A245474 a(n) = smallest positive integer s such that s*n - floor(sqrt(s*n))^2 is a square.

1, 1, 1, 3, 1, 1, 6, 7, 1, 1, 1, 11, 3, 1, 14, 3, 1, 1, 2, 19, 1, 21, 22, 23, 6, 1, 1, 3, 7, 1, 3, 31, 2, 33, 1, 35, 1, 1, 38, 6, 1, 2, 42, 43, 11, 1, 46, 47, 3, 1, 1, 3, 2, 1, 6, 55, 14, 57, 1, 59, 6, 2, 62, 7, 1, 1, 66, 67, 1, 69, 35, 71, 2, 1, 2, 3, 19

Offset: 0

Author

Thomas Ordowski, Jul 23 2014

Keywords

• nonn

Comments

a(n) <= n for n > 0. If prime p == 3 (mod 4) then a(p) = p.

Conjecture: a(p) < p for prime p == 1 (mod 4).

Outline of proof of conjecture: write p = x^2 + y^2. Since gcd(x,y) = 1, there are u,v with x*u + y*v = 1, u^2 + v^2 < y^2 + x^2 = p. Taking s = u^2 + v^2, s*p = (u*y+v*x)^2 + 1^2, and |u*y+v*x| = floor(sqrt(s*p)). - Robert Israel, Aug 04 2014

For the first 100000 primes p == 1 (mod 4), a(p) < sqrt(p)/2. - Robert Israel, Aug 03 2014

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A145022, A145023, A363340.

Programs

Maple

A:= proc(n) local s,a;
     for s from 1 do
       a:= floor(sqrt(s*n));
       if issqr(s*n-a^2) then return s fi
     od
end proc:
seq(A(n),n=0..1000); # Robert Israel, Jul 23 2014

Mathematica

a245474[n_Integer] := Catch[
  Do[
   If[IntegerQ[Sqrt[(s*n - Floor[Sqrt[s*n]]^2)]] == True, Throw[s]],
   {s, n}]
  ]; Map[a245474, Range[100]] (* Michael De Vlieger, Aug 03 2014 *)

PARI

a(n) = s=1; while(!issquare(s*n-sqrtint(s*n)^2), s++); s \\ Colin Barker, Jul 23 2014

Extensions

More terms from Colin Barker, Jul 23 2014