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.
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 >>> def nonsqr(n): return n + int(0.5 + 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>