A306431 Least number x > 1 such that n*x divides 1 + Sum_{k=1..x-1} k^(x-1).

2, 3, 13, 7, 19, 31, 41, 31, 13, 19, 43, 31, 23, 83, 139, 31, 61, 67, 113, 79, 251, 43, 19, 31, 199, 23, 13, 167, 53, 139, 83, 127, 157, 67, 293, 431, 443, 151, 103, 79, 61, 251, 113, 47, 337, 19, 179, 31, 41, 199, 67, 23, 19, 499, 181, 367, 607, 139, 257, 359

Offset

1,1

Comments

If n = 1, all the solutions of x | 1 + Sum_{k=1..x-1} k^(x-1) should be prime numbers, according to Giuga's conjecture.

If n*x | 1 + Sum_{k=1..x-1} k^(x-1), then certainly x does, so Giuga's conjecture would say x must be prime. Similarly if x^n divides it, so does x, so again Giuga would say x is prime. - Robert Israel, Apr 26 2019

E.g., the first solution for x^2 | 1 + Sum_{k=1..x-1} k^(x-1) is x = 1277, that is prime.

Links

Robert Israel, Table of n, a(n) for n = 1..2000

Eric Weisstein's World of Mathematics, Giuga's Conjecture

Formula

Least solution of n*x | 1 + Sum_{k=1..x-1} k^(x-1), for n = 1, 2, 3, ...

Example

a(4) = 7 because (1 + 1^6 + 2^6 + 3^6 + 4^6 + 5^6 + 6^6) / (4*7) = 67172 / 28 = 2399 and it is the least prime to have this property.

Maple

P:=proc(j) local k, n; for n from 2 to 10^6 do
if frac((add(k^(n-1), k=1..n-1)+1)/(j*n))=0
then RETURN(n); break; fi; od; end: seq(P(i), i=1..60);

Mathematica

a[n_] := For[x = 2, True, x++, If[Divisible[1+Sum[k^(x-1), {k, x-1}], n x], Return[x]]];
Array[a, 60] (* Jean-François Alcover, Oct 16 2020 *)

Prog

(PARI) a(n) = my(x=2); while (((1 + sum(k=1, x-1, k^(x-1))) % (n*x)), x++); x; \\ Michel Marcus, Apr 27 2019

Crossrefs

Cf. A191677. All the solutions for n = m: A000040 (m=1), A002145 (m=2), A007522 (m=4), A127576 (m=8), A141887 (m=10), A127578 (m=16), A142198 (m=20), A127579 (m=32), A095995 (m=50).

Sequence in context: A128369 A087568 A087564 * A192362 A057776 A110362

Adjacent sequences: A306428 A306429 A306430 * A306432 A306433 A306434

Keyword

nonn,easy

Author

Paolo P. Lava, Apr 05 2019

Status

approved