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'Floor (0.5 + 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))) 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

Python

<python> >>> from math import sqrt,floor >>> def nonsqr(n): return n + floor(0.5 + sqrt(n))

>>> # first 22 values (as a list) has no squares: >>> [nonsqr(i) for i in xrange(1,23)] [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] >>> # 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>