A350037 a(n) = n^2 mod round(sqrt(n)).

0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 4, 4, 1, 0, 1, 4, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4

Offset

1,22

Links

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

Formula

a(n) = A000290(n) mod A000194(n). - Michel Marcus, Dec 14 2021

Example

a(5) = 5^2 mod round(sqrt(5)) = 25 mod 2 = 1.

Mathematica

a[n_] := Mod[n^2, Round @ Sqrt[n]]; Array[a, 100, 2] (* Amiram Eldar, Dec 10 2021 *)
Table[PowerMod[n, 2, Round[Sqrt[n]]], {n, 2, 101}] (* Stefano Spezia, Dec 15 2021 *)

Prog

(Java)
import java.util.Arrays;
public class modulus_sequence {
  static int[] solutions = new int[480];
  static int a_of_n (double n) {
    int z = (int)Math.round(Math.sqrt(n));
    int w = (int)(Math.pow(n, 2));
    int k = w%z;
    return k;
  }
  public static void main(String[] args) {
    for (double j = 2; j < 482; j++) {
      int h = a_of_n(j);
      solutions[(int) (j-2)]=h;
    }
    System.out.println(Arrays.toString(solutions));
  }
}
(PARI) a(n) = n^2 % round(sqrt(n)); \\ Michel Marcus, Dec 14 2021
(PARI) a(n) = lift(Mod(n, ((sqrtint(4*n) + 1)\2))^2); \\ Michel Marcus, Dec 14 2021
(Python)
from math import isqrt
def A350037(n): return pow(n, 2, (m:=isqrt(n))+int(4*n>=(2*m+1)**2)) # Chai Wah Wu, Jan 10 2022

Crossrefs

Cf. A336302 (with ceiling instead of round).

Sequence in context: A185057 A343720 A048152 * A070430 A336302 A163353

Adjacent sequences: A350034 A350035 A350036 * A350038 A350039 A350040

Keyword

nonn,easy

Author

Sebastian F. Orellana, Dec 09 2021

Status

approved