Sequence of non-squares

Sequence of non-squares
You are encouraged to solve this task according to the task description, using any language you may know.

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

Ada

<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; </ada> Sample output:

 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27

BASIC

Works with: FreeBASIC

Works with: RapidQ

<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" </freebasic>

C

<c>

1. include <math.h>
2. include <stdio.h>
3. 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>

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

Works with: Fortran version 90 and later

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

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

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 i.&.<:1e6   NB.  Number of square RFs < 1e6
0  

Java

<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);
       }
   }

} </java>

OCaml

<ocaml># let nonsqr n = n + truncate (0.5 +. sqrt (float n));; val nonsqr : int -> int = <fun>

1. (* 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 = ()

1. (* The following check shows no squares up to one million: *)

 for i = 1 to 1000000 do
   let j = sqrt (float (nonsqr i)) in
     assert (j <> floor j)
   done;;

- : unit = ()</ocaml>

Python

<python> >>> from math import sqrt >>> # (Using the fact that round(X) is equivalent to floor(0.5+X) for our range of X) >>> def nonsqr(n): return n + int(round(sqrt(n)))

>>> # first 22 values (as a list) has no squares: >>> [nonsqr(i) for i in xrange(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: >>> for i in xrange(1,1000000): j = sqrt(nonsqr(i)) assert j != int(j), "Found a square in the sequence: %i" % i

>>> </python>