Sequence of non-squares
(Redirected from Sequance of Non-squares)
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Show that the following remarkable formula gives the sequence of non-square natural numbers:
n + floor(1/2 + sqrt(n))
- Print out the values for n in the range 1 to 22
- Show that no squares occur for n less than one million
This is sequence A000037 in the OEIS database.
11l
F non_square(Int n)
R n + Int(floor(1/2 + sqrt(n)))
print_elements((1..22).map(non_square))
F is_square(n)
R fract(sqrt(n)) == 0
L(i) 1 .< 10 ^ 6
I is_square(non_square(i))
print(‘Square found ’i)
L.break
L.was_no_break
print(‘No squares found’)- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares found
ABC
HOW TO RETURN non.square n:
RETURN n + floor (1/2 + root n)
HOW TO REPORT square n:
REPORT n = (floor root n)**2
FOR n IN {1..22}: WRITE non.square n
WRITE /
IF NO n IN {1..1000000} HAS square non.square n:
WRITE "No squares occur for n < 1.000.000"- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares occur for n < 1.000.000
Ada
with Ada.Numerics.Long_Elementary_Functions;
with Ada.Text_IO; use Ada.Text_IO;
procedure Sequence_Of_Non_Squares_Test is
use Ada.Numerics.Long_Elementary_Functions;
function Non_Square (N : Positive) return Positive is
begin
return N + Positive (Long_Float'Rounding (Sqrt (Long_Float (N))));
end Non_Square;
I : Positive;
begin
for N in 1..22 loop -- First 22 non-squares
Put (Natural'Image (Non_Square (N)));
end loop;
New_Line;
for N in 1..1_000_000 loop -- Check first million of
I := Non_Square (N);
if I = Positive (Sqrt (Long_Float (I)))**2 then
Put_Line ("Found a square:" & Positive'Image (N));
end if;
end loop;
end Sequence_Of_Non_Squares_Test;
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
ALGOL 68
PROC non square = (INT n)INT: n + ENTIER(0.5 + sqrt(n));
main: (
# first 22 values (as a list) has no squares: #
FOR i TO 22 DO
print((whole(non square(i),-3),space))
OD;
print(new line);
# The following check shows no squares up to one million: #
FOR i TO 1 000 000 DO
REAL j = sqrt(non square(i));
IF j = ENTIER j THEN
put(stand out, ("Error: number is a square:", j, new line));
stop
FI
OD
)- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
ALGOL W
begin
% check values of the function: f(n) = n + floor(1/2 + sqrt(n)) %
% are not squares %
integer procedure f ( integer value n ) ;
begin
n + entier( 0.5 + sqrt( n ) )
end f ;
logical noSquares;
% first 22 values of f %
for n := 1 until 22 do writeon( i_w := 1, f( n ) );
% check f(n) does not produce a square for n in 1..1 000 000 %
noSquares := true;
for n := 1 until 1000000 do begin
integer fn, rn;
fn := f( n );
rn := round( sqrt( fn ) );
if ( rn * rn ) = fn then begin
write( "Found square at: ", n );
noSquares := false
end if_fn_is_a_square
end for_n ;
if noSquares then write( "f(n) did not produce a square in 1 .. 1 000 000" )
else write( "f(n) produced a square" )
end.- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 f(n) did not produce a square in 1 .. 1 000 000
APL
Generate the first 22 numbers:
NONSQUARE←{(⍳⍵)+⌊0.5+(⍳⍵)*0.5}
NONSQUARE 22
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Show there are no squares in the first million:
HOWMANYSQUARES←{+⌿⍵=(⌊⍵*0.5)*2}
HOWMANYSQUARES NONSQUARE 1000000
0
AppleScript
on task()
set values to {}
set squareCount to 0
repeat with n from 1 to (1000000 - 1)
set v to n + (0.5 + n ^ 0.5) div 1
if (n ≤ 22) then set end of values to v
set sqrt to v ^ 0.5
if (sqrt = sqrt as integer) then set squareCount to squareCount + 1
end repeat
return "Values (n = 1 to 22): " & join(values, ", ") & (linefeed & ¬
"Number of squares (n < 1000000): " & squareCount)
end task
on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join
task()
- Output:
"Values (n = 1 to 22): 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27
Number of squares (n < 1000000): 0"
Arturo
f: function [n]->
n + floor 0.5 + sqrt n
loop 1..22 'i ->
print [i "->" f i]
i: new 1, nonSquares: new []
while [i<1000000][ 'nonSquares ++ f i, inc 'i]
squares: map 1..1001 'x -> x ^ 2
if? empty? intersection squares nonSquares -> print "Didn't find any squares!"
else -> print "Ooops! Something went wrong!"
- Output:
1 -> 2 2 -> 3 3 -> 5 4 -> 6 5 -> 7 6 -> 8 7 -> 10 8 -> 11 9 -> 12 10 -> 13 11 -> 14 12 -> 15 13 -> 17 14 -> 18 15 -> 19 16 -> 20 17 -> 21 18 -> 22 19 -> 23 20 -> 24 21 -> 26 22 -> 27 Didn't find any squares!
AutoHotkey
ahk forum: discussion
Loop 22
t .= (A_Index + floor(0.5 + sqrt(A_Index))) " "
MsgBox %t%
s := 0
Loop 1000000
x := A_Index + floor(0.5 + sqrt(A_Index)), s += x = round(sqrt(x))**2
Msgbox Number of bad squares = %s% ; 0
AWK
$ awk 'func f(n){return(n+int(.5+sqrt(n)))}BEGIN{for(i=1;i<=22;i++)print i,f(i)}'
1 2
2 3
3 5
4 6
5 7
6 8
7 10
8 11
9 12
10 13
11 14
12 15
13 17
14 18
15 19
16 20
17 21
18 22
19 23
20 24
21 26
22 27
$ awk 'func f(n){return(n+int(.5+sqrt(n)))}BEGIN{for(i=1;i<100000;i++){n=f(i);r=int(sqrt(n));if(r*r==n)print n"is square"}}'
$
BASIC
DIM i AS Integer
DIM j AS Double
DIM found AS Integer
FUNCTION nonsqr (n AS Integer) AS Integer
nonsqr = n + INT(0.5 + SQR(n))
END FUNCTION
' Display first 22 values
FOR i = 1 TO 22
PRINT nonsqr(i); " ";
NEXT i
PRINT
' Check for squares up to one million
found = 0
FOR i = 1 TO 1000000
j = SQR(nonsqr(i))
IF j = INT(j) THEN
found = 1
PRINT "Found square: "; i
EXIT FOR
END IF
NEXT i
IF found=0 THEN PRINT "No squares found"
BASIC256
# Display first 22 values
print "The first 22 numbers generated by the sequence are : "
for i = 1 to 22
print nonSquare(i); " ";
next i
print
# Check for squares up to one million
found = false
for i = 1 to 1e6
j = sqrt(nonSquare(i))
if j = int(j) then
found = true
print i, " square numbers found"
exit for
end if
next i
if not found then print "No squares found"
end
function nonSquare (n)
return n + int(0.5 + sqrt(n))
end function
BBC BASIC
FOR N% = 1 TO 22
S% = N% + SQR(N%) + 0.5
PRINT S%
NEXT
PRINT '"Checking...."
FOR N% = 1 TO 999999
S% = N% + SQR(N%) + 0.5
R% = SQR(S%)
IF S%/R% = R% STOP
NEXT
PRINT "No squares occur for n < 1000000"
- Output:
2
3
5
6
7
8
10
11
12
13
14
15
17
18
19
20
21
22
23
24
26
27
Checking....
No squares occur for n < 1000000
Bc
Since BC is an arbitrary precision calculator, there are no issues in sqrt (it is enough to increase the scale variable upto the desired precision), nor there are limits (but time) to how many non-squares we can compute.
#! /usr/bin/bc
scale = 20
define ceil(x) {
auto intx
intx=int(x)
if (intx<x) intx+=1
return intx
}
define floor(x) {
return -ceil(-x)
}
define int(x) {
auto old_scale, ret
old_scale=scale
scale=0
ret=x/1
scale=old_scale
return ret
}
define round(x) {
if (x<0) x-=.5 else x+=.5
return int(x)
}
define nonsqr(n) {
return n + round(sqrt(n))
}
for(i=1; i < 23; i++) {
print nonsqr(i), "\n"
}
for(i=1; i < 1000000; i++) {
j = sqrt(nonsqr(i))
if ( j == floor(j) ) {
print i, " square in the seq\n"
}
}
quit
The functions int, round, floor, ceil are taken from here (int is slightly modified) (Here he states the license is GPL).
BQN
NonSquare ← +⟜(⌊0.5+√)
IsSquare ← =⟜⌊√
NonSquare 1+↕22
⟨ 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 ⟩
+´ IsSquare NonSquare 1+↕1e6
0
Burlesque
1 22r@{?s0.5?+av?+}[mC
#include <math.h>
#include <stdio.h>
#include <assert.h>
int nonsqr(int n) {
return n + (int)(0.5 + sqrt(n));
/* return n + (int)round(sqrt(n)); in C99 */
}
int main() {
int i;
/* first 22 values (as a list) has no squares: */
for (i = 1; i < 23; i++)
printf("%d ", nonsqr(i));
printf("\n");
/* The following check shows no squares up to one million: */
for (i = 1; i < 1000000; i++) {
double j = sqrt(nonsqr(i));
assert(j != floor(j));
}
return 0;
}
C#
using System;
using System.Diagnostics;
namespace sons
{
class Program
{
static void Main(string[] args)
{
for (int i = 1; i < 23; i++)
Console.WriteLine(nonsqr(i));
for (int i = 1; i < 1000000; i++)
{
double j = Math.Sqrt(nonsqr(i));
Debug.Assert(j != Math.Floor(j),"Square");
}
}
static int nonsqr(int i)
{
return (int)(i + Math.Floor(0.5 + Math.Sqrt(i)));
}
}
}
C++
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <boost/bind.hpp>
#include <iterator>
double nextNumber( double number ) {
return number + floor( 0.5 + sqrt( number ) ) ;
}
int main( ) {
std::vector<double> non_squares ;
typedef std::vector<double>::iterator SVI ;
non_squares.reserve( 1000000 ) ;
//create a vector with a million sequence numbers
for ( double i = 1.0 ; i < 100001.0 ; i += 1 )
non_squares.push_back( nextNumber( i ) ) ;
//copy the first numbers to standard out
std::copy( non_squares.begin( ) , non_squares.begin( ) + 22 ,
std::ostream_iterator<double>(std::cout, " " ) ) ;
std::cout << '\n' ;
//find if floor of square root equals square root( i. e. it's a square number )
SVI found = std::find_if ( non_squares.begin( ) , non_squares.end( ) ,
boost::bind( &floor, boost::bind( &sqrt, _1 ) ) == boost::bind( &sqrt, _1 ) ) ;
if ( found != non_squares.end( ) ) {
std::cout << "Found a square number in the sequence!\n" ;
std::cout << "It is " << *found << " !\n" ;
}
else {
std::cout << "Up to 1000000, found no square number in the sequence!\n" ;
}
return 0 ;
}
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 Up to 1000000, found no square number in the sequence!
Alternatively, without using an external library
#include <cmath>
#include <cstdint>
#include <iostream>
uint32_t non_square(const uint32_t& n) {
return n + static_cast<uint32_t>(0.5 + sqrt(n));
}
int main() {
std::cout << "The first 22 non-square numbers:" << std::endl;
for ( uint32_t i = 1; i <= 22; ++i ) {
std::cout << non_square(i) << " ";
}
std::cout << std::endl << std::endl;
uint32_t count = 0;
for ( uint32_t i = 1; i < 1'000'000; ++i ) {
double square_root = sqrt(non_square(i));
if ( square_root == floor(square_root) ) {
count++;
}
}
std::cout << "Number of squares less than 1'000'000 produced by the formula: " << count << std::endl;
}
- Output:
The first 22 non-square numbers: 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 Number of squares less than 1'000'000 produced by the formula: 0
Chipmunk Basic
10 rem Sequence of non-squares
20 cls
30 ' Display first 22 values
40 for i = 1 to 22
50 print nonsqr(i) " ";
60 next i
70 print
80 ' Check for squares up to one million
90 found = 0
100 for i = 1 to 1000000
110 j = sqr(nonsqr(i))
120 if j = int(j) then
130 found = 1
140 print "Found square: " i
150 exit for
160 endif
170 next i
180 if found = 0 then print "No squares occur for n < 1000000"
190 end
200 sub nonsqr(n)
210 nonsqr = n+int(0.5+sqr(n))
220 return
Clojure
;; provides floor and sqrt, but we use Java's sqrt as it's faster
;; (Clojure's is more exact)
(use 'clojure.contrib.math)
(defn nonsqr [#^Integer n] (+ n (floor (+ 0.5 (Math/sqrt n)))))
(defn square? [#^Double n]
(let [r (floor (Math/sqrt n))]
(= (* r r) n)))
(doseq [n (range 1 23)] (printf "%s -> %s\n" n (nonsqr n)))
(defn verify [] (not-any? square? (map nonsqr (range 1 1000000))) )
CLU
non_square = proc (n: int) returns (int)
return(n + real$r2i(0.5 + real$i2r(n)**0.5))
end non_square
is_square = proc (n: int) returns (bool)
return(n = real$r2i(real$i2r(n)**0.5))
end is_square
start_up = proc()
po: stream := stream$primary_output()
for n: int in int$from_to(1, 22) do
stream$puts(po, int$unparse(non_square(n)) || " ")
end
stream$putl(po, "")
begin
for n: int in int$from_to(1, 1000000) do
if is_square(non_square(n)) then exit square(n) end
end
stream$putl(po, "No squares found up to 1000000.")
end
except when square(n: int):
stream$putl(po, "Found square " || int$unparse(non_square(n))
|| " at n = " || int$unparse(n))
end
end start_up- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares found up to 1000000.
COBOL
IDENTIFICATION DIVISION.
PROGRAM-ID. NONSQR.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 NEWTON.
03 SQR-INP PIC 9(7)V9(5).
03 SQUARE-ROOT PIC 9(7)V9(5).
03 FILLER REDEFINES SQUARE-ROOT.
05 FILLER PIC 9(7).
05 FILLER PIC 9(5).
88 SQUARE VALUE ZERO.
03 SQR-TEMP PIC 9(7)V9(5).
01 SEQUENCE-VARS.
03 N PIC 9(7).
03 SEQ PIC 9(7).
01 SMALL-FMT.
03 N-O PIC Z9.
03 FILLER PIC XX VALUE ": ".
03 SEQ-O PIC Z9.
PROCEDURE DIVISION.
BEGIN.
DISPLAY "Sequence of non-squares from 1 to 22:"
PERFORM SMALL-NUMS VARYING N FROM 1 BY 1
UNTIL N IS GREATER THAN 22.
DISPLAY SPACES.
DISPLAY "Checking items up to 1 million..."
PERFORM CHECK-NONSQUARE VARYING N FROM 1 BY 1
UNTIL SQUARE OR N IS GREATER THAN 1000000.
IF SQUARE, DISPLAY "Square found at N = " N,
ELSE, DISPLAY "No squares found up to 1 million.".
STOP RUN.
SMALL-NUMS.
PERFORM NONSQUARE.
MOVE N TO N-O.
MOVE SEQ TO SEQ-O.
DISPLAY SMALL-FMT.
CHECK-NONSQUARE.
PERFORM NONSQUARE.
MOVE SEQ TO SQR-INP.
PERFORM SQRT.
NONSQUARE.
MOVE N TO SQR-INP.
PERFORM SQRT.
ADD 0.5, SQUARE-ROOT GIVING SEQ.
ADD N TO SEQ.
SQRT.
MOVE SQR-INP TO SQUARE-ROOT.
COMPUTE SQR-TEMP =
(SQUARE-ROOT + SQR-INP / SQUARE-ROOT) / 2.
PERFORM SQRT-LOOP UNTIL SQUARE-ROOT IS EQUAL TO SQR-TEMP.
SQRT-LOOP.
MOVE SQR-TEMP TO SQUARE-ROOT.
COMPUTE SQR-TEMP =
(SQUARE-ROOT + SQR-INP / SQUARE-ROOT) / 2.
- Output:
1: 2 2: 3 3: 5 4: 6 5: 7 6: 8 7: 10 8: 11 9: 12 10: 13 11: 14 12: 15 13: 17 14: 18 15: 19 16: 20 17: 21 18: 22 19: 23 20: 24 21: 26 22: 27 Checking items up to 1 million... No squares found up to 1 million.
CoffeeScript
non_square = (n) -> n + Math.floor(1/2 + Math.sqrt(n))
is_square = (n) ->
r = Math.floor(Math.sqrt(n))
r * r is n
do ->
first_22_non_squares = (non_square i for i in [1..22])
console.log first_22_non_squares
# test is_square has no false negatives:
for i in [1..10000]
throw Error("is_square broken") unless is_square i*i
# test non_square is valid for first million values of n
for i in [1..1000000]
throw Error("non_square broken") if is_square non_square(i)
console.log "success"
- Output:
> coffee foo.coffee [ 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27 ] success
Common Lisp
(defun non-square-sequence ()
(flet ((non-square (n)
"Compute the N-th number of the non-square sequence"
(+ n (floor (+ 1/2 (sqrt n)))))
(squarep (n)
"Tests, whether N is a square"
(let ((r (floor (sqrt n))))
(= (* r r) n))))
(loop
:for n :upfrom 1 :to 22
:do (format t "~2D -> ~D~%" n (non-square n)))
(loop
:for n :upfrom 1 :to 1000000
:when (squarep (non-square n))
:do (format t "Found a square: ~D -> ~D~%"
n (non-square n)))))
D
import std.stdio, std.math, std.algorithm, std.range;
int nonSquare(in int n) pure nothrow @safe @nogc {
return n + cast(int)(0.5 + real(n).sqrt);
}
void main() {
iota(1, 23).map!nonSquare.writeln;
foreach (immutable i; 1 .. 1_000_000) {
immutable ns = i.nonSquare;
assert(ns != (cast(int)real(ns).sqrt) ^^ 2);
}
}
- Output:
[2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27]
Delphi
Small variation of C#
program Sequence_of_non_squares;
uses
System.SysUtils, System.Math;
function nonsqr(i: Integer): Integer;
begin
Result := Trunc(i + Floor(0.5 + Sqrt(i)));
end;
var
i: Integer;
j: Double;
begin
for i := 1 to 22 do
write(nonsqr(i), ' ');
Writeln;
for i := 1 to 999999 do
begin
j := Sqrt(nonsqr(i));
if (j = Floor(j)) then
Writeln(i, 'Is Square');
end;
end.
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
EasyLang
func nonSqu n .
return n + floor (0.5 + sqrt n)
.
for i = 1 to 22
print nonSqu i
.
for i = 1 to 1e6
j = sqrt nonSqu i
if j = floor j
found = 1
.
.
if found = 0
print "No squares found"
.EchoLisp
(lib 'sequences)
(define (a n) (+ n (floor (+ 0.5 (sqrt n)))))
(define A000037 (iterator/n a 1))
(take A000037 22)
→ (2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27)
(filter square? (take A000037 1000000))
→ null
Eiffel
class
APPLICATION
create
make
feature
make
do
sequence_of_non_squares (22)
io.new_line
sequence_of_non_squares (1000000)
end
sequence_of_non_squares (n: INTEGER)
-- Sequence of non-squares up to the n'th member.
require
n_positive: n >= 1
local
non_sq, part: REAL_64
math: DOUBLE_MATH
square: BOOLEAN
do
create math
across
1 |..| (n) as c
loop
part := (0.5 + math.sqrt (c.item.to_double))
non_sq := c.item + part.floor
io.put_string (non_sq.out + "%N")
if math.sqrt (non_sq) - math.sqrt (non_sq).floor = 0 then
square := True
end
end
if square = True then
io.put_string ("There are squares for n equal to " + n.out + ".")
else
io.put_string ("There are no squares for n equal to " + n.out + ".")
end
end
end
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 There are no squares for n equal to 22. 2 3 5 6 ... 1000999 1001000 There are no squares for n equal to 1000000.
Elixir
f = fn n -> n + trunc(0.5 + :math.sqrt(n)) end
IO.inspect for n <- 1..22, do: f.(n)
n = 1_000_000
non_squares = for i <- 1..n, do: f.(i)
m = :math.sqrt(f.(n)) |> Float.ceil |> trunc
squares = for i <- 1..m, do: i*i
case Enum.find_value(squares, fn i -> i in non_squares end) do
nil -> IO.puts "No squares found below #{n}"
val -> IO.puts "Error: number is a square: #{val}"
end
- Output:
[2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27] No squares found below 1000000
Erlang
% Implemented by Arjun Sunel
-module(non_squares).
-export([main/0]).
main() ->
lists:foreach(fun(X) -> io:format("~p~n",[non_square(X)] ) end, lists:seq(1,22)), % First 22 non-squares.
lists:foreach(fun(X) -> io:format("~p~n",[non_square(X)] ) end, lists:seq(1,1000000)). % First 1 million non-squares.
non_square(N) ->
N+trunc(1/2+ math:sqrt(N)).
Euphoria
This is based on the BASIC and Go examples.
function nonsqr( atom n)
return n + floor( 0.5 + sqrt( n ) )
end function
puts( 1, " n r(n)\n" )
puts( 1, "--- ---\n" )
for i = 1 to 22 do
printf( 1, "%3d %3d\n", { i, nonsqr(i) } )
end for
atom j
atom found
found = 0
for i = 1 to 1000000 do
j = sqrt(nonsqr(i))
if integer(j) then
found = 1
printf( 1, "Found square: %d\n", i )
exit
end if
end for
if found = 0 then
puts( 1, "No squares found\n" )
end ifF#
open System
let SequenceOfNonSquares =
let nonsqr n = n+(int(0.5+Math.Sqrt(float (n))))
let isqrt n = int(Math.Sqrt(float(n)))
let IsSquare n = n = (isqrt n)*(isqrt n)
{1 .. 999999}
|> Seq.map(fun f -> (f, nonsqr f))
|> Seq.filter(fun f -> IsSquare(snd f))
;;
Executing the code gives:
> SequenceOfNonSquares;;
val it : seq<int * int> = seq []
Factor
USING: kernel math math.functions math.ranges prettyprint
sequences ;
: non-sq ( n -- m ) dup sqrt 1/2 + floor + >integer ;
: print-first22 ( -- ) 22 [1,b] [ non-sq ] map . ;
: check-for-sq ( -- ) 1,000,000 [1,b)
[ non-sq sqrt dup floor = [ "Square found." throw ] when ]
each ;
print-first22 check-for-sq
- Output:
{ 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 }
Fantom
class Main
{
static Float fn (Int n)
{
n + (0.5f + (n * 1.0f).sqrt).floor
}
static Bool isSquare (Float n)
{
n.sqrt.floor == n.sqrt
}
public static Void main ()
{
(1..22).each |n|
{
echo ("$n is ${fn(n)}")
}
echo ((1..1000000).toList.any |n| { isSquare (fn(n)) } )
}
}Forth
: u>f 0 d>f ;
: f>u f>d drop ;
: fn ( n -- n ) dup u>f fsqrt fround f>u + ;
: test ( n -- ) 1 do i fn . loop ;
23 test \ 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 ok
: square? ( n -- ? ) u>f fsqrt fdup fround f- f0= ;
: test ( n -- ) 1 do i fn square? if cr i . ." fn was square" then loop ;
1000000 test \ ok
Fortran
PROGRAM NONSQUARES
IMPLICIT NONE
INTEGER :: m, n, nonsqr
DO n = 1, 22
nonsqr = n + FLOOR(0.5 + SQRT(REAL(n))) ! or could use NINT(SQRT(REAL(n)))
WRITE(*,*) nonsqr
END DO
DO n = 1, 1000000
nonsqr = n + FLOOR(0.5 + SQRT(REAL(n)))
m = INT(SQRT(REAL(nonsqr)))
IF (m*m == nonsqr) THEN
WRITE(*,*) "Square found, n=", n
END IF
END DO
END PROGRAM NONSQUARES
FreeBASIC
' FB 1.05.0 Win64
Function nonSquare (n As UInteger) As UInteger
Return CUInt(n + Int(0.5 + Sqr(n)))
End Function
Function isSquare (n As UInteger) As Boolean
Dim As UInteger r = CUInt(Sqr(n))
Return n = r * r
End Function
Print "The first 22 numbers generated by the sequence are :"
For i As Integer = 1 To 22
Print nonSquare(i); " ";
Next
Print : Print
' Test numbers generated for n less than a million to see if they're squares
For i As UInteger = 1 To 999999
If isSquare(nonSquare(i)) Then
Print "The number generated by the sequence for n ="; i; " is square!"
Goto finish
End If
Next
Print "None of the numbers generated by the sequence for n < 1000000 are square"
finish:
Print
Print "Press any key to quit"
Sleep
- Output:
The first 22 numbers generated by the sequence are : 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 None of the numbers generated by the sequence for n < 1000000 are square
GAP
# Here we use generators : the given formula doesn't need one, but the alternate
# non-squares function is better done with a generator.
# The formula is implemented with exact floor(sqrt(n)), so we use
# a trick: multiply by 100 to get the first decimal digit of the
# square root of n, then add 5 (that's 1/2 multiplied by 10).
# Then just divide by 10 to get floor(1/2 + sqrt(n)) exactly.
# It looks weird, but unlike floating point, it will do the job
# for any n.
NonSquaresGen := function()
local ns, n;
n := 0;
ns := function()
n := n + 1;
return n + QuoInt(5 + RootInt(100*n), 10);
end;
return ns;
end;
NonSquaresAlt := function()
local ns, n, q, k;
n := 1;
q := 4;
k := 3;
ns := function()
n := n + 1;
if n = q then
n := n + 1;
k := k + 2;
q := q + k;
fi;
return n;
end;
return ns;
end;
gen := NonSquaresGen();
List([1 .. 22] i -> gen());
# [ 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27 ]
a := NonSquaresGen();
b := NonSquaresAlt();
ForAll([1 .. 1000000], i -> a() = b());
# true
Go
I assume it's obvious that the function monotonically increases, thus it's enough to just watch for the next possible square. If a square is found, the panic will cause an ugly stack trace.
package main
import (
"fmt"
"math"
)
func remarkable(n int) int {
return n + int(.5+math.Sqrt(float64(n)))
}
func main() {
// task 1
fmt.Println(" n r(n)")
fmt.Println("--- ---")
for n := 1; n <= 22; n++ {
fmt.Printf("%3d %3d\n", n, remarkable(n))
}
// task 2
const limit = 1e6
fmt.Println("\nChecking for squares for n <", limit)
next := 2
nextSq := 4
for n := 1; n < limit; n++ {
r := remarkable(n)
switch {
case r == nextSq:
panic(n)
case r > nextSq:
fmt.Println(nextSq, "didn't occur")
next++
nextSq = next * next
}
}
fmt.Println("No squares occur for n <", limit)
}
- Output:
n r(n) --- --- 1 2 2 3 3 5 4 6 5 7 6 8 7 10 8 11 9 12 10 13 11 14 12 15 13 17 14 18 15 19 16 20 17 21 18 22 19 23 20 24 21 26 22 27 Checking for squares for n < 1e+06 4 didn't occur 9 didn't occur 16 didn't occur ... 996004 didn't occur 998001 didn't occur 1000000 didn't occur No squares occur for n < 1e+06
Groovy
Solution:
def nonSquare = { long n -> n + ((1/2 + n**0.5) as long) }
Test Program:
(1..22).each { println nonSquare(it) }
(1..1000000).each { assert ((nonSquare(it)**0.5 as long)**2) != nonSquare(it) }
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Haskell
nonsqr :: Integral a => a -> a
nonsqr n = n + round (sqrt (fromIntegral n))
> map nonsqr [1..22] [2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27] > any (\j -> j == fromIntegral (floor j)) $ map (sqrt . fromIntegral . nonsqr) [1..1000000] False
Or, in a point-free variation, defining a 'main' for the compiler (rather than interpreter)
import Control.Monad (join)
----------------------- NON SQUARES ----------------------
notSquare :: Int -> Bool
notSquare = (/=) <*> (join (*) . floor . root)
nonSqr :: Int -> Int
nonSqr = (+) <*> (round . root)
root :: Int -> Float
root = sqrt . fromIntegral
-------------------------- TESTS -------------------------
main :: IO ()
main =
mapM_
putStrLn
[ "First 22 members of the series:",
unwords $ show . nonSqr <$> [1 .. 22],
"",
"All first 10E6 members non square:",
(show . and) $
notSquare . nonSqr <$> [1 .. 1000000]
]
- Output:
First 22 members of the series: 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 All first 10E6 members non square: True
HicEst
REAL :: n=22, nonSqr(n)
nonSqr = $ + FLOOR(0.5 + $^0.5)
WRITE() nonSqr
squares_found = 0
DO i = 1, 1E6
non2 = i + FLOOR(0.5 + i^0.5)
root = FLOOR( non2^0.5 )
squares_found = squares_found + (non2 == root*root)
ENDDO
WRITE(Name) squares_found
END2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 squares_found=0;
Icon and Unicon
IDL
n = lindgen(1000000)+1 ; Take a million numbers
f = n+floor(.5+sqrt(n)) ; Apply formula
print,f[0:21] ; Output first 22
print,where(sqrt(f) eq fix(sqrt(f))) ; Test for squares
- Output:
2 3 5 6 7 8 10 11 12
13 14 15 17 18 19 20 21 22
23 24 26 27
-1
J
rf=: + 0.5 <.@+ %: NB. Remarkable formula
rf 1+i.22 NB. Results from 1 to 22
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
+/ (rf e. *:) 1+i.1e6 NB. Number of square RFs <= 1e6
0
Java
public class SeqNonSquares {
public static int nonsqr(int n) {
return n + (int)Math.round(Math.sqrt(n));
}
public static void main(String[] args) {
// first 22 values (as a list) has no squares:
for (int i = 1; i < 23; i++)
System.out.print(nonsqr(i) + " ");
System.out.println();
// The following check shows no squares up to one million:
for (int i = 1; i < 1000000; i++) {
double j = Math.sqrt(nonsqr(i));
assert j != Math.floor(j);
}
}
}
JavaScript
ES5
Iterative
var a = [];
for (var i = 1; i < 23; i++) a[i] = i + Math.floor(1/2 + Math.sqrt(i));
console.log(a);
for (i = 1; i < 1000000; i++) if (Number.isInteger(i + Math.floor(1/2 + Math.sqrt(i))) === false) {
console.log("The ",i,"th element of the sequence is a square");
}
ES6
By functional composition
(() => {
'use strict';
// ------------------ OEIS A000037 -------------------
// nonSquare :: Int -> Int
const nonSquare = n =>
// Nth term in the OEIS A000037 series.
n + Math.floor(1 / 2 + Math.sqrt(n));
// isPerfectSquare :: Int -> Bool
const isPerfectSquare = n => {
const root = Math.sqrt(n);
return root === Math.floor(root);
};
// ---------------------- TEST -----------------------
const main = () =>
// First 22 terms, and test of first million.
[
Tuple('First 22 terms:')(
take(22)(
fmapGen(nonSquare)(
enumFrom(1)
)
)
),
Tuple(
'Any perfect squares in 1st 1E6 terms ?'
)(
Array.from({
length: 1E6
})
.map(nonSquare)
.some(isPerfectSquare)
)
]
.map(kv => `${fst(kv)} -> ${snd(kv)}`)
.join('\n\n');
// --------------------- GENERAL ---------------------
// Tuple (,) :: a -> b -> (a, b)
const Tuple = a =>
b => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// enumFrom :: Enum a => a -> [a]
function* enumFrom(x) {
// A non-finite succession of enumerable
// values, starting with the value x.
let v = x;
while (true) {
yield v;
v = 1 + v;
}
}
// fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
const fmapGen = f =>
function* (gen) {
let v = take(1)(gen);
while (0 < v.length) {
yield(f(v[0]));
v = take(1)(gen);
}
};
// fst :: (a, b) -> a
const fst = tpl =>
// First member of a pair.
tpl[0];
// snd :: (a, b) -> b
const snd = tpl =>
// Second member of a pair.
tpl[1];
// 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)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
return main()
})();
- Output:
First 22 terms: -> 2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27 Any perfect squares in 1st 1E6 terms ? -> false
jq
def A000037: . + (0.5 + sqrt | floor);
def is_square: sqrt | . == floor;
"For n up to and including 22:",
(range(1;23) | A000037),
"Check for squares for n up to 1e6:",
(range(1;1e6+1) | A000037 | select( is_square ))- Output:
$ jq -n -r -f sequence_of_non-squares.jq
For n up to and including 22:
2
3
5
6
7
8
10
11
12
13
14
15
17
18
19
20
21
22
23
24
26
27
Check for squares for n up to 1e6:
$
Julia
nonsquare(n::Real) = n + floor(typeof(n), 0.5 + sqrt(n))
@show nonsquare.(1:1_000_000) ∩ collect(1:1000) .^ 2
- Output:
nonsquare.(1:1000000) ∩ collect(1:1000) .^ 2 = Int64[]
So the set of squares of integers between 1 and 1000 and the first 1000000 terms of the given sequence is empty. Note that the given sequence is increasing and that its last term has a square root slightly less than 1000.5.
K
nonsquare:{x+_.5+%x}
nonsquare[1_!23]
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
issquare:{(%x)=_%x}
+/issquare[nonsquare[1_!1000001]] / Number of squares in first million results
- Output:
0
Kotlin
// version 1.1
fun f(n: Int) = n + Math.floor(0.5 + Math.sqrt(n.toDouble())).toInt()
fun main(args: Array<String>) {
println(" n f")
val squares = mutableListOf<Int>()
for (n in 1 until 1000000) {
val v1 = f(n)
val v2 = Math.sqrt(v1.toDouble()).toInt()
if (v1 == v2 * v2) squares.add(n)
if (n < 23) println("${"%2d".format(n)} : $v1")
}
println()
if (squares.size == 0) println("There are no squares for n less than one million")
else println("Squares are generated for the following values of n: $squares")
}
- Output:
n f 1 : 2 2 : 3 3 : 5 4 : 6 5 : 7 6 : 8 7 : 10 8 : 11 9 : 12 10 : 13 11 : 14 12 : 15 13 : 17 14 : 18 15 : 19 16 : 20 17 : 21 18 : 22 19 : 23 20 : 24 21 : 26 22 : 27 There are no squares for n less than one million
Lambdatalk
{def nosquare {lambda {:n} {+ :n {floor {+ 0.5 {sqrt :n}}}}}}
-> nosquare
{def issquare {lambda {:n} {= {sqrt :n} {round {sqrt :n}}}}}
-> issquare
{S.map nosquare {S.serie 1 22}}
-> 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
{S.replace false by in
{S.map issquare _
{S.map nosquare
{S.serie 1 1000000}}}}
-> true
Liberty BASIC
for i = 1 to 22
print nonsqr( i); " ";
next i
print
found = 0
for i = 1 to 1000000
j = ( nonsqr( i))^0.5
if j = int( j) then
found = 1
print "Found square: "; i
exit for
end if
next i
if found =0 then print "No squares found"
end
function nonsqr( n)
nonsqr = n +int( 0.5 +n^0.5)
end function2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares found
Logo
repeat 22 [print sum # round sqrt #]Lua
function nonSquare (n)
return n + math.floor(1/2 + math.sqrt(n))
end
for n = 1, 22 do
io.write(nonSquare(n) .. " ")
end
print()
local sr
for n = 1, 10^6 do
sr = math.sqrt(nonSquare(n))
if sr == math.floor(sr) then
print("Result for n = " .. n .. " is square!")
os.exit()
end
end
print("No squares found")
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares found
MAD
NORMAL MODE IS INTEGER
BOOLEAN FOUND
FOUND = 0B
R SEQUENCE OF NON-SQUARES FORMULA
R FLOOR IS AUTOMATIC DUE TO INTEGER MATH
INTERNAL FUNCTION NONSQR.(N) = N+(.5+SQRT.(N))
R PRINT VALUES FOR 1..N..22
THROUGH SHOW, FOR N=1, 1, N.G.22
SHOW PRINT FORMAT OUTFMT,N,NONSQR.(N)
VECTOR VALUES OUTFMT = $I2,2H: ,I2*$
R CHECK FOR NO SQUARES UP TO ONE MILLION
THROUGH CHECK, FOR N=1, 1, N.GE.1000000
X=NONSQR.(N)
Y=SQRT.(X)
WHENEVER Y*Y.E.X
PRINT FORMAT FINDSQ,N,X
FOUND = 1B
CHECK END OF CONDITIONAL
WHENEVER .NOT. FOUND, PRINT FORMAT NOSQ
VECTOR VALUES FINDSQ = $5HELEM ,I5,2H, ,I5,11H, IS SQUARE*$
VECTOR VALUES NOSQ = $16HNO SQUARES FOUND*$
END OF PROGRAM- Output:
1: 2 2: 3 3: 5 4: 6 5: 7 6: 8 7: 10 8: 11 9: 12 10: 13 11: 14 12: 15 13: 17 14: 18 15: 19 16: 20 17: 21 18: 22 19: 23 20: 24 21: 26 22: 27 NO SQUARES FOUND
Maple
with(NumberTheory):
nonSquareSequence := proc(n::integer)
return n + floor(1 / 2 + sqrt(n));
end proc:
seq(nonSquareSequence(i), i = 1..22);
for number from 1 to 10^6 while not issqr(nonSquareSequence(number)) do end;
number;- Output:
2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26,
271000001
Mathematica /Wolfram Language
nonsq = (# + Floor[0.5 + Sqrt[#]]) &;
nonsq@Range[22]
If[! Or @@ (IntegerQ /@ Sqrt /@ nonsq@Range[10^6]),
Print["No squares for n <= ", 10^6]
]
- Output:
{2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27}
No squares for n <= 1000000
MATLAB
function nonSquares(i)
for n = (1:i)
generatedNumber = n + floor(1/2 + sqrt(n));
if mod(sqrt(generatedNumber),1)==0 %Check to see if the sqrt of the generated number is an integer
fprintf('\n%d generates a square number: %d\n', [n,generatedNumber]);
return
else %If it isn't then the generated number is a square number
if n<=22
fprintf('%d ',generatedNumber);
end
end
end
fprintf('\nNo square numbers were generated for n <= %d\n',i);
end
Solution:
>> nonSquares(1000000)
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
No square numbers were generated for n <= 1000000
No loops
sum(ismember((1:1:sqrt(1e6-1)).^2,(1:1e6-1) + floor(1/2 + sqrt((1:1e6-1)))))
Maxima
nonsquare(n) := n + quotient(isqrt(100 * n) + 5, 10);
makelist(nonsquare(n), n, 1, 20);
[2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24]
not_square(n) := isqrt(n)^2 # n$
m: 10^6$
u: makelist(i, i, 1, m)$
is(sublist(u, not_square) = sublist(map(nonsquare, u), lambda([x], x <= m)));
true
min
(dup sqrt 0.5 + int +) :non-sq
(sqrt dup floor - 0 ==) :sq?
(:n =q 1 'dup q concat 'succ concat n times pop) :upto
(non-sq print! " " print!) 22 upto newline
"Squares for n below one million:" puts!
(non-sq 'sq? 'puts when pop) 999999 upto- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 Squares for n below one million:
Miranda
main :: [sys_message]
main = [Stdout (lay [first22, hassquare])]
first22 :: [char]
first22 = show (take 22 nonsqrseq)
hassquare :: [char]
hassquare = "Square found", if or [issquare n | n<-take 1000000 nonsqrseq]
= "No square found", otherwise
issquare :: num->bool
issquare n = n == (entier (sqrt n))^2
nonsqrseq :: [num]
nonsqrseq = map nonsqr [1..]
nonsqr :: num->num
nonsqr n = n + entier (0.5 + sqrt n)- Output:
[2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27] No square found
МК-61/52
1 П4 ИП4 0 , 5 ИП4 КвКор + [x]
+ С/П КИП4 БП 02
MMIX
LOC Data_Segment
GREG @
buf OCTA 0,0
GREG @
NL BYTE #a,0
errh BYTE "Sorry, number ",0
errt BYTE "is a quare.",0
prtOk BYTE "No squares found below 1000000.",0
i IS $1 % loop var.
x IS $2 % computations
y IS $3 % ..
z IS $4 % ..
t IS $5 % temp
Ja IS $127 % return address
LOC #100 % locate program
GREG @
// print integer of max. 7 digits to StdOut
// primarily used to show the first 22 non squares
// in advance the end of the buffer is filled with ' 0 '
// reg x contains int to be printed
bp IS $71
0H GREG #0000000000203020
prtInt STO 0B,buf % initialize buffer
LDA bp,buf+7 % points after LSD
% REPEAT
1H SUB bp,bp,1 % move buffer pointer
DIV x,x,10 % divmod (x,10)
GET t,rR % get remainder
INCL t,'0' % make char digit
STB t,bp % store digit
PBNZ x,1B % UNTIL no more digits
LDA $255,bp
TRAP 0,Fputs,StdOut % print integer
GO Ja,Ja,0 % 'return'
// function calculates non square
// x = RF ( i )
RF FLOT x,i % convert i to float
FSQRT x,0,x % x = floor ( 0.5 + sqrt i )
FIX x,x % convert float to int
ADD x,x,i % x = i + floor ( 0.5 + sqrt i )
GO Ja,Ja,0 % 'return'
% main (argc, argv) {
// generate the first 22 non squares
Main SET i,1 % for ( i=1; i<=22; i++){
1H GO Ja,RF % x = RF (i)
GO Ja,prtInt % print non square
INCL i,1 % i++
CMP t,i,22 % i<=22 ?
PBNP t,1B % }
LDA $255,NL
TRAP 0,Fputs,StdOut
// check if RF (i) is a square for 0 < i < 1000000
SET i,1000
MUL i,i,i
SUB i,i,1 % for ( i = 999999; i>0; i--)
3H GO Ja,RF % x = RF ( i )
// square test
FLOT y,x % convert int x to float
FSQRT z,3,y % z = floor ( sqrt ( int (x) ) )
FIX z,z % z = cint z
MUL z,z,z % z = z^2
CMP t,x,z % x != (int sqrt x)^2 ?
PBNZ t,2F % if yes then continue
// it should not happen, but if a square is found
LDA $255,errh % else print err-message
TRAP 0,Fputs,StdOut
GO Ja,prtInt % show trespasser
LDA $255,errt
TRAP 0,Fputs,StdOut
LDA $255,NL
TRAP 0,Fputs,StdOut
TRAP 0,Halt,0
2H SUB i,i,1 % i--
PBNZ i,3B % i>0? }
LDA $255,prtOk %
TRAP 0,Fputs,StdOut
LDA $255,NL
TRAP 0,Fputs,StdOut
TRAP 0,Halt,0 % }
- Output:
~/MIX/MMIX/Rosetta> mmix SoNS 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares found below 1000000.
Modula-3
MODULE NonSquare EXPORTS Main;
IMPORT IO, Fmt, Math;
VAR i: INTEGER;
PROCEDURE NonSquare(n: INTEGER): INTEGER =
BEGIN
RETURN n + FLOOR(0.5D0 + Math.sqrt(FLOAT(n, LONGREAL)));
END NonSquare;
BEGIN
FOR n := 1 TO 22 DO
IO.Put(Fmt.Int(NonSquare(n)) & " ");
END;
IO.Put("\n");
FOR n := 1 TO 1000000 DO
i := NonSquare(n);
IF i = FLOOR(Math.sqrt(FLOAT(i, LONGREAL))) THEN
IO.Put("Found square: " & Fmt.Int(n) & "\n");
END;
END;
END NonSquare.
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Nim
import math, strutils
func nosqr(n: int): seq[int] =
result = newSeq[int](n)
for i in 1..n:
result[i - 1] = i + i.float.sqrt.toInt
func issqr(n: int): bool =
sqrt(float(n)).splitDecimal().floatpart < 1e-7
echo "Sequence for n = 22:"
echo nosqr(22).join(" ")
for i in nosqr(1_000_000 - 1):
assert not issqr(i)
echo "\nNo squares were found for n less than 1_000_000."
- Output:
Sequence for n = 22: 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 No squares were found for n less than 1_000_000.
OCaml
# let nonsqr n = n + truncate (0.5 +. sqrt (float n));;
val nonsqr : int -> int = <fun>
# (* first 22 values (as a list) has no squares: *)
for i = 1 to 22 do
Printf.printf "%d " (nonsqr i)
done;
print_newline ();;
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
- : unit = ()
# (* The following check shows no squares up to one million: *)
for i = 1 to 1_000_000 do
let j = sqrt (float (nonsqr i)) in
assert (j <> floor j)
done;;
- : unit = ()
Oforth
22 seq map(#[ dup sqrt 0.5 + floor + ]) println
1000000 seq map(#[ dup sqrt 0.5 + floor + ]) conform(#[ sqrt dup floor <>]) println- Output:
[2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27] 1
Ol
(import (lib math))
(print
; sequence for 1 .. 22
(map (lambda (n)
(+ n (floor (+ 1/2 (exact (sqrt n))))))
(iota 22 1)))
; ==> (2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27)
(print
; filter out non squares
(filter
(lambda (x)
(let ((s (floor (exact (sqrt x)))))
(= (* s s) x)))
(map (lambda (n)
(+ n (floor (+ 1/2 (exact (sqrt n))))))
(iota 1000000 1))))
; ==> ()
Oz
declare
fun {NonSqr N}
N + {Float.toInt {Floor 0.5 + {Sqrt {Int.toFloat N}}}}
end
fun {SqrtInt N}
{Float.toInt {Sqrt {Int.toFloat N}}}
end
fun {IsSquare N}
{Pow {SqrtInt N} 2} == N
end
Ns = {Map {List.number 1 999999 1} NonSqr}
in
{Show {List.take Ns 22}}
{Show {Some Ns IsSquare}}PARI/GP
[vector(22,n,n + floor(1/2 + sqrt(n))), sum(n=1,1e6,issquare(n + floor(1/2 + sqrt(n))))]Pascal
Program SequenceOfNonSquares(output);
uses
Math;
var
m, n, test: longint;
begin
for n := 1 to 22 do
begin
test := n + floor(0.5 + sqrt(n));
write(test, ' ');
end;
writeln;
for n := 1 to 1000000 do
begin
test := n + floor(0.5 + sqrt(n));
m := round(sqrt(test));
if (m*m = test) then
writeln('square found for n = ', n);
end;
end.
- Output:
:> ./SequenceOfNonSquares 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
a little speedup in testing upto 1 billion. 5 secs instead of 21 secs using fpc2.6.4
program seqNonSq;
//sequence of non-squares
//n = i + floor(1/2 + sqrt(i))
function NonSquare(i: LongInt): LongInt;
Begin
NonSquare := i+trunc(sqrt(i) + 0.5);
end;
procedure First22;
var
i : integer;
begin
For i := 1 to 21 do
write(NonSquare(i):3,',');
writeln(NonSquare(22):3);
end;
procedure OutSquare(i: integer);
var
n : LongInt;
begin
n := NonSquare(i);
writeln('Square ',n,' found at ',i);
end;
procedure Test(Limit: LongWord);
var
i ,n,sq,sn : LongWord;
Begin
sn := 1;
sq := 1;
For i := 1 to Limit do
begin
n := NonSquare(i);
if n >= sq then
begin
if n > sq then
begin
sq := sq+2*sn+1; inc(sn);
end
else
OutSquare(i);
end;
end;
end;
Begin
First22;
Test(1000*1000*1000);
end.
Perl
sub nonsqr { my $n = shift; $n + int(0.5 + sqrt $n) }
print join(' ', map nonsqr($_), 1..22), "\n";
foreach my $i (1..1_000_000) {
my $root = sqrt nonsqr($i);
die "Oops, nonsqr($i) is a square!" if $root == int $root;
}
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Phix
with javascript_semantics sequence s = repeat(0,22) for n=1 to length(s) do s[n] = n + floor(1/2 + sqrt(n)) end for printf(1,"%V\n",{s}) integer nxt = 2, snxt = nxt*nxt, k for n=1 to 1000000 do k = n + floor(1/2 + sqrt(n)) if k>snxt then -- printf(1,"%d didn't occur\n",snxt) nxt += 1 snxt = nxt*nxt end if if k=snxt then puts(1,"error!!\n") end if end for puts(1,"none found ") ?{nxt,snxt}
- Output:
{2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27}
none found {1001,1002001}
Phixmonti
include ..\Utilitys.pmt
def non-sq dup sqrt 0.5 + int + enddef
22 for dup print ", " print non-sq ? endfor
1000000 for
non-sq sqrt dup int == if "Square found." ? exitfor endif
endfor- Output:
1, 2 2, 3 3, 5 4, 6 5, 7 6, 8 7, 10 8, 11 9, 12 10, 13 11, 14 12, 15 13, 17 14, 18 15, 19 16, 20 17, 21 18, 22 19, 23 20, 24 21, 26 22, 27 === Press any key to exit ===
PHP
<?php
//First Task
for($i=1;$i<=22;$i++){
echo($i + floor(1/2 + sqrt($i)) . "\n");
}
//Second Task
$found_square=False;
for($i=1;$i<=1000000;$i++){
$non_square=$i + floor(1/2 + sqrt($i));
if(sqrt($non_square)==intval(sqrt($non_square))){
$found_square=True;
}
}
echo("\n");
if($found_square){
echo("Found a square number, so the formula does not always work.");
} else {
echo("Up to 1000000, found no square number in the sequence!");
}
?>
- Output:
>php nsqrt.php 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 Up to 1000000, found no square number in the sequence! >
Picat
go =>
println([f(I) : I in 1..22]),
nl,
check(1_000_000),
nl.
% The formula
f(N) = N + floor(1/2 + sqrt(N)).
check(Limit) =>
Squares = new_map([I*I=1:I in 1..sqrt(Limit)]),
Check = [[I,T] : I in 1..Limit-1, T=f(I), Squares.has_key(T)],
println(check=Check.len).- Output:
[2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27] check = 0
PicoLisp
(de sqfun (N)
(+ N (sqrt N T)) ) # 'sqrt' rounds when called with 'T'
(for I 22
(println I (sqfun I)) )
(for I 1000000
(let (N (sqfun I) R (sqrt N))
(when (= N (* R R))
(prinl N " is square") ) ) )- Output:
1 2 2 3 3 5 4 6 5 7 6 8 7 10 8 11 9 12 10 13 11 14 12 15 13 17 14 18 15 19 16 20 17 21 18 22 19 23 20 24 21 26 22 27
PL/I
put skip edit ((n, n + floor(sqrt(n) + 0.5) do n = 1 to n))
(skip, 2 f(5));Results:
1 2
2 3
3 5
4 6
5 7
6 8
7 10
8 11
9 12
10 13
11 14
12 15
13 17
14 18
15 19
16 20
17 21
18 22
19 23
20 24
21 26
Test 1,000,000 values:
test: proc options (main);
declare n fixed (15);
do n = 1 to 1000000;
if perfect_square (n + fixed(sqrt(n) + 0.5, 15)) then
do; put skip list ('formula fails for n = ', n); stop; end;
end;
perfect_square: procedure (N) returns (bit (1) aligned);
declare N fixed (15);
declare K fixed (15);
k = sqrt(N)+0.1;
return ( k*k = N );
end perfect_square;
end test;
PostScript
/nonsquare { dup sqrt .5 add floor add } def
/issquare { dup sqrt floor dup mul eq } def
1 1 22 { nonsquare = } for
1 1 1000 {
dup nonsquare issquare {
(produced a square!) = = exit
} if pop
} for
- Output:
(lack of error message shows none below 1000 produced a square)
2.0 3.0 5.0 6.0 7.0 8.0 10.0 11.0 12.0 13.0 14.0 15.0 17.0 18.0 19.0 20.0 21.0 22.0 23.0 24.0 26.0 27.0
PowerShell
Implemented as a filter here, which can be used directly on the pipeline.
filter Get-NonSquare {
return $_ + [Math]::Floor(1/2 + [Math]::Sqrt($_))
}
Printing out the first 22 values is straightforward, then:
1..22 | Get-NonSquare
If there were any squares for n up to one million, they would be printed with the following, but there is no output:
1..1000000 `
| Get-NonSquare `
| Where-Object {
$r = [Math]::Sqrt($_)
[Math]::Truncate($r) -eq $r
}
PureBasic
OpenConsole()
For a = 1 To 22
; Integer, so no floor needed
tmp = 1 / 2 + Sqr(a)
Print(Str(a + tmp) + ", ")
Next
PrintN("")
PrintN("Starting check till one million")
For a = 1 To 1000000
value.d = a + Round((1 / 2 + Sqr(a)), #PB_Round_Down)
root = Sqr(value)
If value - root*root = 0
found + 1
If found < 20
PrintN("Found a square! " + Str(value))
ElseIf found = 20
PrintN("And more")
EndIf
EndIf
Next
If found
PrintN(Str(found) + " Squares found, see above")
Else
PrintN("No squares, all ok")
EndIf
; Wait for enter
Input()
Python
>>> from math import floor, sqrt
>>> def non_square(n):
return n + floor(1/2 + sqrt(n))
>>> # first 22 values has no squares:
>>> print(*map(non_square, range(1, 23)))
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
>>> # The following check shows no squares up to one million:
>>> def is_square(n):
return sqrt(n).is_integer()
>>> non_squares = map(non_square, range(1, 10 ** 6))
>>> next(filter(is_square, non_squares))
StopIteration Traceback (most recent call last)
<ipython-input-45-f32645fc1c0a> in <module>()
1 non_squares = map(non_square, range(1, 10 ** 6))
----> 2 next(filter(is_square, non_squares))
StopIteration:
Or, defining OEIS A000037 as a non-finite series:
'''Sequence of non-squares'''
from itertools import count, islice
from math import floor, sqrt
# A000037 :: [Int]
def A000037():
'''A non-finite series of integers.'''
return map(nonSquare, count(1))
# nonSquare :: Int -> Int
def nonSquare(n):
'''Nth term in the OEIS A000037 series.'''
return n + floor(1 / 2 + sqrt(n))
# --------------------------TEST---------------------------
# main :: IO ()
def main():
'''OEIS A000037'''
def first22():
'''First 22 terms'''
return take(22)(A000037())
def squareInFirstMillion():
'''True if any of the first 10^6 terms are perfect squares'''
return any(map(
isPerfectSquare,
take(10 ** 6)(A000037())
))
print(
fTable(main.__doc__)(
lambda f: '\n' + f.__doc__
)(lambda x: ' ' + showList(x))(
lambda f: f()
)([first22, squareInFirstMillion])
)
# -------------------------DISPLAY-------------------------
# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
return s + '\n' + '\n'.join(map(
lambda x, y: y + ':\n' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
# -------------------------GENERAL-------------------------
# isPerfectSquare :: Int -> Bool
def isPerfectSquare(n):
'''True if n is a perfect square.'''
return sqrt(n).is_integer()
# showList :: [a] -> String
def showList(xs):
'''Compact stringification of any list value.'''
return '[' + ','.join(repr(x) for x in xs) + ']' if (
isinstance(xs, list)
) else repr(xs)
# take :: Int -> [a] -> [a]
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
return lambda xs: list(islice(xs, n))
# MAIN ---
if __name__ == '__main__':
main()
- Output:
OEIS A000037
First 22 terms:
[2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27]
True if any of the first 10^6 terms are perfect squares:
False
Quackery
$ "bigrat.qky" loadfile
[ dup n->v 2 vsqrt
drop 1 2 v+ / + ] is nonsquare ( n --> n )
[ sqrt nip 0 = ] is squarenum ( n --> b )
say "Non-squares: "
22 times [ i^ 1+ nonsquare echo sp ]
cr cr
0
999999 times
[ i^ 1+ nonsquare
squarenum if 1+ ]
echo say " square numbers found"- Output:
Non-squares: 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 0 square numbers found
R
Printing the first 22 nonsquares.
nonsqr <- function(n) n + floor(1/2 + sqrt(n))
nonsqr(1:22)
[1] 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Testing the first million nonsquares.
is.square <- function(x)
{
sqrx <- sqrt(x)
err <- abs(sqrx - round(sqrx))
err < 100*.Machine$double.eps
}
any(is.square(nonsqr(1:1e6)))
[1] FALSE
Racket
#lang racket
(define (non-square n)
(+ n (exact-floor (+ 1/2 (sqrt n)))))
(map non-square (range 1 23))
(define (square? n) (integer? (sqrt n)))
(for/or ([n (in-range 1 1000001)])
(square? (non-square n)))
- Output:
'(2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27) #f
Raku
(formerly Perl 6)
sub nth-term (Int $n) { $n + round sqrt $n }
# Print the first 22 values of the sequence
say (nth-term $_ for 1 .. 22);
# Check that the first million values of the sequence are indeed non-square
for 1 .. 1_000_000 -> $i {
say "Oops, nth-term($i) is square!" if (sqrt nth-term $i) %% 1;
}
- Output:
(2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27)
Red
Red ["Sequence of non-squares"]
repeat i 999'999 [
n: i + round/floor 0.5 + sqrt i
if i < 23 [prin [to-integer n ""]]
if equal? round/floor n sqrt n [
print "Square found!"
break
]
]
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
REXX
REXX has no native support for floor or sqrt, so these subroutines (functionsa) are written in REXX and are included below.
The iSqrt is a special integer square root function, it returns the integer root (and uses no floating point).
- 7 = iSqrt(63)
- 8 = iSqrt(64)
- 8 = iSqrt(65)
/*REXX pgm displays some non─square numbers, & also displays a validation check up to 1M*/
parse arg N M . /*obtain optional arguments from the CL*/
if N=='' | N=="," then N= 22 /*Not specified? Then use the default.*/
if M=='' | M=="," then M= 1000000 /* " " " " " " */
say 'The first ' N " non─square numbers:" /*display a header of what's to come. */
say /* [↑] default for M is one million.*/
say center('index', 20) center("non─square numbers", 20)
say center('' , 20, "═") center('' , 20, "═")
do j=1 for N
say center(j, 20) center(j +floor(1/2 +sqrt(j)), 20)
end /*j*/
#= 0
do k=1 for M /*have it step through a million of 'em*/
$= k + floor( sqrt(k) + .5 ) /*use the specified formula (algorithm)*/
iRoot= iSqrt($) /*··· and also use the ISQRT function.*/
if iRoot * iRoot == $ then #= # + 1 /*have we found a mistook? (sic) */
end /*k*/
say; if #==0 then #= 'no' /*use gooder English for display below.*/
say 'Using the formula: floor[ 1/2 + sqrt(n) ], ' # " squares found up to " M'.'
/* [↑] display (possible) error count.*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
floor: parse arg floor_; return trunc( floor_ - (floor_ < 0) )
/*──────────────────────────────────────────────────────────────────────────────────────*/
iSqrt: procedure; parse arg x; #=1; r= 0; do while # <= x; #= #*4; end
do while #>1; #=#%4; _=x-r-#; r=r%2; if _<0 then iterate; x=_; r=r+#; end; return r
/*──────────────────────────────────────────────────────────────────────────────────────*/
sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); m.=9; numeric form; h=d+6
numeric digits; parse value format(x,2,1,,0) 'E0' with g 'E' _ .; g=g *.5'e'_%2
do j=0 while h>9; m.j= h; h= h % 2 + 1; end /*j*/
do k=j+5 to 0 by -1; numeric digits m.k; g= (g+x/g)*.5; end /*k*/; return g
- output:
The first 22 non─square numbers:
index non─square numbers
════════════════════ ════════════════════
1 2
2 3
3 5
4 6
5 7
6 8
7 10
8 11
9 12
10 13
11 14
12 15
13 17
14 18
15 19
16 20
17 21
18 22
19 23
20 24
21 26
22 27
Using the formula: floor[ 1/2 + sqrt(n) ], no squares found up to 1000000.
Ring
for n=1 to 22
x = n + floor(1/2 + sqrt(n))
see "" + x + " "
next
see nlRPL
≪ DUP √ 0.5 + FLOOR + ≫ ‘A0037’ STO
≪ 0 ROT ROT FOR n
IF n A0037 √ FP NOT THEN 1 + END
NEXT →STR " square(s) found" +
≫ ‘TEST’ STO
2 runs were necessary to test one million numbers without waking emulator's timedog up.
≪ 1 22 FOR n n A0037 + NEXT ≫ EVAL 1 500000 TEST 500001 1000000 TEST
- Output:
3: { 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 }
2: “0 square(s) found“
1: “0 square(s) found“
Ruby
def f(n)
n + (0.5 + Math.sqrt(n)).floor
end
(1..22).each { |n| puts "#{n} #{f(n)}" }
non_squares = (1..1_000_000).map { |n| f(n) }
squares = (1..1001).map { |n| n**2 } # Note: 1001*1001 = 1_002_001 > 1_001_000 = f(1_000_000)
(squares & non_squares).each do |n|
puts "Oops, found a square f(#{non_squares.index(n)}) = #{n}"
end
Rust
fn f(n: i64) -> i64 {
n + (0.5 + (n as f64).sqrt()) as i64
}
fn is_sqr(n: i64) -> bool {
let a = (n as f64).sqrt() as i64;
n == a * a || n == (a+1) * (a+1) || n == (a-1) * (a-1)
}
fn main() {
println!( "{:?}", (1..23).map(|n| f(n)).collect::<Vec<i64>>() );
let count = (1..1_000_000).map(|n| f(n)).filter(|&n| is_sqr(n)).count();
println!("{} unexpected squares found", count);
}
Scala
def nonsqr(n:Int)=n+math.round(math.sqrt(n)).toInt
for(n<-1 to 22) println(n + " "+ nonsqr(n))
val test=(1 to 1000000).exists{n =>
val j=math.sqrt(nonsqr(n))
j==math.floor(j)
}
println("squares up to one million="+test)
Scheme
(define non-squares
(lambda (index)
(+ index (inexact->exact (floor (+ (/ 1 2) (sqrt index)))))))
(define sequence
(lambda (function)
(lambda (start)
(lambda (stop)
(if (> start stop)
(list)
(cons (function start)
(((sequence function) (+ start 1)) stop)))))))
(define square?
(lambda (number)
((lambda (root)
(= (* root root) number))
(floor (sqrt number)))))
(define any?
(lambda (predicate?)
(lambda (list)
(and (not (null? list))
(or (predicate? (car list))
((any? predicate?) (cdr list)))))))
(display (((sequence non-squares) 1) 22))
(newline)
(display ((any? square?) (((sequence non-squares) 1) 999999)))
(newline)
- Output:
(2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27) #f
Seed7
$ include "seed7_05.s7i";
include "float.s7i";
include "math.s7i";
const func integer: nonsqr (in integer: n) is
return n + trunc(0.5 + sqrt(flt(n)));
const proc: main is func
local
var integer: i is 0;
var float: j is 0.0;
begin
# First 22 values (as a list) has no squares:
for i range 1 to 22 do
write(nonsqr(i) <& " ");
end for;
writeln;
# The following check shows no squares up to one million:
for i range 1 to 1000000 do
j := sqrt(flt(nonsqr(i)));
if j = floor(j) then
writeln("Found square for nonsqr(" <& i <& ")");
end if;
end for;
end func;SETL
program sequence_of_non_squares;
print([nonsquare n : n in [1..22]]);
if exists n in [1..1000000] | is_square nonsquare n then
print("Found square", nonsquare n, "at", n);
else
print("No squares found up to 1 million");
end if;
op is_square(n);
return (floor sqrt n)**2 = n;
end op;
op nonsquare(n);
return n + floor(0.5 + sqrt n);
end op;
end program;- Output:
[2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27] No squares found up to 1 million
Sidef
func nonsqr(n) { 0.5 + n.sqrt -> floor + n }
{|i| nonsqr(i) }.map(1..22).join(' ').say
{ |i|
if (nonsqr(i).is_sqr) {
die "Found a square in the sequence: #{i}"
}
} << 1..1e6
Smalltalk
| nonSquare isSquare squaresFound |
nonSquare := [:n |
n + (n sqrt) rounded
].
isSquare := [:n |
n = (((n sqrt) asInteger) raisedTo: 2)
].
Transcript show: 'The first few non-squares:'; cr.
1 to: 22 do: [:n |
Transcript show: (nonSquare value: n) asString; cr
].
squaresFound := 0.
1 to: 1000000 do: [:n |
(isSquare value: (nonSquare value: n)) ifTrue: [
squaresFound := squaresFound + 1
]
].
Transcript show: 'Squares found for values up to 1,000,000: ';
show: squaresFound asString; cr
SparForte
As a structured script.
#!/usr/local/bin/spar
pragma annotate( summary, "nonsquares" );
pragma annotate( description, "Show that the following remarkable formula gives the" );
pragma annotate( description, "sequence of non-square natural numbers: n +" );
pragma annotate( description, "floor(1/2 + sqrt(n)). Print out the values for n in" );
pragma annotate( description, "the range 1 to 22. Show that no squares occur for n" );
pragma annotate( description, "less than one million." );
pragma annotate( see_also, "http://rosettacode.org/wiki/Sequence_of_non-squares" );
pragma annotate( author, "Ken O. Burtch" );
pragma license( unrestricted );
pragma restriction( no_external_commands );
procedure nonsquares is
function is_non_square (n : positive) return positive is
begin
return n + positive (numerics.rounding(numerics.sqrt (long_float (n))));
end is_non_square;
i : positive;
begin
for n in 1..22 loop -- First 22 non-squares
put (strings.image (is_non_square (n)));
end loop;
new_line;
for n in 1..1_000_000 loop -- Check first million of
i := is_non_square (n);
if i = positive (numerics.rounding(numerics.sqrt (long_float (i))))**2 then
put_line ("Found a square:" & strings.image (n));
end if;
end loop;
end nonsquares;
Standard ML
- fun nonsqr n = n + round (Math.sqrt (real n));
val nonsqr = fn : int -> int
- List.tabulate (23, nonsqr);
val it = [0,2,3,5,6,7,8,10,11,12,13,14,...] : int list
- let fun loop i = if i = 1000000 then true
else let val j = Math.sqrt (real (nonsqr i)) in
Real.!= (j, Real.realFloor j) andalso
loop (i+1)
end in
loop 1
end;
val it = true : bool
Tcl
package require Tcl 8.5
set f {n {expr {$n + floor(0.5 + sqrt($n))}}}
for {set x 1} {$x <= 22} {incr x} {
puts [format "%d\t%s" $x [apply $f $x]]
}
puts "looking for a square..."
for {set x 1} {$x <= 1000000} {incr x} {
set y [apply $f $x]
set s [expr {sqrt($y)}]
if {$s == int($s)} {
error "found a square in the sequence: $x -> $y"
}
}
puts "done"
- Output:
1 2.0 2 3.0 3 5.0 4 6.0 5 7.0 6 8.0 7 10.0 8 11.0 9 12.0 10 13.0 11 14.0 12 15.0 13 17.0 14 18.0 15 19.0 16 20.0 17 21.0 18 22.0 19 23.0 20 24.0 21 26.0 22 27.0 looking for a square... done
TI-89 BASIC
Definition and 1 to 22, interactively:
■ n+floor(1/2+√(n)) → f(n)
Done
■ seq(f(n),n,1,22)
{2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27}Program testing up to one million:
test()
Prgm
Local i, ns
For i, 1, 10^6
f(i) → ns
If (floor(√(ns)))^2 = ns Then
Disp "Oops: " & string(ns)
EndIf
EndFor
Disp "Done"
EndPrgm(This program has not been run to completion.)
Transd
#lang transd
MainModule: {
nonsqr: (λ i Int()
(ret (+ i (to-Int (floor (+ 0.5 (sqrt i))))))),
_start: (lambda locals: d Double()
(for i in Range(1 23) do
(textout (nonsqr i) " "))
(for i in Range(1 1000001) do
(= d (sqrt (nonsqr i)))
(if (eq d (floor d))
(throw String("Square: " i))))
(textout "\n\nUp to 1 000 000 - no squares found.")
)
}
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 Up to 1 000 000 - no squares found.
True BASIC
FUNCTION nonSquare (n)
LET nonSquare = n + INT(0.5 + SQR(n))
END FUNCTION
! Display first 22 values
PRINT "The first 22 numbers generated by the sequence are : "
FOR i = 1 TO 22
PRINT nonSquare(i); " ";
NEXT i
PRINT
! Check FOR squares up TO one million
LET found = 0
FOR i = 1 TO 1e6
LET j = SQR(nonSquare(i))
IF j = INT(j) THEN
LET found = 1
PRINT i, " square numbers found"
EXIT FOR
END IF
NEXT i
IF found = 0 THEN PRINT "No squares found"
END
Ursala
#import nat
#import flo
nth_non_square = float; plus^/~& math..trunc+ plus/0.5+ sqrt
is_square = sqrt; ^E/~& math..trunc
#show+
examples = %neALP ^(~&,nth_non_square)*t iota23
check = (is_square*~+nth_non_square*t; ~&i&& %eLP)||-[no squares found]-! iota 1000000- Output:
< 1: 2.000000e+00, 2: 3.000000e+00, 3: 5.000000e+00, 4: 6.000000e+00, 5: 7.000000e+00, 6: 8.000000e+00, 7: 1.000000e+01, 8: 1.100000e+01, 9: 1.200000e+01, 10: 1.300000e+01, 11: 1.400000e+01, 12: 1.500000e+01, 13: 1.700000e+01, 14: 1.800000e+01, 15: 1.900000e+01, 16: 2.000000e+01, 17: 2.100000e+01, 18: 2.200000e+01, 19: 2.300000e+01, 20: 2.400000e+01, 21: 2.600000e+01, 22: 2.700000e+01> no squares found
VBA
Sub Main()
Dim i&, c&, j#, s$
Const N& = 1000000
s = "values for n in the range 1 to 22 : "
For i = 1 To 22
s = s & ns(i) & ", "
Next
For i = 1 To N
j = Sqr(ns(i))
If j = CInt(j) Then c = c + 1
Next
Debug.Print s
Debug.Print c & " squares less than " & N
End Sub
Private Function ns(l As Long) As Long
ns = l + Int(1 / 2 + Sqr(l))
End Function
- Output:
values for n in the range 1 to 22 : 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 0 squares less than 1000000
Wren
import "./fmt" for Fmt
System.print("The first 22 numbers in the sequence are:")
System.print(" n term")
for (n in 1...1e6) {
var s = n + (0.5 + n.sqrt).floor
var ss = s.sqrt.round
if (ss * ss == s) {
Fmt.print("The $r number in the sequence $d = $d x $d is a square.", n, s, ss, ss)
return
}
if (n <= 22) Fmt.print(" $2d $2d", n, s)
}
System.print("\nNo squares were found in the first 999,999 terms.")
- Output:
The first 22 numbers in the sequence are: n term 1 2 2 3 3 5 4 6 5 7 6 8 7 10 8 11 9 12 10 13 11 14 12 15 13 17 14 18 15 19 16 20 17 21 18 22 19 23 20 24 21 26 22 27 No squares were found in the first 999,999 terms.
XLISP
(defun non-square (n)
(+ n (floor (+ 0.5 (sqrt n)))))
(defun range (x y)
(if (< x y)
(cons x (range (+ x 1) y))))
(defun squarep (x)
(= x (expt (floor (sqrt x)) 2)))
(defun count-squares (x y)
(define squares 0)
(if (squarep (non-square x))
(define squares (+ squares 1)))
(if (= x y)
squares
(count-squares (+ x 1) y)))
(print (mapcar non-square (range 1 23)))
(print `(number of squares for values less than 1000000 = ,(count-squares 1 1000000)))
- Output:
(2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27) (NUMBER OF SQUARES FOR VALUES LESS THAN 1000000 = 0)
XPL0
include c:\cxpl\codes; \intrinsic 'code' declarations
func real Floor(X); \Truncate X toward - infinity
real X;
return float(fix(X-0.5));
func PerfectSq(N); \Return 'true' if N is a perfect square
int N;
return sqrt(N)*sqrt(N) = N;
int N, M, M0;
[for N:= 1 to 22 do
[IntOut(0, fix(float(N) + Floor(0.5 + sqrt(float(N))))); ChOut(0,^ )];
CrLf(0);
M0:= 1;
for N:= 1 to 999_999 do
[M:= fix(float(N) + Floor(0.5 + sqrt(float(N))));
if PerfectSq(M) then [IntOut(0, M); Crlf(0)]; \error: have square
if M#M0+1 and not PerfectSq(M0+1) then \error: not sequential
[IntOut(0, M); Crlf(0)];
M0:= M;
];
]- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Yabasic
// Display first 22 values
print "The first 22 numbers generated by the sequence are : "
for i = 1 to 22
print nonSquare(i), " ";
next i
print
// Check for squares up to one million
found = false
for i = 1 to 1e6
j = sqrt(nonSquare(i))
if j = int(j) then
found = true
print i, " square numbers found" //print "Found square: ", i
break
end if
next i
if not found print "No squares found"
end
sub nonSquare (n)
return n + int(0.5 + sqrt(n))
end sub
zkl
fcn seq(n){n + (0.5+n.toFloat().sqrt()).floor()}
[1..22].apply(seq).toString(*).println();
fcn isSquare(n){n.toFloat().sqrt().modf()[1]==0.0}
isSquare(25) //-->True
isSquare(26) //-->False
[2..0d1_000_000].filter(fcn(n){isSquare(seq(n))}).println();modf returns the integer and fractional parts of a float
- Output:
L(2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27) L()
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