Sequence 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
<lang 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’)</lang>
- 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
Ada
<lang 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;</lang>
- 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
<lang algol68>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
)</lang>
- 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
<lang algolw>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.</lang>
- 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: <lang apl> 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</lang> Show there are no squares in the first million: <lang apl> HOWMANYSQUARES←{+⌿⍵=(⌊⍵*0.5)*2}
HOWMANYSQUARES NONSQUARE 1000000
0</lang>
Arturo
<lang rebol>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!"</lang>
- 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 <lang AutoHotkey>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</lang>
AWK
<lang 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"}}' $</lang>
BASIC
<lang freebasic>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"</lang>
BBC BASIC
<lang bbcbasic> 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"</lang>
- 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.
<lang bc>#! /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</lang>
The functions int, round, floor, ceil are taken from here (int is slightly modified) (Here he states the license is GPL).
Burlesque
<lang burlesque> 1 22r@{?s0.5?+av?+}[m </lang>
C
<lang c>#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;
}</lang>
C#
<lang csharp>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))); } }
}</lang>
C++
<lang cpp>#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 ;
}</lang>
- 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!
Clojure
<lang 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))) )</lang>
CoffeeScript
<lang 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"
</lang>
- 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
<lang 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)))))</lang>
D
<lang 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); }
}</lang>
- Output:
[2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27]
EchoLisp
<lang scheme> (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
</lang>
Eiffel
<lang 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
</lang>
- 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
<lang 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</lang>
- 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
<lang 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)). </lang>
Euphoria
This is based on the BASIC and Go examples. <lang Euphoria>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 if</lang>
F#
<lang fsharp>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))
- </lang>
Executing the code gives:<lang fsharp> > SequenceOfNonSquares;; val it : seq<int * int> = seq []</lang>
Factor
<lang 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</lang>
- Output:
{ 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 }
Fantom
<lang 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)) } ) }
} </lang>
Forth
<lang 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</lang>
Fortran
<lang 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</lang>
FreeBASIC
<lang 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</lang>
- 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
<lang># 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</lang>
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. <lang go>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)
}</lang>
- 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: <lang groovy> def nonSquare = { long n -> n + ((1/2 + n**0.5) as long) }</lang>
Test Program: <lang groovy>(1..22).each { println nonSquare(it) } (1..1000000).each { assert ((nonSquare(it)**0.5 as long)**2) != nonSquare(it) }</lang>
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Haskell
<lang haskell>nonsqr :: Integral a => a -> a nonsqr n = n + round (sqrt (fromIntegral n))</lang>
> 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)
<lang haskell>import Control.Monad (join)
root :: Int -> Float root = sqrt . fromIntegral
nonSqr :: Int -> Int nonSqr = (+) <*> (round . root)
notSquare :: Int -> Bool notSquare = (/=) <*> (join (*) . floor . root)
main :: IO () main =
mapM_ putStrLn [ "First 22 members of the series:" , unwords $ (show . nonSqr) <$> [1 .. 22] , "" , "All first 10E6 members non square:" , show $ all (== True) $ (notSquare . nonSqr) <$> [1 .. 1000000] ]</lang>
- 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
<lang 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 END</lang>
2 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
<lang Icon>link numbers
procedure main()
every n := 1 to 22 do
write("nsq(",n,") := ",nsq(n))
every x := sqrt(nsq(n := 1 to 1000000)) do
if x = floor(x)^2 then write("nsq(",n,") = ",x," is a square.")
write("finished.") end
procedure nsq(n) # return non-squares return n + floor(0.5 + sqrt(n)) end</lang>
IDL
<lang 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</lang>
- 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
<lang 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</lang>
Java
<lang 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); } }
}</lang>
JavaScript
ES5
Iterative
<lang javascript>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");
}</lang>
ES6
By functional composition <lang JavaScript>(() => {
'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()
})();</lang>
- 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
<lang 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 ))</lang>
- Output:
<lang sh>$ 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: $</lang>
Julia
<lang julia>nonsquare(n::Real) = n + floor(typeof(n), 0.5 + sqrt(n)) @show nonsquare.(1:1_000_000) ∩ collect(1:1000) .^ 2</lang>
- Output:
nonsquare.(1:1000000) ∩ collect(1:1000) .^ 2 = Int64[]
So the set of squares of integers between 1 and 1000 and the first 1000000 terms of the given sequence is empty. Note that the given sequence is increasing and that its last term has a square root slightly less than 1000.5.
K
<lang k> nonsquare:{x+_.5+%x}
nonsquare[1_!23]</lang>
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
<lang k> issquare:{(%x)=_%x}
+/issquare[nonsquare[1_!1000001]] / Number of squares in first million results</lang>
- Output:
0
Kotlin
<lang scala>// version 1.1
fun f(n: Int) = n + Math.floor(0.5 + Math.sqrt(n.toDouble())).toInt()
fun main(args: Array<String>) {
println(" n f") val squares = mutableListOf<Int>() for (n in 1 until 1000000) { val v1 = f(n) val v2 = Math.sqrt(v1.toDouble()).toInt() if (v1 == v2 * v2) squares.add(n) if (n < 23) println("${"%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")
}</lang>
- 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
Liberty BASIC
<lang lb> 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 function </lang>
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
Logo
<lang logo>repeat 22 [print sum # round sqrt #]</lang>
Lua
<lang 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")</lang>
- 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
<lang 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</lang>
- 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
<lang 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; </lang>
- 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
<lang Mathematica>nonsq = (# + Floor[0.5 + Sqrt[#]]) &; nonsq@Range[22] If[! Or @@ (IntegerQ /@ Sqrt /@ nonsq@Range[10^6]),
Print["No squares for n <= ", 10^6] ]</lang>
- 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
<lang 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</lang> Solution: <lang MATLAB>>> 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</lang>
Maxima
<lang 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</lang>
min
<lang 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</lang>
- 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:
МК-61/52
<lang>1 П4 ИП4 0 , 5 ИП4 КвКор + [x] + С/П КИП4 БП 02</lang>
MMIX
<lang 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 % }</lang>
- 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
<lang modula3>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.</lang>
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Nim
<lang 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."</lang>
- 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
<lang 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 = ()</lang>
Oforth
<lang Oforth>22 seq map(#[ dup sqrt 0.5 + floor + ]) println
1000000 seq map(#[ dup sqrt 0.5 + floor + ]) conform(#[ sqrt dup floor <>]) println</lang>
- 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
<lang scheme> (import (lib math))
; 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)
; 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))))
- ==> ()
</lang>
Oz
<lang 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}}</lang>
PARI/GP
<lang parigp>[vector(22,n,n + floor(1/2 + sqrt(n))), sum(n=1,1e6,issquare(n + floor(1/2 + sqrt(n))))]</lang>
Pascal
<lang 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.</lang>
- 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 <lang pascal>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.</lang>
Perl
<lang 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 $root;
}</lang>
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
Phix
with javascript_semantics sequence s = repeat(0,22) for n=1 to length(s) do s[n] = n + floor(1/2 + sqrt(n)) end for printf(1,"%V\n",{s}) integer nxt = 2, snxt = nxt*nxt, k for n=1 to 1000000 do k = n + floor(1/2 + sqrt(n)) if k>snxt then -- printf(1,"%d didn't occur\n",snxt) nxt += 1 snxt = nxt*nxt end if if k=snxt then puts(1,"error!!\n") end if end for puts(1,"none found ") ?{nxt,snxt}
- Output:
{2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27} none found {1001,1002001}
PHP
<lang php><?php //First Task for($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!"); } ?></lang>
- 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! >
PicoLisp
<lang 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") ) ) )</lang>
- 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
<lang PL/I>
put skip edit ((n, n + floor(sqrt(n) + 0.5) do n = 1 to n)) (skip, 2 f(5));
</lang>
Results:
<lang>
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
</lang>
Test 1,000,000 values:
<lang> 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; </lang>
PostScript
<lang>/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 </lang>
- 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. <lang powershell>filter Get-NonSquare {
return $_ + [Math]::Floor(1/2 + [Math]::Sqrt($_))
}</lang> Printing out the first 22 values is straightforward, then: <lang powershell>1..22 | Get-NonSquare</lang> If there were any squares for n up to one million, they would be printed with the following, but there is no output: <lang powershell>1..1000000 `
| Get-NonSquare ` | Where-Object { $r = [Math]::Sqrt($_) [Math]::Truncate($r) -eq $r }</lang>
PureBasic
<lang 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()</lang>
Python
<lang 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: </lang>
Or, defining OEIS A000037 as a non-finite series:
<lang python>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()</lang>
- 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
<lang 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"</lang>
- 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. <lang R>nonsqr <- function(n) n + floor(1/2 + sqrt(n)) nonsqr(1:22)</lang>
[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. <lang R>is.square <- function(x) {
sqrx <- sqrt(x) err <- abs(sqrx - round(sqrx)) err < 100*.Machine$double.eps
} any(is.square(nonsqr(1:1e6)))</lang>
[1] FALSE
Racket
<lang 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)))
</lang>
- 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)
<lang perl6>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;
}</lang>
- 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)
<lang rexx>/*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</lang>
- 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
<lang ring> for n=1 to 22
x = n + floor(1/2 + sqrt(n)) see "" + x + " "
next see nl </lang>
Ruby
<lang 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</lang>
Rust
<lang 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);
} </lang>
Scala
<lang 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)</lang>
Scheme
<lang 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)</lang>
- 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
<lang 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;</lang>
Sidef
<lang ruby>func nonsqr(n) { 0.5 + n.sqrt -> floor + n }
{ |i| if (nonsqr(i).is_sqr) { die "Found a square in the sequence: #{i}" } } << 1..1e6</lang>Smalltalk
<lang 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</lang>
Standard ML
<lang sml>- 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</lang>
Tcl
<lang 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 -> $y" }
} puts "done"</lang>
- Output:
1 2.0 2 3.0 3 5.0 4 6.0 5 7.0 6 8.0 7 10.0 8 11.0 9 12.0 10 13.0 11 14.0 12 15.0 13 17.0 14 18.0 15 19.0 16 20.0 17 21.0 18 22.0 19 23.0 20 24.0 21 26.0 22 27.0 looking for a square... done
TI-89 BASIC
Definition and 1 to 22, interactively:
<lang ti89b>■ n+floor(1/2+√(n)) → f(n)
Done
■ seq(f(n),n,1,22)
{2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27}</lang>
Program testing up to one million:
<lang ti89b>test() Prgm
Local i, ns For i, 1, 10^6 f(i) → ns If (floor(√(ns)))^2 = ns Then Disp "Oops: " & string(ns) EndIf EndFor Disp "Done"
EndPrgm</lang>
(This program has not been run to completion.)
Ursala
<lang Ursala>#import nat
- import flo
nth_non_square = float; plus^/~& math..trunc+ plus/0.5+ sqrt is_square = sqrt; ^E/~& math..trunc
- show+
examples = %neALP ^(~&,nth_non_square)*t iota23 check = (is_square*~+nth_non_square*t; ~&i&& %eLP)||-[no squares found]-! iota 1000000</lang>
- Output:
< 1: 2.000000e+00, 2: 3.000000e+00, 3: 5.000000e+00, 4: 6.000000e+00, 5: 7.000000e+00, 6: 8.000000e+00, 7: 1.000000e+01, 8: 1.100000e+01, 9: 1.200000e+01, 10: 1.300000e+01, 11: 1.400000e+01, 12: 1.500000e+01, 13: 1.700000e+01, 14: 1.800000e+01, 15: 1.900000e+01, 16: 2.000000e+01, 17: 2.100000e+01, 18: 2.200000e+01, 19: 2.300000e+01, 20: 2.400000e+01, 21: 2.600000e+01, 22: 2.700000e+01> no squares found
VBA
<lang vb> Sub Main() Dim i&, c&, j#, s$ 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</lang>
- 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
<lang ecmascript>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.")</lang>
- 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
<lang lisp>(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)))</lang>
- 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
<lang 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; ];
]</lang>
- Output:
2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27
zkl
<lang 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();</lang> 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()
- Programming Tasks
- Arithmetic operations
- 11l
- Ada
- ALGOL 68
- ALGOL W
- APL
- Arturo
- AutoHotkey
- AWK
- BASIC
- BBC BASIC
- Bc
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- D
- EchoLisp
- Eiffel
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- Erlang
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- F Sharp
- Factor
- Fantom
- Forth
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- FreeBASIC
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- HicEst
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- Icon Programming Library
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- J
- Java
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- Kotlin
- Liberty BASIC
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