2674440
15 9694845 9694845 9694845
</pre>
 
==={{header|FutureBasic}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="futurebasic">
include "NSLog.incl"
 
local fn Factorial( n as NSInteger ) as UInt64
UInt64 sum = 0
if n = 0 then sum = 1 : exit fn
sum = n * fn Factorial( n - 1 )
end fn = sum
 
local fn Catalan1( n as NSInteger ) as UInt64
UInt64 product = 1, result
NSUInteger i
for i = n + 2 to 2 * n
product = product * i
next
result = product / fn Factorial( n )
end fn = result
 
local fn Catalan2( n as NSInteger ) as UInt64
UInt64 sum = 0
NSUInteger i
if n = 0 then sum = 1 : exit fn
for i = 0 to n - 1
sum += fn Catalan2(i) * fn Catalan2( n - 1 - i )
next
end fn = sum
 
local fn Catalan3( n as NSInteger ) as UInt64
UInt64 result
if n = 0 then result = 1 : exit fn
result = fn Catalan3( n - 1 ) * 2 * ( 2 * n - 1 ) / ( n + 1 )
end fn = result
 
NSUInteger i
 
for i = 0 to 19
if( i < 16 )
NSLog( @"%3d.\t\t%7llu\t\t%12llu\t\t%12llu", i, fn Catalan1( i ), fn Catalan2( i ), fn Catalan3( i ) )
else
NSLog( @"%3d.\t\t%@\t\t%12llu\t\t%12llu", i, @"[-err-]", fn Catalan2( i ), fn Catalan3( i ) )
end if
next
HandleEvents
</syntaxhighlight>
{{output}}
<pre>
0. 1 1 1
1. 1 1 1
2. 2 2 2
3. 5 5 5
4. 14 14 14
5. 42 42 42
6. 132 132 132
7. 429 429 429
8. 1430 1430 1430
9. 4862 4862 4862
10. 16796 16796 16796
11. 58786 58786 58786
12. 208012 208012 208012
13. 742900 742900 742900
14. 2674440 2674440 2674440
15. 9694845 9694845 9694845
16. [-err-] 35357670 35357670
17. [-err-] 129644790 129644790
18. [-err-] 477638700 477638700
19. [-err-] 1767263190 1767263190
</pre>
 
==={{header|GW-BASIC}}===
{{works with|BASICA}}
<lang gwbasic>10 DIM C(15)
<syntaxhighlight lang="gwbasic">
20 C(0) = 1
100 REM Catalan numbers
30 PRINT 0, C(0)
110 DIM C(15)
40 FOR N = 0 to 14
50120 C(N+10) = 01
130 PRINT 0, C(0)
60 FOR I = 0 TO N
140 FOR N = 0 TO 14
70 C(N+1) = C(N+1) + C(I)*C(N-I)
150 C(N + 1) = 0
80 NEXT I
160 FOR I = 0 TO N
90 PRINT N+1, C(N+1)
170 C(N + 1) = C(N + 1) + C(I) * C(N - I)
100 NEXT N
180 NEXT I
</lang>
190 PRINT N + 1, C(N + 1)
200 NEXT N
210 END
</syntaxhighlight>
{{out}}
<pre>
0 1
1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845
</pre>
 
==={{header|Liberty BASIC}}===
{{works with|Just BASIC}}
<syntaxhighlight lang="lb">print "non-recursive version"
print catNonRec(5)
for i = 0 to 15
print i;" = "; catNonRec(i)
next
print
 
print "recursive version"
print catRec(5)
for i = 0 to 15
print i;" = "; catRec(i)
next
print
 
print "recursive with memoisation"
redim cats(20) 'clear the array
print catRecMemo(5)
for i = 0 to 15
print i;" = "; catRecMemo(i)
next
print
 
 
wait
 
function catNonRec(n) 'non-recursive version
catNonRec=1
for i=1 to n
catNonRec=((2*((2*i)-1))/(i+1))*catNonRec
next
end function
 
function catRec(n) 'recursive version
if n=0 then
catRec=1
else
catRec=((2*((2*n)-1))/(n+1))*catRec(n-1)
end if
end function
 
function catRecMemo(n) 'recursive version with memoisation
if n=0 then
catRecMemo=1
else
if cats(n-1)=0 then 'call it recursively only if not already calculated
prev = catRecMemo(n-1)
else
prev = cats(n-1)
end if
catRecMemo=((2*((2*n)-1))/(n+1))*prev
end if
cats(n) = catRecMemo 'memoisation for future use
end function</syntaxhighlight>
{{out}}
<pre>
non-recursive version
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
 
recursive version
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
 
recursive with memoisation
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845</pre>
 
==={{header|Minimal BASIC}}===
{{trans|GW-BASIC}}
<langsyntaxhighlight lang="gwbasic">
10 REM Catalan numbers
20 DIM C(15)
Line 492 ⟶ 886:
110 NEXT N
120 END
</syntaxhighlight>
</lang>
 
==={{header|PureBasic}}===
Using the third formula...
<syntaxhighlight lang="purebasic">; saving the division for last ensures we divide the largest
; numerator by the smallest denominator
 
Procedure.q CatalanNumber(n.q)
If n<0:ProcedureReturn 0:EndIf
If n=0:ProcedureReturn 1:EndIf
ProcedureReturn (2*(2*n-1))*CatalanNumber(n-1)/(n+1)
EndProcedure
 
ls=25
rs=12
 
a.s=""
a.s+LSet(RSet("n",rs),ls)+"CatalanNumber(n)"
; cw(a.s)
Debug a.s
 
For n=0 to 33 ;33 largest correct quad for n
a.s=""
a.s+LSet(RSet(Str(n),rs),ls)+Str(CatalanNumber(n))
; cw(a.s)
Debug a.s
Next</syntaxhighlight>
{{out|Sample Output}}
<pre>
n CatalanNumber(n)
0 1
1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845
16 35357670
17 129644790
18 477638700
19 1767263190
20 6564120420
21 24466267020
22 91482563640
23 343059613650
24 1289904147324
25 4861946401452
26 18367353072152
27 69533550916004
28 263747951750360
29 1002242216651368
30 3814986502092304
31 14544636039226909
32 55534064877048198
33 212336130412243110
</pre>
 
==={{header|Run BASIC}}===
<syntaxhighlight lang="runbasic">FOR i = 1 TO 15
PRINT i;" ";catalan(i)
NEXT
FUNCTION catalan(n)
catalan = 1
if n <> 0 then catalan = ((2 * ((2 * n) - 1)) / (n + 1)) * catalan(n - 1)
END FUNCTION</syntaxhighlight>
<pre>1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845</pre>
 
==={{header|Sinclair ZX81 BASIC}}===
Line 499 ⟶ 982:
The specification asks for the first 15 Catalan numbers. A lot of the other implementations produce either C(0) to C(15), which is 16 numbers, or else C(1) to C(15)—which is 15 numbers, but I'm not convinced they're the first 15. This program produces C(0) to C(14).
 
<langsyntaxhighlight lang="basic"> 10 FOR N=0 TO 14
20 LET X=N
30 GOSUB 130
Line 515 ⟶ 998:
150 LET FX=FX*I
160 NEXT I
170 RETURN</langsyntaxhighlight>
{{out}}
<pre>0 1
Line 532 ⟶ 1,015:
13 742900
14 2674440</pre>
 
==={{header|smart BASIC}}===
<syntaxhighlight lang="qbasic">PRINT "Recursive:"!PRINT
FOR n = 0 TO 15
PRINT n,"#######":catrec(n)
NEXT n
PRINT!PRINT
 
PRINT "Non-recursive:"!PRINT
FOR n = 0 TO 15
PRINT n,"#######":catnonrec(n)
NEXT n
 
END
DEF catrec(x)
IF x = 0 THEN
temp = 1
ELSE
n = x
temp = ((2*((2*n)-1))/(n+1))*catrec(n-1)
END IF
catrec = temp
END DEF
 
DEF catnonrec(x)
temp = 1
FOR n = 1 TO x
temp = (2*((2*n)-1))/(n+1)*temp
NEXT n
catnonrec = temp
END DEF</syntaxhighlight>
 
==={{header|TI-83 BASIC}}===
This problem is perfectly suited for a TI calculator.
<syntaxhighlight lang="ti-83 basic">:For(I,1,15
:Disp (2I)!/((I+1)!I!
:End</syntaxhighlight>
{{out}}
<pre> 1
2
4
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done</pre>
 
==={{Header|Tiny BASIC}}===
Integers are limited to 32767 so only the first ten Catalan numbers can be represented. And even then one has to do some finagling to avoid internal overflows.
{{works with|TinyBasic}}
<lang Tiny BASIC>
<syntaxhighlight lang="basic">
10 LET N = 0
10 REM Catalan numbers
20 LET C = 1
3020 printLET N," ",C= 0
30 LET C = 1
40 IF N > 9 THEN END
40 PRINT N," ",C
50 LET N = N + 1
50 IF N > 9 THEN END
60 GOSUB 100
7060 LET CN = (2*N -+ 1)*C
70 GOSUB 200
80 LET C = 2*C/(N+1) + 2*I*(2*N-1)
80 LET C = (2 * N - 1) * C
90 GOTO 30
90 LET C = 2 * C / (N + 1)
100 LET I = 0 REM to avoid internal overflow, I subtract something clever from
110100 IFLET C <= 0C THEN+ RETURN REM C2 and* thenI add* it(2 back* atN the- end1)
110 GOTO 40
120 LET C = C - (N+1)
130200 LET I = I + 10
210 IF C <= 0 THEN RETURN
140 GOTO 110</lang>
220 LET C = C - (N + 1)
230 LET I = I + 1
240 GOTO 210
250 REM To avoid internal overflow, I subtract something clever from
260 REM C and then add it back at the end.</syntaxhighlight>
{{out}}
<pre>0 1
1 1
1 1
2 2
2 2
3 5
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796</pre>
 
==={{header|BBC BASICVBA}}===
<syntaxhighlight lang="vb">Public Sub Catalan1(n As Integer)
'Computes the first n Catalan numbers according to the first recursion given
Dim Cat() As Long
Dim sum As Long
 
ReDim Cat(n)
<lang bbcbasic> 10 FOR i% = 1 TO 15
Cat(0) = 1
20 PRINT FNcatalan(i%)
For i = 0 To n 30- NEXT1
sum = 40 END0
For j = 0 50To DEF FNcatalan(n%)i
sum = 60sum + Cat(j) IF* n%Cat(i = 0 THEN =- 1j)
Next j
70 = 2 * (2 * n% - 1) * FNcatalan(n% - 1) / (n% + 1)</lang>
Cat(i + 1) = sum
Next i
Debug.Print
For i = 0 To n
Debug.Print i, Cat(i)
Next
End Sub
 
Public Sub Catalan2(n As Integer)
'Computes the first n Catalan numbers according to the second recursion given
Dim Cat() As Long
 
ReDim Cat(n)
Cat(0) = 1
For i = 1 To n
Cat(i) = 2 * Cat(i - 1) * (2 * i - 1) / (i + 1)
Next i
Debug.Print
For i = 0 To n
Debug.Print i, Cat(i)
Next
End Sub</syntaxhighlight>
{{out|Result}}
<pre>
Catalan1 15
 
0 1
1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845
</pre>
(Expect same result with "Catalan2 15")
 
==={{header|VBScript}}===
<syntaxhighlight lang="vb">
Function catalan(n)
catalan = factorial(2*n)/(factorial(n+1)*factorial(n))
End Function
 
Function factorial(n)
If n = 0 Then
Factorial = 1
Else
For i = n To 1 Step -1
If i = n Then
factorial = n
Else
factorial = factorial * i
End If
Next
End If
End Function
 
'Find the first 15 Catalan numbers.
For j = 1 To 15
WScript.StdOut.Write j & " = " & catalan(j)
WScript.StdOut.WriteLine
Next
</syntaxhighlight>
 
{{Out}}
<pre>
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
</pre>
 
==={{header|Visual Basic .NET}}===
{{trans|C#}}
<syntaxhighlight lang="vbnet">Module Module1
 
Function Factorial(n As Double) As Double
If n < 1 Then
Return 1
End If
 
Dim result = 1.0
For i = 1 To n
result = result * i
Next
 
Return result
End Function
 
Function FirstOption(n As Double) As Double
Return Factorial(2 * n) / (Factorial(n + 1) * Factorial(n))
End Function
 
Function SecondOption(n As Double) As Double
If n = 0 Then
Return 1
End If
 
Dim sum = 0
For i = 0 To n - 1
sum = sum + SecondOption(i) * SecondOption((n - 1) - i)
Next
Return sum
End Function
 
Function ThirdOption(n As Double) As Double
If n = 0 Then
Return 1
End If
 
Return ((2 * (2 * n - 1)) / (n + 1)) * ThirdOption(n - 1)
End Function
 
Sub Main()
Const MaxCatalanNumber = 15
 
Dim initial As DateTime
Dim final As DateTime
Dim ts As TimeSpan
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, FirstOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, SecondOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, ThirdOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
End Sub
 
End Module</syntaxhighlight>
{{out}}
<pre> CatalanNumber(0:1)
CatalanNumber(1:1)
1
CatalanNumber(2:2)
2
CatalanNumber(3:5)
5
CatalanNumber(4:14)
14
CatalanNumber(5:42)
42
CatalanNumber(6:132)
132
CatalanNumber(7:429)
429
CatalanNumber(8:1430)
1430
CatalanNumber(9:4862)
4862
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440</pre>)
CatalanNumber(15:9694845)
It took 0.19 to execute
 
CatalanNumber(0:1)
=={{header|Befunge}}==
CatalanNumber(1:1)
CatalanNumber(2:2)
CatalanNumber(3:5)
CatalanNumber(4:14)
CatalanNumber(5:42)
CatalanNumber(6:132)
CatalanNumber(7:429)
CatalanNumber(8:1430)
CatalanNumber(9:4862)
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440)
CatalanNumber(15:9694845)
It took 0.831 to execute
 
CatalanNumber(0:1)
CatalanNumber(1:1)
CatalanNumber(2:2)
CatalanNumber(3:5)
CatalanNumber(4:14)
CatalanNumber(5:42)
CatalanNumber(6:132)
CatalanNumber(7:429)
CatalanNumber(8:1430)
CatalanNumber(9:4862)
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440)
CatalanNumber(15:9694845)
It took 0.8 to execute</pre>
 
==={{header|Yabasic}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="yabasic">print " n First Second Third"
print " - ----- ------ -----"
print
for i = 0 to 15
print i using "###", catalan1(i) using "########", catalan2(i) using "########", catalan3(i) using "########"
next i
end
 
sub factorial(n)
if n = 0 return 1
return n * factorial(n - 1)
end sub
 
sub catalan1(n)
local proc, i
 
prod = 1
for i = n + 2 to 2 * n
prod = prod * i
next i
return int(prod / factorial(n))
end sub
 
sub catalan2(n)
local sum, i
 
if n = 0 return 1
sum = 0
for i = 0 to n - 1
sum = sum + catalan2(i) * catalan2(n - 1 - i)
next i
return sum
end sub
 
sub catalan3(n)
if n = 0 return 1
return ((2 * ((2 * n) - 1)) / (n + 1)) * catalan3(n - 1)
end sub</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
 
=={{header|Befunge}}==
{{trans|Ada}}
<langsyntaxhighlight lang="befunge">0>:.:000p1>\:00g-#v_v
v 2-1*2p00 :+1g00\< $
> **00g1+/^v,*84,"="<
_^#<`*53:+1>#,.#+5< @</langsyntaxhighlight>
 
{{out}}
Line 614 ⟶ 1,408:
14 = 2674440
15 = 9694845</pre>
=={{header|BQN}}==
<syntaxhighlight lang="bqn">Cat←{ 0⊸<◶⟨1, (𝕊-⟜1)×(¯2+4×⊢)÷1+⊢⟩ 𝕩 }
Fact ← ×´1+↕
Cat1 ← { # direct formula
⌊0.5 + (Fact 2×𝕩) ÷ (Fact 𝕩+1) × Fact 𝕩
}
Cat2 ← { # header based recursion
0: 1;
(𝕊 𝕩-1)×2×(1-˜2×𝕩)÷𝕩+1
}
 
Cat¨ ↕15
Cat1¨ ↕15
Cat2¨ ↕15</syntaxhighlight>
{{out}}
<pre>⟨ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 ⟩</pre>
=={{header|Bracmat}}==
<langsyntaxhighlight lang="bracmat">( out$straight
& ( C
=
Line 672 ⟶ 1,481:
& out$(1+(1+-1*tay$((1+-4*X)^1/2,X,16))*(2*X)^-1+-1)
& out$
);</langsyntaxhighlight>
{{out}}
<langsyntaxhighlight lang="bracmat">straight
C0 = 1
C1 = 1
Line 742 ⟶ 1,551:
+ 2674440*X^14
+ 9694845*X^15
</syntaxhighlight>
</lang>
 
=={{header|Brat}}==
<langsyntaxhighlight lang="brat">catalan = { n |
true? n == 0
{ 1 }
Line 753 ⟶ 1,561:
0.to 15 { n |
p "#{n} - #{catalan n}"
}</langsyntaxhighlight>
{{out}}
<pre>0 - 1
Line 772 ⟶ 1,580:
15 - 9694845
</pre>
 
=={{header|BQN}}==
<lang bqn>Cat←{ 0⊸<◶⟨1, (𝕊-⟜1)×(¯2+4×⊢)÷1+⊢⟩ 𝕩 }
Fact ← ×´1+↕
Cat1 ← { # direct formula
⌊0.5 + (Fact 2×𝕩) ÷ (Fact 𝕩+1) × Fact 𝕩
}
Cat2 ← { # header based recursion
0: 1;
(𝕊 𝕩-1)×2×(1-˜2×𝕩)÷𝕩+1
}
 
Cat¨ ↕15
Cat1¨ ↕15
Cat2¨ ↕15</lang>
{{out}}
<pre>⟨ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 ⟩</pre>
 
=={{header|C}}==
All three methods mentioned in the task:
<langsyntaxhighlight lang="c">#include <stdio.h>
 
typedef unsigned long long ull;
Line 837 ⟶ 1,627:
 
return 0;
}</langsyntaxhighlight>
{{out}}
direct summing frac
Line 856 ⟶ 1,646:
14 2674440 2674440 2674440
15 9694845 9694845 9694845
 
=={{header|C sharp|C#}}==
<langsyntaxhighlight lang="csharp">namespace CatalanNumbers
{
/// <summary>
Line 971 ⟶ 1,760:
}
}
}</langsyntaxhighlight>
{{out}}
<pre>
Line 1,028 ⟶ 1,817:
It took 0.3 to execute
</pre>
 
=={{header|C++}}==
===4 Classes===
We declare 4 classes representing the four different algorithms for calculating Catalan numbers as given in the description of the task. In addition, we declare two supporting classes for the calculation of factorials and binomial coefficients. Because these two are only internal supporting code they are hidden in namespace 'detail'. Overloading the function call operator to execute the calculation is an obvious decision when using C++. (algorithms.h)
<langsyntaxhighlight lang="cpp">#if !defined __ALGORITHMS_H__
#define __ALGORITHMS_H__
 
Line 1,091 ⟶ 1,879:
} //namespace rosetta
 
#endif //!defined __ALGORITHMS_H__</langsyntaxhighlight>
Here is the implementation of the algorithms. The c'tor of each class tells us the algorithm which will be used. (algorithms.cpp)
<langsyntaxhighlight lang="cpp">#include <iostream>
using std::cout;
using std::endl;
Line 1,193 ⟶ 1,981:
product *= (double(n - (k - i)) / i);
return (unsigned long long)(floor(product + 0.5));
}</langsyntaxhighlight>
In order to test what we have done, a class Test is created. Using the template parameters N (number of Catalan numbers to be calculated) and A (the kind of algorithm to be used) the compiler will create code for all the test cases we need. What would C++ be without templates ;-) (tester.h)
<langsyntaxhighlight lang="cpp">#if !defined __TESTER_H__
#define __TESTER_H__
 
Line 1,220 ⟶ 2,008:
} //namespace rosetta
 
#endif //!defined __TESTER_H__</langsyntaxhighlight>
Finally, we test the four different algorithms. Note that the first one (direct calculation using the factorial) only works up to N = 10 because some intermediate result (namely (2n)! with n = 11) exceeds the boundaries of an unsigned 64 bit integer. (catalanNumbersTest.cpp)
<langsyntaxhighlight lang="cpp">#include "algorithms.h"
#include "tester.h"
using namespace rosetta::catalanNumbers;
Line 1,233 ⟶ 2,021:
Test<15, CatalanNumbersRecursiveSum>::Do();
return 0;
}</langsyntaxhighlight>
{{out}}
(source code is compiled both by MS Visual C++ 10.0 (WinXP 32 bit) and GNU g++ 4.4.3 (Ubuntu 10.04 64 bit) compilers)
Line 1,301 ⟶ 2,089:
C(15) = 9694845
</pre>
 
=={{header|Clojure}}==
<langsyntaxhighlight Clojurelang="clojure">(def ! (memoize #(apply * (range 1 (inc %)))))
 
(defn catalan-numbers-direct []
Line 1,319 ⟶ 2,106:
 
user> (take 15 (catalan-numbers-recursive))
(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)</langsyntaxhighlight>
=={{header|CLU}}==
<syntaxhighlight lang="clu">catalan = iter (amount: int) yields (int)
c: int := 1
for n: int in int$from_to(1, amount) do
yield(c)
c := (4*n-2)*c/(n+1)
end
end catalan
 
start_up = proc ()
po: stream := stream$primary_output()
for n: int in catalan(15) do
stream$putl(po, int$unparse(n))
end
end start_up</syntaxhighlight>
{{out}}
<pre>1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440</pre>
 
=={{header|Common Lisp}}==
With all three methods defined.
<langsyntaxhighlight lang="lisp">(defun catalan1 (n)
;; factorial. CLISP actually has "!" defined for this
(labels ((! (x) (if (zerop x) 1 (* x (! (1- x))))))
Line 1,350 ⟶ 2,168:
for i from 1 to 3 do
(format t "~%Method ~d:~%" i)
(dotimes (i 16) (format t "C(~2d) = ~d~%" i (funcall f i))))</langsyntaxhighlight>
 
=={{header|Cowgol}}==
<syntaxhighlight lang="cowgol">include "cowgol.coh";
 
sub catalan(n: uint32): (c: uint32) is
c := 1;
var i: uint32 := 1;
while i <= n loop
c := (4*i-2)*c/(i+1);
i := i+1;
end loop;
end sub;
 
var i: uint8 := 0;
while i < 15 loop
print("catalan(");
print_i8(i);
print(") = ");
print_i32(catalan(i as uint32));
print_nl();
i := i+1;
end loop;</syntaxhighlight>
{{out}}
<pre>catalan(0) = 1
catalan(1) = 1
catalan(2) = 2
catalan(3) = 5
catalan(4) = 14
catalan(5) = 42
catalan(6) = 132
catalan(7) = 429
catalan(8) = 1430
catalan(9) = 4862
catalan(10) = 16796
catalan(11) = 58786
catalan(12) = 208012
catalan(13) = 742900
catalan(14) = 2674440</pre>
 
=={{header|Crystal}}==
{{trans|Ruby}}
<langsyntaxhighlight lang="ruby">require "big"
require "benchmark"
 
Line 1,395 ⟶ 2,251:
puts "\n direct rec1 rec2"
16.times { |n| puts "%2d :%9d%9d%9d" % [n, catalan_direct(n), catalan_rec1(n), catalan_rec2(n)] }
</syntaxhighlight>
</lang>
 
{{out}}
Line 1,426 ⟶ 2,282:
 
=={{header|D}}==
<langsyntaxhighlight lang="d">import std.stdio, std.algorithm, std.bigint, std.functional, std.range;
 
auto product(R)(R r) { return reduce!q{a * b}(1.BigInt, r); }
Line 1,445 ⟶ 2,301:
foreach (cats; TypeTuple!(cats1, cats2, cats3))
cats.take(15).writeln;
}</langsyntaxhighlight>
{{out}}
<pre>[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440]
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440]
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440]</pre>
 
=={{header|Delphi}}==
See [https://www.rosettacode.org/wiki/Catalan_numbers#Pascal Pascal].
=={{header|EasyLang}}==
<syntaxhighlight lang="text">
<lang>func catalan n . ans .
func ifcatalan n = 0.
if ansn = 10
return 1
else
.
call catalan n - 1 h
ans =return 2 * (2 * n - 1) * hcatalan (n - 1) div (1 + n)
.
for i = 0 to 14
print catalan i
.
</syntaxhighlight>
for i range 15
call catalan i h
print h
.</lang>
{{out}}
<pre>
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
</pre>
 
=={{header|EchoLisp}}==
{{incorrect|Echolisp|series starts 1, 1, 2, ...}}
<langsyntaxhighlight lang="scheme">
(lib 'sequences)
(lib 'bigint)
Line 1,525 ⟶ 2,361:
(for ((c C3) (i 15)) (write c))
→ 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
</syntaxhighlight>
</lang>
 
=={{header|EDSAC order code}}==
The Catalan numbers are here computed by the second method, as a sum of products.
Line 1,535 ⟶ 2,370:
But if we multiply two 35-bit integers A and B, the result is really A*B*(2^-68),
and needs to be multiplied by 2^34 to get the same scaling as for A and B.
<langsyntaxhighlight lang="edsac">
[Calculation of Catalan numbers.
EDSAC program, Initial Orders 2.]
Line 1,635 ⟶ 2,470:
E 12 Z [define entry point]
P F [acc = 0 on entry]
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 1,655 ⟶ 2,490:
15 9694845
</pre>
 
=={{header|Eiffel}}==
<syntaxhighlight lang="eiffel">
<lang Eiffel>
class
APPLICATION
Line 1,695 ⟶ 2,529:
 
 
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 1,714 ⟶ 2,548:
2674440
</pre>
 
=={{header|Elixir}}==
{{trans|Erlang}}
<langsyntaxhighlight lang="elixir">defmodule Catalan do
def cat(n), do: div( factorial(2*n), factorial(n+1) * factorial(n) )
Line 1,739 ⟶ 2,572:
end
 
Catalan.test</langsyntaxhighlight>
 
{{out}}
Line 1,750 ⟶ 2,583:
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
</pre>
 
=={{header|Erlang}}==
<langsyntaxhighlight lang="erlang">-module(catalan).
 
-export([test/0]).
Line 1,781 ⟶ 2,613:
io:format("Directly:\n~p\n",[[cat(N) || N <- TestList]]),
io:format("1st recusive method:\n~p\n",[[cat_r1(N) || N <- TestList]]),
io:format("2nd recusive method:\n~p\n",[[cat_r2(N) || N <- TestList]]).</langsyntaxhighlight>
{{out}}
<pre>
Line 1,791 ⟶ 2,623:
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
</pre>
 
=={{header|ERRE}}==
<syntaxhighlight lang="text">PROGRAM CATALAN
 
PROCEDURE CATALAN(N->RES)
Line 1,808 ⟶ 2,639:
END FOR
END PROGRAM
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 1,828 ⟶ 2,659:
15 = 9694845
</pre>
 
=={{header|Euphoria}}==
<langsyntaxhighlight Euphorialang="euphoria">--Catalan number task from Rosetta Code wiki
--User:Lnettnay
 
Line 1,852 ⟶ 2,682:
for i = 0 to 15 do
? catalan(i)
end for</langsyntaxhighlight>
{{out}}
<pre>
Line 1,872 ⟶ 2,702:
9694845
</pre>
 
=={{header|F_Sharp|F#}}==
<langsyntaxhighlight lang="fsharp">
Seq.unfold(fun (c,n) -> let cc = 2*(2*n-1)*c/(n+1) in Some(c,(cc,n+1))) (1,1) |> Seq.take 15 |> Seq.iter (printf "%i, ")
</syntaxhighlight>
</lang>
{{out}}
<pre>
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440,
</pre>
 
=={{header|Factor}}==
The first method:
<langsyntaxhighlight lang="factor">USING: kernel math math.combinatorics prettyprint ;
 
: catalan ( n -- n ) [ 1 + recip ] [ 2 * ] [ nCk * ] tri ;
 
15 [ catalan . ] each-integer</langsyntaxhighlight>
{{out}}
<pre>
Line 1,909 ⟶ 2,737:
 
The last method, memoized by using arrays.
<langsyntaxhighlight lang="factor">USING: kernel math prettyprint sequences ;
 
: next ( seq -- newseq )
Line 1,918 ⟶ 2,746:
: Catalan ( n -- seq ) V{ 1 } swap 1 - [ next ] times ;
 
15 Catalan .</langsyntaxhighlight>
{{out}}
Similar to above.
 
=={{header|Fantom}}==
{{incorrect|Fantom|series starts 1, 1, 2, ...}}
<langsyntaxhighlight lang="fantom">class Main
{
static Int factorial (Int n)
Line 1,975 ⟶ 2,802:
}
}
}</langsyntaxhighlight>
22! exceeds the range of Fantom's Int class, so the first technique fails afer n=10
<pre>
Line 1,994 ⟶ 2,821:
15 -2 9694845 9694845
</pre>
 
=={{header|Fermat}}==
<langsyntaxhighlight lang="fermat">Func Catalan(n)=(2*n)!/((n+1)!*n!).;
for i=1 to 15 do !Catalan(i);!' ' od;</langsyntaxhighlight>
{{out}}<pre>
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
</pre>
 
=={{header|Forth}}==
<langsyntaxhighlight lang="forth">: catalan ( n -- ) 1 swap 1+ 1 do dup cr . i 2* 1- 2* i 1+ */ loop drop ;</langsyntaxhighlight>
 
=={{header|Fortran}}==
{{works with|Fortran|90 and later}}
<langsyntaxhighlight lang="fortran">program main
!=======================================================================================
implicit none
Line 2,050 ⟶ 2,874:
 
!=======================================================================================
end function catalan_numbers</langsyntaxhighlight>
{{out}}
<pre>
Line 2,073 ⟶ 2,897:
===============
</pre>
 
=={{header|FreeBASIC}}==
<lang freebasic>' FB 1.05.0 Win64
 
Function factorial(n As UInteger) As UInteger
If n = 0 Then Return 1
Return n * factorial(n - 1)
End Function
 
Function catalan1(n As UInteger) As UInteger
Dim prod As UInteger = 1
For i As UInteger = n + 2 To 2 * n
prod *= i
Next
Return prod / factorial(n)
End Function
 
Function catalan2(n As UInteger) As UInteger
If n = 0 Then Return 1
Dim sum As UInteger = 0
For i As UInteger = 0 To n - 1
sum += catalan2(i) * catalan2(n - 1 - i)
Next
Return sum
End Function
 
Function catalan3(n As UInteger) As UInteger
If n = 0 Then Return 1
Return catalan3(n - 1) * 2 * (2 * n - 1) \ (n + 1)
End Function
 
Print "n", "First", "Second", "Third"
Print "-", "-----", "------", "-----"
Print
For i As UInteger = 0 To 15
Print i, catalan1(i), catalan2(i), catalan3(i)
Next
Print
Print "Press any key to quit"
Sleep</lang>
 
{{out}}
<pre>
n First Second Third
- ----- ------ -----
 
0 1 1 1
1 1 1 1
2 2 2 2
3 5 5 5
4 14 14 14
5 42 42 42
6 132 132 132
7 429 429 429
8 1430 1430 1430
9 4862 4862 4862
10 16796 16796 16796
11 58786 58786 58786
12 208012 208012 208012
13 742900 742900 742900
14 2674440 2674440 2674440
15 9694845 9694845 9694845
</pre>
 
=={{header|Frink}}==
Frink includes efficient algorithms for calculating arbitrarily-large binomial coefficients and automatically caches factorials.
<langsyntaxhighlight lang="frink">catalan[n] := binomial[2n,n]/(n+1)
for n = 0 to 15
println[catalan[n]]</langsyntaxhighlight>
 
=={{header|FunL}}==
<langsyntaxhighlight lang="funl">import integers.choose
import util.TextTable
 
Line 2,165 ⟶ 2,924:
t.row( i, catalan(i), catalan2(i), catalan3(i) )
println( t )</langsyntaxhighlight>
 
{{out}}
Line 2,194 ⟶ 2,953:
=={{header|Fōrmulæ}}==
 
{{FormulaeEntry|page=https://formulae.org/?script=examples/Catalan_numbers}}
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 &mdash;i.e. XML, JSON&mdash; they are intended for storage and transfer purposes more than visualization and edition.
 
'''Solution'''
 
'''Direct definition'''
 
[[File:Fōrmulæ - Catalan numbers 01.png]]
 
[[File:Fōrmulæ - Catalan numbers 02.png]]
 
'''Direct definition (alternative)'''
 
The expression <math>\frac{(2n)!}{(n+1)!\,n!}</math> turns out to be equals to <math>\prod_{k=2}^{n}\frac{n + k}{k}</math>
 
[[File:Fōrmulæ - Catalan numbers 03.png]]
 
(same result)
 
'''No directly defined'''
 
Recursive definitions are easy to write, but extremely inefficient (specially the first one).
 
Because a list is intended to be get, the list of previous values can be used as a form of memoization, avoiding recursion.
 
The next function make use of the "second" form of recursive definition (without recursion):
 
[[File:Fōrmulæ - Catalan numbers 04.png]]
 
[[File:Fōrmulæ - Catalan numbers 05.png]]
Programs in Fōrmulæ are created/edited online in its [https://formulae.org website], However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
 
(same result)
In '''[https://formulae.org/?example=Catalan_numbers this]''' page you can see the program(s) related to this task and their results.
 
=={{header|GAP}}==
<langsyntaxhighlight lang="gap">Catalan1 := n -> Binomial(2*n, n) - Binomial(2*n, n - 1);
 
Catalan2 := n -> Binomial(2*n, n)/(n + 1);
Line 2,231 ⟶ 3,016:
List([0 .. 14], Catalan4);
# Same output for all four:
# [ 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440 ]</langsyntaxhighlight>
 
=={{header|Go}}==
Direct:
<langsyntaxhighlight lang="go">package main
 
import (
Line 2,247 ⟶ 3,031:
fmt.Println(c.Div(b.Binomial(n*2, n), c.SetInt64(n+1)))
}
}</langsyntaxhighlight>
{{out}}
<pre>
Line 2,267 ⟶ 3,051:
</pre>
Recursive (alternative):
<syntaxhighlight lang="go">
<lang go>
package main
 
Line 2,295 ⟶ 3,079:
}
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 2,314 ⟶ 3,098:
9694845
</pre>
 
=={{header|Groovy}}==
<syntaxhighlight lang="groovy">
<lang Groovy>
class Catalan
{
Line 2,340 ⟶ 3,123:
}
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 2,359 ⟶ 3,142:
9694845
</pre>
 
=={{header|Harbour}}==
<langsyntaxhighlight lang="visualfoxpro">
PROCEDURE Main()
LOCAL i
Line 2,379 ⟶ 3,161:
 
RETURN nCatalan
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 2,399 ⟶ 3,181:
5: 9694845
</pre>
 
=={{header|Haskell}}==
<langsyntaxhighlight lang="haskell">-- Three infinite lists, corresponding to the three
-- definitions in the problem statement.
 
Line 2,424 ⟶ 3,205:
 
main :: IO ()
main = mapM_ (print . take 15) [cats1, cats2, cats3]</langsyntaxhighlight>
{{out}}
<pre>[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]</pre>
 
=={{header|Icon}} and {{header|Unicon}}==
<langsyntaxhighlight Iconlang="icon">procedure main()
every writes(catalan(0 to 14)," ")
end
Line 2,442 ⟶ 3,222:
if n > 0 then
return (n = 1) | \M[n] | ( M[n] := (2*(2*n-1)*catalan(n-1))/(n+1))
end</langsyntaxhighlight>
{{out}}
<pre>1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440</pre>
 
=={{header|J}}==
<langsyntaxhighlight lang="j"> ((! +:) % >:) i.15x
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440</langsyntaxhighlight>
 
=={{header|Java}}==
Replace double inexact computations with BigInteger implementation.
<langsyntaxhighlight lang="java">
import java.math.BigInteger;
import java.util.ArrayList;
Line 2,561 ⟶ 3,339:
 
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 2,582 ⟶ 3,360:
C(15) = 9,694,845 9,694,845 9,694,845
</pre>
 
=={{header|JavaScript}}==
 
<lang javascript><html><head><title>Catalan</title></head>
===Procedural===
<syntaxhighlight lang="javascript"><html><head><title>Catalan</title></head>
<body><pre id='x'></pre><script type="application/javascript">
function disp(x) {
Line 2,612 ⟶ 3,391:
disp(i + '\t' + cata1(i) + '\t' + cata2(i) + '\t' + cata3(i));
 
</script></body></html></langsyntaxhighlight>
{{out}}
<pre> meth1 meth2 meth3
Line 2,631 ⟶ 3,410:
14 2674440 2674440 2674440
15 9694845 9694845 9694845</pre>
 
===Functional===
 
Defining an infinite list:
 
<syntaxhighlight lang="javascript">(() => {
"use strict";
 
// ----------------- CATALAN NUMBERS -----------------
 
// catalansDefinitionThree :: [Int]
const catalansDefinitionThree = () =>
// An infinite sequence of Catalan numbers.
scanlGen(
c => n => Math.floor(
(2 * c * pred(2 * n)) / succ(n)
)
)(1)(
enumFrom(1)
);
 
 
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () =>
take(15)(
catalansDefinitionThree()
);
 
 
// --------------------- GENERIC ---------------------
 
// enumFrom :: Enum a => a -> [a]
const enumFrom = function* (n) {
// An infinite sequence of integers,
// starting with n.
let v = n;
 
while (true) {
yield v;
v = 1 + v;
}
};
 
 
// pred :: Int -> Int
const pred = x =>
x - 1;
 
 
// scanlGen :: (b -> a -> b) -> b -> Gen [a] -> [b]
const scanlGen = f =>
// The series of interim values arising
// from a catamorphism over an infinite list.
startValue => function* (gen) {
let
a = startValue,
x = gen.next();
 
yield a;
while (!x.done) {
a = f(a)(x.value);
yield a;
x = gen.next();
}
};
 
 
// succ :: Int -> Int
const succ = x =>
1 + x;
 
 
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => Array.from({
length: n
}, () => {
const x = xs.next();
 
return x.done ? [] : [x.value];
}).flat();
 
 
// MAIN ---
return JSON.stringify(main(), null, 2);
})();</syntaxhighlight>
{{Out}}
<pre>[
1,
1,
2,
5,
14,
42,
132,
429,
1430,
4862,
16796,
58786,
208012,
742900,
2674440
]</pre>
 
=={{header|jq}}==
Line 2,638 ⟶ 3,525:
that formula is used (a) to define a function for computing a single Catalan number; (b) to define a function for generating a sequence of Catalan numbers; and (c) to write a single expression for generating a sequence of Catalan numbers using jq's builtin "recurse/1" filter.
==== Compute a single Catalan number====
<langsyntaxhighlight lang="jq">def catalan:
if . == 0 then 1
elif . < 0 then error("catalan is not defined on \(.)")
else (2 * (2*. - 1) * ((. - 1) | catalan)) / (. + 1)
end;</langsyntaxhighlight>
'''Example 1'''
<langsyntaxhighlight lang="jq">(range(0; 16), 100) as $i | $i | catalan | [$i, .]</langsyntaxhighlight>
{{Out}}
<div style="overflow:scroll; height:150px;">
<langsyntaxhighlight lang="sh">$ jq -M -n -c -f Catalan_numbers.jq
[0,1]
[1,1]
Line 2,665 ⟶ 3,552:
[15,9694845]
[100,8.96519947090131e+56]
</langsyntaxhighlight></div>
 
==== Generate a sequence of Catalan numbers ====
<langsyntaxhighlight lang="jq">def catalan_series(max):
def _catalan: # state: [n, catalan(n)]
if .[0] > max then empty
Line 2,676 ⟶ 3,563:
end;
[0,1] | _catalan;
</syntaxhighlight>
</lang>
'''Example 2''':
<syntaxhighlight lang ="jq">catalan_series(15)</langsyntaxhighlight>
{{Out}}
As above for 0 to 15.
==== An expression to generate Catalan numbers ====
<syntaxhighlight lang="jq">
<lang jq>
[0,1]
| recurse( if .[0] == 15 then empty
else .[1] as $c | (.[0] + 1) | [ ., (2 * (2*. - 1) * $c) / (. + 1) ]
end )</langsyntaxhighlight>
{{out}}
As above for 0 to 15.
 
=={{header|Julia}}==
{{works with|Julia|0.6}}
 
<langsyntaxhighlight lang="julia">catalannum(n::Integer) = binomial(2n, n) ÷ (n + 1)
 
@show catalannum.(1:15)
@show catalannum(big(100))</langsyntaxhighlight>
 
{{out}}
Line 2,703 ⟶ 3,589:
 
(In the second example, we have used arbitrary-precision integers to avoid overflow for large Catalan numbers.)
 
=={{header|K}}==
<langsyntaxhighlight lang="k"> catalan: {_{*/(x-i)%1+i:!y-1}[2*x;x+1]%x+1}
catalan'!:15
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440</langsyntaxhighlight>
 
=={{header|Kotlin}}==
{{works with|Java|1.7.0}}
{{works with|Kotlin|1.1.4}}
<langsyntaxhighlight lang="scala">abstract class Catalan {
abstract operator fun invoke(n: Int) : Double
 
Line 2,764 ⟶ 3,648:
println()
}
}</langsyntaxhighlight>
{{out}}
<pre> 1 1 1
Line 2,782 ⟶ 3,666:
2674440 2674440 2674440
9694845 9694845 9694845</pre>
 
=={{header|Lambdatalk}}==
{{trans|Javascript}}
 
<h3>1) catalan1</h3>
 
<syntaxhighlight lang="scheme">
{def catalan1
{def fac {lambda {:n} {* {S.serie 1 :n}}}}
{lambda {:n}
{floor {+ {/ {fac {* 2 :n}} {fac {+ :n 1}} {fac :n}} 0.5}}}}
-> catalan1
 
{S.map catalan1 {S.serie 1 15}}
-> 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
</syntaxhighlight>
 
<h3>2) catalan2</h3>
 
<syntaxhighlight lang="scheme">
{def catalan2
{def catalan2.sum
{lambda {:n :a :s :i}
{if {= :i :n}
then {A.set! :n :s :a}
else {catalan2.sum :n
:a
{+ :s {* {catalan2.loop :i :a}
{catalan2.loop {- :n :i 1} :a}}}
{+ :i 1}} }}}
{def catalan2.loop
{lambda {:n :a}
{if {= :n 0}
then 1
else {if {W.equal? {A.get :n :a} undefined}
then {A.get :n {catalan2.sum :n :a 0 0}}
else {A.get :n :a} }}}}
{lambda {:n}
{catalan2.loop :n {A.new}} }}
-> catalan2
 
{S.map catalan2 {S.serie 0 15}}
-> 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
 
</syntaxhighlight>
 
<h3>3) catalan3</h3>
 
<syntaxhighlight lang="scheme">
{def catalan3
{def catalan3.loop
{lambda {:n :a}
{if {= :n 0}
then 1
else {if {W.equal? {A.get :n :a} undefined}
then {A.get :n
{A.set! :n
{/ {* {- {* 4 :n} 2}
{catalan3.loop {- :n 1} :a}}
{+ :n 1}}
:a}}
else {A.get :n :a}
}}}}
{lambda {:n}
{catalan3.loop :n {A.new}}}}
-> catalan3
 
{S.map catalan3 {S.serie 0 15}}
-> 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
</syntaxhighlight>
 
<h3>4) Alternative for a vertical diplay</h3>
 
<syntaxhighlight lang="scheme">
 
{style
td { text-align:right;
font-family:monospace;
}
}
 
{table
{tr {td} {td cat1} {td cat2} {td cat3}}
{S.map {lambda {:i} {tr {td :i}
{td {catalan1 :i}}
{td {catalan2 :i}}
{td {catalan3 :i}}}}
{S.serie 0 15}}
}
 
cat1 cat2 cat3
0 1 1 1
1 1 1 1
2 2 2 2
3 5 5 5
4 14 14 14
5 42 42 42
6 132 132 132
7 429 429 429
8 1430 1430 1430
9 4862 4862 4862
10 16796 16796 16796
11 58786 58786 58786
12 208012 208012 208012
13 742900 742900 742900
14 2674440 2674440 2674440
15 9694845 9694845 9694845
 
</syntaxhighlight>
 
=={{header|langur}}==
{{trans|Perl}}
<langsyntaxhighlight lang="langur">val .factorial = ffn(.x) { if(.x < 2: 1; .x x self(.x - 1)) }
 
val .catalan = f(.n) .factorial(2 x .n) / .factorial(.n+1) / .factorial(.n)
val .catalan = fn(.n) { .factorial(2 x .n) / .factorial(.n+1) / .factorial(.n) }
 
for .i in 0..15 {
Line 2,792 ⟶ 3,786:
}
 
writeln "10000: ", .catalan(10000)</langsyntaxhighlight>
 
{{out}}
Line 2,813 ⟶ 3,807:
10000: 224537812493385215633593584257360578701103586219365887773293713835854436588700534490998102719114320210209905393799589701149327326500953702713977513001838761306936534407802585494454599941773729984591764542782202886796997833276495496514760245912220654267091568311812071300891219894022165175451441066691435091975969499731921675488934120638046514134965974069039677192984714638704528752769863567952620334847707274529741976558104236293861846622622783294667505268651205024766408784881872997404042356319626323351089169906635603513309014645157443570842822082866699012415455339518777770781742052837799476906230350785959040487158118992753484022865373274100095762968510625236915280143408460651206678398725681703811505423791566261735329550627967717189932855983913468867794806585863794483869239933179341394259456515091026456652770409848702116046445406995085092488210998732255656992243441519938747425554228724734242623566663631968254490897214106655375215196762710825001305055093871863518797311135688370964194817463890187212845332422257193414201244344808864449873736345425670715824582633802476282521798739438044652622163657359012681653473214512797365047989922327391063907061792126264420963262176161781711086630089636828211837643128677915076724947168653050318426339007489738275045346257959685376480042860870398232333705506506342394485443047987642390287346746539674780326188825579548593281319807827279403944008553690033855132088140116099772393778770685018936338194366302053586633406848404622048675525765095697363909787189635178694239275237185046710057476484117945279786897787624602379494797322427251542758312638233073625857897083435831841717971137851874666094337671443717108457737153283641719103639784923520519013700030680553564442331411313831920775983175313709250333784211385811480015293165463406576311626295629412110652218717603537723650144357966952842696678735624157616428716812764985074925414219421312810089785108621126934245959900367104035334200067714905754827856122801987429837706493130435832752072139392743006620396370486473952500144779413596417260472218266529167783118015414918168260722824885550181735638670588682513610805160133611349864194033776132438535863120087679096358696928233598996870302136347936567444208209125300149683552369341937471817860835774359234009557030148123353114950735217736514617017504851011193104728986836180908987352236659629183725016607437110422583156042941955830763092095074443334625318588569114114087985404048889671202396824806275701581378689568449507132793603852731445602923990458926101180821029108808623323378547869169352237448925371763574346501610378415722137519019474474794069155118626291447578558908522430436148987521551911541787974276591708584289036595642180860178815462862735993859177180582760389253540408842580225467216988321950591728369194164290645992782274919561096308372635908842325870580231011459216934235078490764707633348336131667313582584404397290232519769625777374165187949140092779343812345117947306771376053099536367169631889642304360871187460737580808157222861127968703067542270175460553478533349238111434409526724363429611803844595968793121871649699680963646793415774160274520010905236593324062464542927011227158945796188186430711399250096518886617184049325827319276468018789191520522185358895653192882843061349706085770767046601045697944646638311930027354235643643713545212361580694059553720806659066661496416423676930095857438882302891350789287291844752601744462789158506243012088536936184422120232369244564444689340142897415432231452353338115944183447986470689449043710051589958391273681116292415738776171575775695905846247205522469202801517417551374761549677412720803623129527503286287755308576386461385928958587649159872019202866614901547860974883963007792442796064165417207167072370586790722366932349325253877744621251386864069101337572557790214048760202008337611577675840153696735860276810033694744314488435390547908483357054897387317002405793108554524629034558098886977538473481750772616164313845337139245688079995996839933620829828339492800825536599964878893947278408890351634126931068657027524005795713514365098086505030570362785115155293306343520969872400876180105031975302255898787642403303027682634969586730202117121076117629457710028105378124677420093990476071697970354661002217702623344454780740808459286778553016318604430682610618871098652904537323336381304469735192868285840882036271136058499391069436145426450229039329475974178236465920534171895204155964515055983303017823692138977622016292722019365841360360274557488926673754175222061483328914099598663902320310143583379354121664996173733086613692927391384486261610892314450463841637667054196985332620403539011932606618414419229492637564924726411270720189611019154677281846409387514072618176832310721327819277699943226895919915049652045449281057471199978267843961724883768772155477073354744908923995448752333726740642292872107500458349718026322755698226793850983280706045951407323891263270928264657562125955511946782954645656015480418543664557515041692091317941000997342935512311493290722434384401250133402934163457264794261787386862382738330195237770190998115114193014769006071380834085352290585937952429981509893303796306071520571655936820282768086579891336876000368502562579738337809071051261343359121744773055264455701014137255399929760233753812017596045145926791136761130783810840502248142803073720015451941006030172192834375431286154255159659778817089767964922549014569972777126726537787896968876337799235679125368824867754881036161730805613471278633981478858113141202728303435218970292775366288829203013873713349923690394124920402725698544786016048685431525811047414746045227535216327530901827040588505255466803793791888002231571686068617764292584075135236237044383334893874602177596602979234717936820827427229615827657960492946059695301906791494260652411424538532836730097985187522379068364429583532675896349363295120431429006688249818006722311568902288350452581968418068616818268667067741994472455501649753611708445979082338902214467454627107888156489438584617793175431865532382711812960546611287516640
</pre>
 
=={{header|Liberty BASIC}}==
<lang lb>print "non-recursive version"
print catNonRec(5)
for i = 0 to 15
print i;" = "; catNonRec(i)
next
print
 
print "recursive version"
print catRec(5)
for i = 0 to 15
print i;" = "; catRec(i)
next
print
 
print "recursive with memoisation"
redim cats(20) 'clear the array
print catRecMemo(5)
for i = 0 to 15
print i;" = "; catRecMemo(i)
next
print
 
 
wait
 
function catNonRec(n) 'non-recursive version
catNonRec=1
for i=1 to n
catNonRec=((2*((2*i)-1))/(i+1))*catNonRec
next
end function
 
function catRec(n) 'recursive version
if n=0 then
catRec=1
else
catRec=((2*((2*n)-1))/(n+1))*catRec(n-1)
end if
end function
 
function catRecMemo(n) 'recursive version with memoisation
if n=0 then
catRecMemo=1
else
if cats(n-1)=0 then 'call it recursively only if not already calculated
prev = catRecMemo(n-1)
else
prev = cats(n-1)
end if
catRecMemo=((2*((2*n)-1))/(n+1))*prev
end if
cats(n) = catRecMemo 'memoisation for future use
end function</lang>
{{out}}
<pre>
non-recursive version
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
 
recursive version
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
 
recursive with memoisation
42
0 = 1
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845</pre>
 
=={{header|Logo}}==
<langsyntaxhighlight lang="logo">to factorial :n
output ifelse [less? :n 1] 1 [product :n factorial difference :n 1]
end
Line 2,938 ⟶ 3,819:
end
 
repeat 15 [print catalan repcount]</langsyntaxhighlight>
{{out}}
<pre>
Line 2,956 ⟶ 3,837:
742900
2674440</pre>
=={{header|Logtalk}}==
 
For this task it is instructive to show a more general-purpose interface for sequences and an implementation of it for Catalan numbers.
 
First, <code>loader.lgt</code>:
 
<syntaxhighlight lang="logtalk">
:- initialization((
% libraries
logtalk_load(dates(loader)),
logtalk_load(meta(loader)),
logtalk_load(types(loader)),
% application
logtalk_load(seqp),
logtalk_load(catalan),
logtalk_load(catalan_test)
)).
</syntaxhighlight>
 
The interface is defined in <code>seqp.lgt</code> as a protocol:
 
<syntaxhighlight lang="logtalk">
:- protocol(seqp).
 
:- public(init/0). % reset to a beginning state if meaningful
 
:- public(nth/2). % get the nth value of the sequence
 
:- public(to_nth/2). % get from the start to the nth value of the sequence as a list
 
:- end_protocol.
</syntaxhighlight>
 
The implementation of a Catalan sequence generator is in <code>catalan.lgt</code>:
 
<syntaxhighlight lang="logtalk">
:- object(catalan, implements(seqp)).
 
:- private(catalan/2).
:- dynamic(catalan/2).
 
% Public interface.
 
init :- retractall(catalan(_,_)). % flush any memoized results
 
nth(N, V) :- \+ catalan(N, V), catalan_(N, V), !. % generate iff it's not been memoized
nth(N, V) :- catalan(N, V), !. % otherwise use the memoized version
 
to_nth(N, L) :-
integer::sequence(0, N, S), % generate a list of 0 to N
meta::map(nth, S, L). % map the nth/2 predicate to the list for all Catalan numbers up to N
 
% Local helper predicates.
 
catalan_(N, V) :-
N > 0, % calculate
N1 is N - 1,
N2 is N + 1,
catalan_(N1, V1), % via a recursive call
V is V1 * 2 * (2 * N - 1) // N2,
assertz(catalan(N, V)). % and memoize the result
catalan_(0, 1).
 
:- end_object.
</syntaxhighlight>
 
This is a memoizing implementation whose impact we will check in the test. The <code>init/0</code> predicate flushes any memoized results.
 
The test driver is a simple one that generates the first fifteen Catalan numbers four times, comparing times with and without memoization. From <code>catalan_test.lgt</code>:
 
<syntaxhighlight lang="logtalk">
:- object(catalan_test).
 
:- public(run/0).
 
run :-
% put the object into a known initial state
catalan::init,
 
% first 15 Catalan numbers, record duration.
time_operation(catalan::to_nth(15, C1), D1),
 
% first 15 Catalan numbers again, twice, recording duration.
time_operation(catalan::to_nth(15, C2), D2),
time_operation(catalan::to_nth(15, C3), D3),
 
% reset the object again
catalan::init,
 
% first 15 Catalan numbers, record duration.
time_operation(catalan::to_nth(15, C4), D4),
 
% ensure the results were the same each time
C1 = C2, C2 = C3, C3 = C4,
 
% write the results and durations of each run
write(C1), write(' '), write(D1), nl,
write(C2), write(' '), write(D2), nl,
write(C3), write(' '), write(D3), nl,
write(C4), write(' '), write(D4), nl.
% visual inspection should show all results the same
% first and final durations should be much larger
 
:- meta_predicate(time_operation(0, *)).
 
time_operation(Goal, Duration) :-
time::cpu_time(Before),
call(Goal),
time::cpu_time(After),
Duration is After - Before.
 
:- end_object.
</syntaxhighlight>
 
 
{{Out}}
 
The session at the top-level looks like this:
<pre>
?- {loader}.
% ... messages elided ...
% (0 warnings)
true.
 
?- catalan_test::run.
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845] 0.001603
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845] 0.000306
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845] 0.00026
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845] 0.001346
true.
</pre>
 
This test shows:
 
# The <code>nth/2</code> predicate works (since <code>to_nth/2</code> is implemented in terms of it).
# The <code>to_nth/2</code> predicate works.
# Memoization generates a speedup of between ~4.5× to ~6.2× over generating from scratch.
 
=={{header|Lua}}==
<langsyntaxhighlight Lualang="lua">-- recursive with memoization
local catalan = { [0] = 1 }
setmetatable(catalan, {
__index = function(c, n)
c[n] = c[n - 1] * 2 * (2 * n - 1) / (n + 1)
return c[n]
end
})
)
 
for i = 0, 14 do
print(string.format("%d", catalan[i]))
end</langsyntaxhighlight>
{{out}}
<pre>1
1
1
2
Line 2,991 ⟶ 4,007:
=={{header|MAD}}==
 
<langsyntaxhighlight MADlang="mad"> NORMAL MODE IS INTEGER
DIMENSION C(15)
Line 3,002 ⟶ 4,018:
 
VECTOR VALUES CFMT=$2HC(,I2,4H) = ,I7*$
END OF PROGRAM </langsyntaxhighlight>
 
{{out}}
Line 3,021 ⟶ 4,037:
C(13) = 742900
C(14) = 2674440</pre>
 
=={{header|Maple}}==
<langsyntaxhighlight Maplelang="maple">CatalanNumbers := proc( n::posint )
return seq( (2*i)!/((i + 1)!*i!), i = 0 .. n - 1 );
end proc:
CatalanNumbers(15);
</syntaxhighlight>
</lang>
Output:
<pre>
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440
</pre>
 
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<langsyntaxhighlight Mathematicalang="mathematica">CatalanN[n_Integer /; n >= 0] := (2 n)!/((n + 1)! n!)</langsyntaxhighlight>
{{out|Sample Output}}
<langsyntaxhighlight Mathematicalang="mathematica">TableForm[CatalanN/@Range[0,15]]
//TableForm=
1
Line 3,053 ⟶ 4,067:
742900
2674440
9694845</langsyntaxhighlight>
 
=={{header|MATLAB}} / {{header|Octave}}==
<langsyntaxhighlight MATLABlang="matlab">function n = catalanNumber(n)
for i = (1:length(n))
n(i) = (1/(n(i)+1))*nchoosek(2*n(i),n(i));
end
end</langsyntaxhighlight>
The following version computes at the same time the n first Catalan numbers (including C0).
<langsyntaxhighlight MATLABlang="matlab">function n = catalanNumbers(n)
n = [1 cumprod((2:4:4*n-6) ./ (2:n))];
end</langsyntaxhighlight>
{{out|Sample Output}}
<langsyntaxhighlight MATLABlang="matlab">>> catalanNumber(14)
 
ans =
Line 3,093 ⟶ 4,106:
9694845
35357670
129644790</langsyntaxhighlight>
 
The following version uses the identity Ln(x!)=Gammaln(x+1) and prod(1:x)=sum(ln(1:x))
<syntaxhighlight lang="matlab">
<lang MATLAB>
CatalanNumber=@(n) round(exp(gammaln(2*n+1)-sum(gammaln([n+2 n+1]))));
</syntaxhighlight>
</lang>
{{out|Sample Output}}
<langsyntaxhighlight MATLABlang="matlab">>>CatalanNumber(10)
 
ans =
Line 3,111 ⟶ 4,124:
 
'6564120420'
</syntaxhighlight>
</lang>
=={{header|Maxima}}==
<langsyntaxhighlight lang="maxima">/* The following is an array function, hence the square brackets. It uses memoization automatically */
cata[n] := sum(cata[i]*cata[n - 1 - i], i, 0, n - 1)$
cata[0]: 1$
Line 3,123 ⟶ 4,136:
makelist(cata2(n), n, 0, 14);
 
/* both return [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440] */</langsyntaxhighlight>
=={{header|Miranda}}==
<syntaxhighlight lang="miranda">main :: [sys_message]
main = [Stdout (lay (map (show . catalan) [0..14]))]
 
catalan :: num->num
catalan 0 = 1
catalan n = (4*n - 2) * catalan (n - 1) div (n + 1)</syntaxhighlight>
{{out}}
<pre>1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440</pre>
 
=={{header|Modula-2}}==
<langsyntaxhighlight lang="modula2">MODULE CatalanNumbers;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
Line 3,190 ⟶ 4,226:
END;
ReadChar
END CatalanNumbers.</langsyntaxhighlight>
 
=={{header|Nim}}==
<langsyntaxhighlight lang="nim">import math
import strformat
 
Line 3,210 ⟶ 4,246:
 
for i in 0..15:
echo &"{i:7} {catalan1(i):7} {catalan2(i):7} {catalan3(i):7}"</langsyntaxhighlight>
 
{{out}}
Line 3,229 ⟶ 4,265:
14 2674440 2674440 2674440
15 9694845 9694845 9694845</pre>
 
=={{header|OCaml}}==
<langsyntaxhighlight OCamllang="ocaml">let imp_catalan n =
let return = ref 1 in
for i = 1 to n do
Line 3,285 ⟶ 4,320:
"memoized", (memoize catalan)];
show imp_catalan catalan memo_catalan (memoize catalan) 15
</syntaxhighlight>
</lang>
{{out}}
<pre>$ ocaml unix.cma catalan.ml
Line 3,316 ⟶ 4,351:
memoized (10000000 runs) : 0.167
...</pre>
 
=={{header|Oforth}}==
 
<langsyntaxhighlight Oforthlang="oforth">: catalan( n -- m )
n ifZero: [ 1 ] else: [ catalan( n 1- ) 2 n * 1- * 2 * n 1+ / ] ;</langsyntaxhighlight>
 
{{out}}
Line 3,328 ⟶ 4,362:
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
</pre>
 
=={{header|ooRexx}}==
Three versions of this.
<langsyntaxhighlight ooRexxlang="oorexx">loop i = 0 to 15
say "catI("i") =" .catalan~catI(i)
say "catR1("i") =" .catalan~catR1(i)
Line 3,411 ⟶ 4,444:
catR2[n] = res
end
return res</langsyntaxhighlight>
{{out}}
<pre>catI(0) = 1
Line 3,461 ⟶ 4,494:
catR1(15) = 9694845
catR2(15) = 9694845</pre>
 
=={{header|PARI/GP}}==
Memoization is not worthwhile; PARI has fast built-in facilities for calculating binomial coefficients and factorials.
<langsyntaxhighlight lang="parigp">catalan(n)=binomial(2*n,n+1)/n</langsyntaxhighlight>
A second version:
<langsyntaxhighlight lang="parigp">catalan(n)=(2*n)!/(n+1)!/n!</langsyntaxhighlight>
Naive version with binary splitting:
<langsyntaxhighlight lang="parigp">catalan(n)=prod(k=n+2,2*n,k)/prod(k=2,n,k)</langsyntaxhighlight>
Naive version:
<langsyntaxhighlight lang="parigp">catalan(n)={
my(t=1);
for(k=n+2,2*n,t*=k);
for(k=2,n,t/=k);
t
};</langsyntaxhighlight>
The first version takes about 1.5 seconds to compute the millionth Catalan number, while the second takes 3.9 seconds. The naive implementations, for comparison, take 21 and 45 minutes. In any case, printing the first 15 is simple:
<langsyntaxhighlight lang="parigp">vector(15,n,catalan(n))</langsyntaxhighlight>
 
=={{header|Pascal}}==
<langsyntaxhighlight lang="pascal">Program CatalanNumbers(output);
 
function catalanNumber1(n: integer): double;
Line 3,498 ⟶ 4,529:
for number := 0 to 14 do
writeln (number:3, round(catalanNumber1(number)):9);
end.</langsyntaxhighlight>
{{out}}
<pre>
Line 3,520 ⟶ 4,551:
14 2674440
</pre>
 
=={{header|Perl}}==
<langsyntaxhighlight lang="perl">sub factorial { my $f = 1; $f *= $_ for 2 .. $_[0]; $f; }
sub catalan {
my $n = shift;
Line 3,528 ⟶ 4,558:
}
 
print "$_\t@{[ catalan($_) ]}\n" for 0 .. 20;</langsyntaxhighlight>
For computing up to 20 ish, memoization is not needed. For much bigger numbers, this is faster:
<langsyntaxhighlight lang="perl">my @c = (1);
sub catalan {
use bigint;
Line 3,537 ⟶ 4,567:
 
# most of the time is spent displaying the long numbers, actually
print "$_\t", catalan($_), "\n" for 0 .. 10000;</langsyntaxhighlight>
 
That has two downsides: high memory use and slow access to an isolated large value. Using a fast binomial function can solve both these issues. The downside here is if the platform doesn't have the GMP library then binomials won't be fast.
{{libheader|ntheory}}
<langsyntaxhighlight lang="perl">use ntheory qw/binomial/;
sub catalan {
my $n = shift;
binomial(2*$n,$n)/($n+1);
}
print "$_\t", catalan($_), "\n" for 0 .. 10000;</langsyntaxhighlight>
 
=={{header|Phix}}==
See also [[Catalan_numbers/Pascal%27s_triangle#Phix]] which may be faster.
<!--<langsyntaxhighlight Phixlang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #000080;font-style:italic;">-- returns inf/-nan for n&gt;85, and needs the rounding for n&gt;=14, accurate to n=29</span>
Line 3,589 ⟶ 4,618:
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"times:%8s %10s %10s\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">times</span><span style="color: #0000FF;">)</span>
<!--</langsyntaxhighlight>-->
{{out}}
<pre>
Line 3,614 ⟶ 4,643:
=== memoized recursive gmp version ===
{{libheader|Phix/mpfr}}
<!--<langsyntaxhighlight Phixlang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">include</span> <span style="color: #000000;">builtins</span><span style="color: #0000FF;">\</span><span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>
Line 3,645 ⟶ 4,674:
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"0..15: %s\n"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">,</span><span style="color: #008000;">","</span><span style="color: #0000FF;">))</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"100: %s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">mpz_get_str</span><span style="color: #0000FF;">(</span><span style="color: #000000;">catalan2m</span><span style="color: #0000FF;">(</span><span style="color: #000000;">100</span><span style="color: #0000FF;">))})</span>
<!--</langsyntaxhighlight>-->
{{out}}
<pre>
Line 3,651 ⟶ 4,680:
100: 896519947090131496687170070074100632420837521538745909320
</pre>
 
=={{header|PHP}}==
<langsyntaxhighlight lang="php"><?php
 
class CatalanNumbersSerie
Line 3,685 ⟶ 4,713:
echo "$i = $r\r\n";
}
?></langsyntaxhighlight>
{{out}}
<pre>
Line 3,705 ⟶ 4,733:
15 = 9694845
</pre>
<langsyntaxhighlight lang="php">
<?php
$n = 15;
Line 3,719 ⟶ 4,747:
print ($t[$i+1]-$t[$i])."\t";
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 35357670
</pre>
=={{header|Picat}}==
{{works with|Picat}}
<syntaxhighlight lang="picat">
table
factorial(0) = 1.
 
factorial(N) = N * factorial(N - 1).
 
catalan1(N) = factorial(2 * N) // (factorial(N + 1) * factorial(N)).
 
catalan2(0) = 1.
 
catalan2(N) = 2 * (2 * N - 1) * catalan2(N - 1) // (N + 1).
 
main =>
foreach (I in 0..14)
printf("%d. %d = %d\n", I, catalan1(I), catalan2(I))
end.
</syntaxhighlight>
{{out}}
<pre>
0. 1 = 1
1. 1 = 1
2. 2 = 2
3. 5 = 5
4. 14 = 14
5. 42 = 42
6. 132 = 132
7. 429 = 429
8. 1430 = 1430
9. 4862 = 4862
10. 16796 = 16796
11. 58786 = 58786
12. 208012 = 208012
13. 742900 = 742900
14. 2674440 = 2674440
</pre>
=={{header|PicoLisp}}==
<langsyntaxhighlight PicoLisplang="picolisp"># Factorial
(de fact (N)
(if (=0 N)
Line 3,758 ⟶ 4,822:
(catalanDir N)
(catalanRec N)
(catalanAlt N) ) )</langsyntaxhighlight>
{{out}}
<pre> 0 => 1 1 1
Line 3,775 ⟶ 4,839:
13 => 742900 742900 742900
14 => 2674440 2674440 2674440</pre>
 
=={{header|PL/0}}==
{{trans|Tiny BASIC}}
Integers are limited to 32767 so only the first ten Catalan numbers can be represented. To avoid internal overflow, the program subtracts something clever from <code>c</code> and then adds it back at the end.
<syntaxhighlight lang="pascal">
var n, c, i;
begin
n := 0; c := 1;
! c;
while n <= 9 do
begin
n := n + 1;
i := 0;
while c > 0 do
begin
c := c - (n + 1);
i := i + 1
end;
c := 2 * (2 * n - 1) * c / (n + 1);
c := c + 2 * i * (2 * n - 1);
! c
end;
end.
</syntaxhighlight>
{{out}}
<pre>
1
1
2
5
14
42
132
429
1430
4862
16796
</pre>
 
=={{header|PL/I}}==
<langsyntaxhighlight PLlang="pl/Ii">catalan: procedure options (main); /* 23 February 2012 */
declare (i, n) fixed;
 
Line 3,795 ⟶ 4,897:
end c;
 
end catalan;</langsyntaxhighlight>
{{out}}
<pre>
Line 3,822 ⟶ 4,924:
6564120420
</pre>
 
=={{header|PlainTeX}}==
<langsyntaxhighlight lang="tex">\newcount\n
\newcount\r
\newcount\x
Line 3,845 ⟶ 4,946:
\advance\x by 1\ifnum\x<15\repeat
 
\bye</langsyntaxhighlight>
 
=={{header|PowerShell}}==
<syntaxhighlight lang="powershell">
<lang PowerShell>
function Catalan([uint64]$m) {
function fact([bigint]$n) {
Line 3,859 ⟶ 4,959:
}
0..15 | foreach {"catalan($_): $(catalan $_)"}
</syntaxhighlight>
</lang>
<b>Output:</b>
<pre>
Line 3,882 ⟶ 4,982:
===An Alternate Version===
This version could easily be modified to work with big integers.
<syntaxhighlight lang="powershell">
<lang PowerShell>
function Get-CatalanNumber
{
Line 3,934 ⟶ 5,034:
}
}
</syntaxhighlight>
</lang>
Get the first fifteen Catalan numbers as a PSCustomObject:
<syntaxhighlight lang="powershell">
<lang PowerShell>
0..14 | Get-CatalanNumber
</syntaxhighlight>
</lang>
{{Out}}
<pre>
Line 3,960 ⟶ 5,060:
</pre>
To return only the array of Catalan numbers:
<syntaxhighlight lang="powershell">
<lang PowerShell>
(0..14 | Get-CatalanNumber).CatalanNumber
</syntaxhighlight>
</lang>
{{Out}}
<pre>
Line 3,981 ⟶ 5,081:
2674440
</pre>
 
=={{header|Prolog}}==
{{Works with|SWI-Prolog}}
<langsyntaxhighlight Prologlang="prolog">catalan(N) :-
length(L1, N),
L = [1 | L1],
Line 4,000 ⟶ 5,099:
 
my_write(N, V) :-
format('~w : ~w~n', [N, V]).</langsyntaxhighlight>
{{out}}
<pre> ?- catalan(15).
Line 4,020 ⟶ 5,119:
15 : 9694845
true .
</pre>
 
=={{header|PureBasic}}==
Using the third formula...
<lang PureBasic>; saving the division for last ensures we divide the largest
; numerator by the smallest denominator
 
Procedure.q CatalanNumber(n.q)
If n<0:ProcedureReturn 0:EndIf
If n=0:ProcedureReturn 1:EndIf
ProcedureReturn (2*(2*n-1))*CatalanNumber(n-1)/(n+1)
EndProcedure
 
ls=25
rs=12
 
a.s=""
a.s+LSet(RSet("n",rs),ls)+"CatalanNumber(n)"
; cw(a.s)
Debug a.s
 
For n=0 to 33 ;33 largest correct quad for n
a.s=""
a.s+LSet(RSet(Str(n),rs),ls)+Str(CatalanNumber(n))
; cw(a.s)
Debug a.s
Next</lang>
{{out|Sample Output}}
<pre>
n CatalanNumber(n)
0 1
1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845
16 35357670
17 129644790
18 477638700
19 1767263190
20 6564120420
21 24466267020
22 91482563640
23 343059613650
24 1289904147324
25 4861946401452
26 18367353072152
27 69533550916004
28 263747951750360
29 1002242216651368
30 3814986502092304
31 14544636039226909
32 55534064877048198
33 212336130412243110
</pre>
 
Line 4,092 ⟶ 5,127:
 
{{Works with|Python|3}}
<langsyntaxhighlight lang="python">from math import factorial
import functools
 
Line 4,140 ⟶ 5,175:
defs = (cat_direct, catR1, catR2)
results = [tuple(c(i) for i in range(15)) for c in defs]
pr(results)</langsyntaxhighlight>
{{out|Sample Output}}
<pre>CAT_DIRECT CATR1 CATR2
Line 4,162 ⟶ 5,197:
 
The third definition is directly expressible, as an infinite series, in terms of '''itertools.accumulate''':
<langsyntaxhighlight lang="python">'''Catalan numbers'''
 
from itertools import accumulate, chain, count, islice
Line 4,208 ⟶ 5,243:
# MAIN ---
if __name__ == '__main__':
main()</langsyntaxhighlight>
{{Out}}
<pre>Catalans 1-15:
Line 4,227 ⟶ 5,262:
742900
2674440</pre>
 
=={{header|Quackery}}==
 
<langsyntaxhighlight Quackerylang="quackery"> [ 1 over times [ over i 1+ + * ] nip ] is 2n!/n! ( n --> n )
 
[ times [ i 1+ / ] ] is /n! ( n --> n )
Line 4,236 ⟶ 5,270:
[ dup 2n!/n! swap 1+ /n! ] is catalan ( n --> n )
 
15 times [ i^ dup echo say " : " catalan echo cr ]</langsyntaxhighlight>
 
{{out}}
Line 4,256 ⟶ 5,290:
14 : 2674440
</pre>
 
=={{header|R}}==
<langsyntaxhighlight lang="r">catalan <- function(n) choose(2*n, n)/(n + 1)
catalan(0:15)
[1] 1 1 2 5 14 42 132 429 1430
[10] 4862 16796 58786 208012 742900 2674440 9694845</langsyntaxhighlight>
 
=={{header|Racket}}==
<langsyntaxhighlight lang="racket">#lang racket
(require planet2)
; (install "this-and-that") ; uncomment to install
Line 4,275 ⟶ 5,307:
(* (catalan i) (catalan (- m i 1))))))
(map catalan (range 1 15))</langsyntaxhighlight>
{{out}}
<pre>
'(1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)
</pre>
 
=={{header|Raku}}==
(formerly Perl 6)
Line 4,286 ⟶ 5,317:
The recursive formulas are easily written into a constant array, either:
 
<syntaxhighlight lang="raku" perl6line>constant Catalan = 1, { [+] @_ Z* @_.reverse } ... *;</langsyntaxhighlight>
 
or
 
<syntaxhighlight lang="raku" perl6line>constant Catalan = 1, |[\*] (2, 6 ... *) Z/ 2 .. *;
 
 
# In both cases, the sixteen first values can be seen with:
.say for Catalan[^16];</langsyntaxhighlight>
{{out}}
<pre>1
Line 4,311 ⟶ 5,342:
742900
2674440</pre>
 
=={{header|REXX}}==
===version 1===
Line 4,321 ⟶ 5,351:
has been rearranged to:
:::::::: <big> (n+1) * [fact(n) **2] </big>
<langsyntaxhighlight lang="rexx">/*REXX program calculates and displays Catalan numbers using four different methods. */
parse arg LO HI . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then do; HI=15; LO=0; end /*No args? Then use a range of 0 ──► 15*/
Line 4,343 ⟶ 5,373:
Cat2: procedure expose c.; parse arg n; $=0; if c.n\==. then return c.n
do k=0 for n; $=$ + Cat2(k) * Cat2(n-k-1); end
c.n=$; return $ /*use a memoization technique.*/</langsyntaxhighlight>
'''output''' &nbsp; when using the input of: &nbsp; <tt> 0 &nbsp; 16 </tt>
<pre>
Line 4,388 ⟶ 5,418:
===version 2===
Implements the 3 methods shown in the task description
<langsyntaxhighlight lang="rexx">/* REXX ---------------------------------------------------------------
* 01.07.2014 Walter Pachl
*--------------------------------------------------------------------*/
Line 4,438 ⟶ 5,468:
f=f*i
End
Return f</langsyntaxhighlight>
{{out}}
<pre> n c1.n c2.n c3.n
Line 4,463 ⟶ 5,493:
20 6564120420 6564120420 6564120420
n c1.n c2.n c3.n</pre>
 
=={{header|Ring}}==
<langsyntaxhighlight lang="ring">
for n = 1 to 15
see catalan(n) + nl
Line 4,474 ⟶ 5,503:
cat = 2 * (2 * n - 1) * catalan(n - 1) / (n + 1)
return cat
</syntaxhighlight>
</lang>
Output:
<pre>
Line 4,492 ⟶ 5,521:
2674440
9694845
</pre>
=={{header|RPL}}==
{{works with|Halcyon Calc|4.2.7}}
{| class="wikitable"
! Code
! Comments
|-
|
≪
'''IFERR''' R→B '''THEN END'''
'''IF''' DUP #1 ≠
'''THEN'''
DUP 2 * 1 - 2 * OVER 1 - →CAT * SWAP 1 + /
'''END'''
≫ ''''→CAT'''' STO
|
''( n -- C(n) )''
Ignore the conversion error if n is already a binary integer
C(1) = 1
Divide by (n+1) at the end to stay in the integer world
|}
To speed up execution, additions can be preferred to multiplications by replacing <code>2 *</code> with <code>DUP +</code>.
The following piece of code will deliver what is required:
≪ 1 { } '''DO''' OVER →CAT B→R + SWAP 1 + SWAP '''UNTIL''' OVER 15 > '''END''' ≫ EVAL
{{out}}
<pre>
1: { 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 }
</pre>
 
=={{header|Ruby}}==
<syntaxhighlight lang="ruby">def factorial(n)
{{libheader|RubyGems}}
<lang ruby>def factorial(n)
(1..n).reduce(1, :*)
end
Line 4,510 ⟶ 5,568:
def catalan_rec1(n)
return 1 if n == 0
(0...n).inject(0) sum{|sum, i| sum + catalan_rec1(i) * catalan_rec1(n-1-i)}
end
 
Line 4,534 ⟶ 5,592:
 
puts "\n direct rec1 rec2"
16.times {|n| puts "%2d :%9d%9d%9d" % [n, catalan_direct(n), catalan_rec1(n), catalan_rec2(n)]}</langsyntaxhighlight>
The output shows the dramatic difference memoizing makes.
<pre>
Line 4,561 ⟶ 5,619:
15 : 9694845 9694845 9694845
</pre>
 
=={{header|Run BASIC}}==
<lang Runbasic>FOR i = 1 TO 15
PRINT i;" ";catalan(i)
NEXT
FUNCTION catalan(n)
catalan = 1
if n <> 0 then catalan = ((2 * ((2 * n) - 1)) / (n + 1)) * catalan(n - 1)
END FUNCTION</lang>
<pre>1 1
2 2
3 5
4 14
5 42
6 132
7 429
8 1430
9 4862
10 16796
11 58786
12 208012
13 742900
14 2674440
15 9694845</pre>
 
=={{header|Rust}}==
<langsyntaxhighlight lang="rust">fn c_n(n: u64) -> u64 {
match n {
0 => 1,
Line 4,599 ⟶ 5,632:
println!("c_n({}) = {}", i, c_n(i));
}
}</langsyntaxhighlight>
 
{{out}}
Line 4,618 ⟶ 5,651:
c(14) = 2674440
c(15) = 9694845</pre>
 
=={{header|Scala}}==
 
Simple and straightforward. Noticeably out of steam without memoizing at about 5000.
<langsyntaxhighlight lang="scala">
object CatalanNumbers {
def main(args: Array[String]): Unit = {
Line 4,634 ⟶ 5,666:
def factorial(n: BigInt): BigInt = BigInt(1).to(n).foldLeft(BigInt(1))(_ * _)
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 4,654 ⟶ 5,686:
catalan(15) = 9694845
</pre>
 
=={{header|Scheme}}==
Tail recursive implementation.
<langsyntaxhighlight lang="scheme">(define (catalan m)
(let loop ((c 1)(n 0))
(if (not (eqv? n m))
Line 4,664 ⟶ 5,695:
(loop (* (/ (* 2 (- (* 2 (+ n 1)) 1)) (+ (+ n 1) 1)) c) (+ n 1) )))))
 
(catalan 15)</langsyntaxhighlight>
{{out}}
<pre>0: 1
Line 4,681 ⟶ 5,712:
13: 742900
14: 2674440</pre>
 
=={{header|Seed7}}==
<langsyntaxhighlight lang="seed7">$ include "seed7_05.s7i";
include "bigint.s7i";
 
Line 4,693 ⟶ 5,723:
writeln((2_ * n) ! n div succ(n));
end for;
end func;</langsyntaxhighlight>
{{out}}
<pre>
Line 4,713 ⟶ 5,743:
9694845
</pre>
 
=={{header|Sidef}}==
<langsyntaxhighlight lang="ruby">func f(i) { i==0 ? 1 : (i * f(i-1)) }
func c(n) { f(2*n) / f(n) / f(n+1) }</langsyntaxhighlight>
With memoization:
<langsyntaxhighlight lang="ruby">func c(n) is cached {
n == 0 ? 1 : (c(n-1) * (4 * n - 2) / (n + 1))
}</langsyntaxhighlight>
 
Calling the function:
<langsyntaxhighlight lang="ruby">15.times { |i|
say "#{i}\t#{c(i)}"
}</langsyntaxhighlight>
{{out}}
<pre>
Line 4,744 ⟶ 5,773:
14 2674440
</pre>
 
=={{header|smart BASIC}}==
 
<lang qbasic>PRINT "Recursive:"!PRINT
FOR n = 0 TO 15
PRINT n,"#######":catrec(n)
NEXT n
PRINT!PRINT
 
PRINT "Non-recursive:"!PRINT
FOR n = 0 TO 15
PRINT n,"#######":catnonrec(n)
NEXT n
 
END
DEF catrec(x)
IF x = 0 THEN
temp = 1
ELSE
n = x
temp = ((2*((2*n)-1))/(n+1))*catrec(n-1)
END IF
catrec = temp
END DEF
 
DEF catnonrec(x)
temp = 1
FOR n = 1 TO x
temp = (2*((2*n)-1))/(n+1)*temp
NEXT n
catnonrec = temp
END DEF</lang>
 
=={{header|Standard ML}}==
<langsyntaxhighlight lang="sml">(*
* val catalan : int -> int
* Returns the nth Catalan number.
Line 4,810 ⟶ 5,806:
* 2674440
* 9694845
*)</langsyntaxhighlight>
 
=={{header|Stata}}==
 
<langsyntaxhighlight lang="stata">clear
set obs 15
gen catalan=1 in 1
replace catalan=catalan[_n-1]*2*(2*_n-3)/_n in 2/l
list, noobs noh</langsyntaxhighlight>
 
'''Output'''
Line 4,841 ⟶ 5,836:
| 2674440 |
+---------+</pre>
 
=={{header|Swift}}==
 
{{trans|Rust}}
<langsyntaxhighlight lang="swift">func catalan(_ n: Int) -> Int {
switch n {
case 0:
Line 4,857 ⟶ 5,851:
print("catalan(\(i)) => \(catalan(i))")
}
</syntaxhighlight>
</lang>
 
{{out}}
Line 4,877 ⟶ 5,871:
catalan(15) => 9694845
</pre>
 
=={{header|Tcl}}==
<langsyntaxhighlight lang="tcl">package require Tcl 8.5
 
# Memoization wrapper
Line 4,897 ⟶ 5,890:
$n == 0 ? 1 : 2 * (2*$n - 1) * catalan($n - 1) / ($n + 1)
}}
}</langsyntaxhighlight>
Demonstration:
<langsyntaxhighlight lang="tcl">for {set i 0} {$i < 15} {incr i} {
puts "C_$i = [expr {catalan($i)}]"
}</langsyntaxhighlight>
{{out}}
<pre>
Line 4,929 ⟶ 5,922:
</pre>
 
== {{header|TI-83 BASICTypeScript}} ==
{{trans|GW-BASIC}}
This problem is perfectly suited for a TI calculator.
<syntaxhighlight lang="javascript">
<lang TI-83 BASIC>:For(I,1,15
// Catalan numbers
:Disp (2I)!/((I+1)!I!
var c: number[] = [1];
:End</lang>
console.log(`${0}\t${c[0]}`);
for (n = 0; n < 15; n++) {
c[n + 1] = 0;
for (i = 0; i <= n; i++)
c[n + 1] = c[n + 1] + c[i] * c[n - i];
console.log(`${n + 1}\t${c[n + 1]}`);
}
</syntaxhighlight>
{{out}}
<pre> 1
0 1
2
1 1
4
2 2
14
3 5
42
4 14
132
5 42
429
6 132
1430
7 429
4862
8 1430
16796
9 4862
58786
10 16796
208012
11 58786
742900
12 208012
2674440
13 742900
9694845
14 2674440
Done</pre>
15 9694845
</pre>
 
=={{header|Ursala}}==
<langsyntaxhighlight lang="ursala">#import std
#import nat
 
Line 4,960 ⟶ 5,963:
#cast %nL
 
t = catalan* iota 16</langsyntaxhighlight>
{{out}}
<pre><
Line 4,979 ⟶ 5,982:
2674440,
9694845></pre>
 
=={{header|Vala}}==
<langsyntaxhighlight lang="vala">namespace CatalanNumbers {
public class CatalanNumberGenerator {
private static double factorial(double n) {
Line 5,057 ⟶ 6,059:
 
}
}</langsyntaxhighlight>
{{out}}
<pre>
Line 5,127 ⟶ 6,129:
</pre>
 
=={{header|VBAV (Vlang)}}==
{{trans|Go}}
<lang vb>Public Sub Catalan1(n As Integer)
<syntaxhighlight lang="v (vlang)">import math.big
'Computes the first n Catalan numbers according to the first recursion given
Dim Cat() As Long
Dim sum As Long
 
fn main() {
ReDim Cat(n)
mut b:= big.zero_int
Cat(0) = 1
For i for n := i64(0 To); n -< 15; n++ 1{
b = big.integer_from_i64(n)
sum = 0
b = (b*big.two_int).factorial()/(b.factorial()*(b*big.two_int-b).factorial())
For j = 0 To i
println(b/big.integer_from_i64(n+1))
sum = sum + Cat(j) * Cat(i - j)
Next j }
}</syntaxhighlight>
Cat(i + 1) = sum
{{out}}
Next i
Debug.Print
For i = 0 To n
Debug.Print i, Cat(i)
Next
End Sub
 
Public Sub Catalan2(n As Integer)
'Computes the first n Catalan numbers according to the second recursion given
Dim Cat() As Long
 
ReDim Cat(n)
Cat(0) = 1
For i = 1 To n
Cat(i) = 2 * Cat(i - 1) * (2 * i - 1) / (i + 1)
Next i
Debug.Print
For i = 0 To n
Debug.Print i, Cat(i)
Next
End Sub</lang>
{{out|Result}}
<pre>
1
Catalan1 15
1
 
2
0 1
5
1 1
14
2 2
42
3 5
132
4 14
429
5 42
1430
6 132
4862
7 429
16796
8 1430
58786
9 4862
208012
10 16796
742900
11 58786
2674440
12 208012
13 742900
14 2674440
15 9694845
</pre>
(Expect same result with "Catalan2 15")
 
=={{header|VBScript}}==
<lang vb>
Function catalan(n)
catalan = factorial(2*n)/(factorial(n+1)*factorial(n))
End Function
 
Function factorial(n)
If n = 0 Then
Factorial = 1
Else
For i = n To 1 Step -1
If i = n Then
factorial = n
Else
factorial = factorial * i
End If
Next
End If
End Function
 
'Find the first 15 Catalan numbers.
For j = 1 To 15
WScript.StdOut.Write j & " = " & catalan(j)
WScript.StdOut.WriteLine
Next
</lang>
 
{{Out}}
<pre>
1 = 1
2 = 2
3 = 5
4 = 14
5 = 42
6 = 132
7 = 429
8 = 1430
9 = 4862
10 = 16796
11 = 58786
12 = 208012
13 = 742900
14 = 2674440
15 = 9694845
</pre>
 
=={{header|Visual Basic .NET}}==
{{trans|C#}}
<lang vbnet>Module Module1
 
Function Factorial(n As Double) As Double
If n < 1 Then
Return 1
End If
 
Dim result = 1.0
For i = 1 To n
result = result * i
Next
 
Return result
End Function
 
Function FirstOption(n As Double) As Double
Return Factorial(2 * n) / (Factorial(n + 1) * Factorial(n))
End Function
 
Function SecondOption(n As Double) As Double
If n = 0 Then
Return 1
End If
 
Dim sum = 0
For i = 0 To n - 1
sum = sum + SecondOption(i) * SecondOption((n - 1) - i)
Next
Return sum
End Function
 
Function ThirdOption(n As Double) As Double
If n = 0 Then
Return 1
End If
 
Return ((2 * (2 * n - 1)) / (n + 1)) * ThirdOption(n - 1)
End Function
 
Sub Main()
Const MaxCatalanNumber = 15
 
Dim initial As DateTime
Dim final As DateTime
Dim ts As TimeSpan
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, FirstOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, SecondOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()
 
initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, ThirdOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
End Sub
 
End Module</lang>
{{out}}
<pre>CatalanNumber(0:1)
CatalanNumber(1:1)
CatalanNumber(2:2)
CatalanNumber(3:5)
CatalanNumber(4:14)
CatalanNumber(5:42)
CatalanNumber(6:132)
CatalanNumber(7:429)
CatalanNumber(8:1430)
CatalanNumber(9:4862)
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440)
CatalanNumber(15:9694845)
It took 0.19 to execute
 
CatalanNumber(0:1)
CatalanNumber(1:1)
CatalanNumber(2:2)
CatalanNumber(3:5)
CatalanNumber(4:14)
CatalanNumber(5:42)
CatalanNumber(6:132)
CatalanNumber(7:429)
CatalanNumber(8:1430)
CatalanNumber(9:4862)
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440)
CatalanNumber(15:9694845)
It took 0.831 to execute
 
CatalanNumber(0:1)
CatalanNumber(1:1)
CatalanNumber(2:2)
CatalanNumber(3:5)
CatalanNumber(4:14)
CatalanNumber(5:42)
CatalanNumber(6:132)
CatalanNumber(7:429)
CatalanNumber(8:1430)
CatalanNumber(9:4862)
CatalanNumber(10:16796)
CatalanNumber(11:58786)
CatalanNumber(12:208012)
CatalanNumber(13:742900)
CatalanNumber(14:2674440)
CatalanNumber(15:9694845)
It took 0.8 to execute</pre>
 
=={{header|Wortel}}==
<langsyntaxhighlight lang="wortel">; the following number expression calculcates the nth Catalan number
#~ddiFSFmSoFSn
; which stands for: dup dup inc fac swap fac mult swap double fac swap divide
; to get the first 15 Catalan numbers we map this function over a list from 0 to 15
!*#~ddiFSFmSoFSn @til 15
; returns [1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674439.9999999995]</langsyntaxhighlight>
 
=={{header|Wren}}==
{{libheader|Wren-fmt}}
{{libheader|Wren-math}}
<langsyntaxhighlight ecmascriptlang="wren">import "./fmt" for Fmt
import "./math" for Int
var catalan = Fn.new { |n|
Line 5,394 ⟶ 6,191:
for (i in 0..15) System.print("%(Fmt.d(2, i)) %(catalan.call(i))")
System.print("\nand again using a recursive function:\n")
for (i in 0..15) System.print("%(Fmt.d(2, i)) %(catalanRec.call(i))")</langsyntaxhighlight>
 
{{out}}
Line 5,436 ⟶ 6,233:
15 9694845
</pre>
 
 
=={{header|XLISP}}==
<langsyntaxhighlight lang="lisp">(defun catalan (n)
(if (= n 0)
1
Line 5,449 ⟶ 6,245:
(range (+ x 1) y) ) ) )
 
(print (mapcar catalan (range 0 14)))</langsyntaxhighlight>
{{out}}
<pre>(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)</pre>
 
=={{header|XPL0}}==
<langsyntaxhighlight XPL0lang="xpl0">code CrLf=9, IntOut=11;
int C, N;
[C:= 1;
Line 5,462 ⟶ 6,257:
IntOut(0, C); CrLf(0);
];
]</langsyntaxhighlight>
{{out}}
<pre>
Line 5,484 ⟶ 6,279:
=={{header|zkl}}==
Uses GMP to calculate big factorials.
<langsyntaxhighlight lang="zkl">var BN=Import("zklBigNum");
fcn catalan(n){
BN(2*n).factorial() / BN(n+1).factorial() / BN(n).factorial();
Line 5,492 ⟶ 6,287:
println("%2d --> %,d".fmt(n, catalan(n)));
}
println("%2d --> %,d".fmt(100, catalan(100)));</langsyntaxhighlight>
And an iterative solution at works up to the limit of 64 bit ints (n=33). Would be 35 but need to avoid factional intermediate results.
<langsyntaxhighlight lang="zkl">fcn catalan(n){ (1).reduce(n,fcn(p,n){ 2*(2*n-1)*p/(n+1) },1) }</langsyntaxhighlight>
{{out}}
<pre>
Line 5,515 ⟶ 6,310:
100 --> 896,519,947,090,131,496,687,170,070,074,100,632,420,837,521,538,745,909,320
</pre>
 
=={{header|ZX Spectrum Basic}}==
{{trans|C}}
<langsyntaxhighlight lang="zxbasic">10 FOR i=0 TO 15
20 LET n=i: LET m=2*n
30 LET r=1: LET d=m-n
Line 5,530 ⟶ 6,324:
110 STOP
120 DEF FN m(a,b)=a-INT (a/b)*b: REM Modulus function
</syntaxhighlight>
</lang>
Langurmonkey
889

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