A354139 a(n) is the least positive integer m such that (k+1)^n + (k+2)^n + ... + (k+m)^n == 0 (mod n) for every positive integer k.

1, 4, 3, 8, 5, 36, 7, 16, 3, 20, 11, 72, 13, 28, 15, 32, 17, 108, 19, 200, 21, 44, 23, 144, 5, 52, 3, 56, 29, 180, 31, 64, 33, 68, 35, 216, 37, 76, 39, 400, 41, 1764, 43, 88, 15, 92, 47, 288, 7, 20, 51, 104, 53, 324, 55, 112, 57, 116, 59, 1800, 61, 124, 21, 128, 65, 396, 67, 136, 69, 140, 71

Offset: 1

Author

Dimitrios T. Tambakos, May 18 2022

Keywords

• nonn

Comments

a(n) divides n * A007947(n).

Examples

			a(2) = 4 because, for every positive integer k, (k+1)^2 + (k+2)^2 + (k+3)^2 + (k+4)^2 == 0 (mod 2), and no smaller positive integer satisfies this condition.
		

Programs

Mathematica

sum[n_, r_] := Mod[Sum[k^r, {k, 1, n}], r];
rad[r_] := Product[i[[1]], {i, FactorInteger[r]}];
seq[r_] := Table[sum[n, r], {n, 1, r*rad[r]}];
A354139[r_] := Piecewise[   {    {rad[r], OddQ[r]},
    {2*r, EvenQ[r] && PrimePowerQ[r]},
    {Length[FindRepeat[seq[r]]], EvenQ[r] && Not[PrimePowerQ[r]]}
    }
   ];
Table[A354139[r], {r, 1, 20}] (* Improved by Dimitrios T. Tambakos, Feb 08 2023 *)

PARI

isok(k, n) = my(p=sum(i=1, k, Mod(i+x, n)^n)); if (p==0, return(1)); for (i=1, n, if (subst(p, x, i) != 0, return(0))); return(1);
a(n) = my(k=1); while (!isok(k,n), k++); k; \\ Michel Marcus, May 21 2022

Formula

a(2^t) = 2^(t+1) for integers t>0.

a(n) = A007947(n) for odd integers n.

Conjecture: a(n) = A007947(n) * A193267(n).