A362502 Least k > 0 such that (floor(sqrt(n*k)) + 1)^2 mod n is a square.

1, 1, 1, 1, 1, 2, 3, 1, 3, 4, 5, 1, 7, 8, 1, 1, 9, 4, 11, 2, 1, 14, 15, 1, 8, 16, 1, 3, 19, 2, 21, 1, 3, 24, 1, 2, 25, 26, 3, 1, 29, 2, 31, 6, 1, 34, 35, 1, 15, 4, 3, 7, 39, 4, 1, 2, 3, 44, 45, 1, 47, 48, 1, 2, 1, 4, 51, 10, 5, 2, 55, 1, 57, 58, 5, 12, 1, 6, 63, 1, 5, 64, 65, 1, 3, 68

Offset: 1

Author

Darío Clavijo, Apr 22 2023

Keywords

• nonn

Links

• Winston de Greef, Table of n, a(n) for n = 1..10000

Programs

Mathematica

nmax=86; a={}; For[n=1, n<=nmax, n++, For[k=1, k>0, k++, If[IntegerQ[Sqrt[Mod[Floor[Sqrt[n k]+1]^2, n]]], AppendTo[a,k]; k=-1]]]; a (* Stefano Spezia, Apr 24 2023 *)

PARI

a(n) = my(k=1); while(!issquare((sqrtint(n*k)+1)^2 % n), k++); k; \\ Michel Marcus, Apr 24 2023

Python

from gmpy2 import is_square, isqrt
def a(n):
  m,k = 2,0
  while not is_square(m):
    k+=1
    m = pow(isqrt(n * k) + 1, 2, n)
  return k