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

Author

Sebastian F. Orellana, Dec 09 2021

Keywords

• nonn
• easy

Examples

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

Links

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

Crossrefs

Cf. A336302 (with ceiling instead of round).

Programs

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));
  }
}

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 *)

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

Formula

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