A095995 -id:A095995 - cp's OEIS frontend

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

Paolo P. Lava, Apr 05 2019

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.

			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.

Robert Israel, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Giuga's Conjecture

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).

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);

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 *)

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

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

A177688 Numbers n such that (n+2)//n - (n+1) is prime, where // represents the concatenation of decimals.

Original entry on oeis.org

0, 1, 4, 6, 7, 9, 12, 13, 15, 18, 19, 22, 25, 28, 31, 33, 39, 46, 48, 49, 52, 60, 61, 64, 67, 73, 75, 84, 85, 88, 90, 99, 100, 103, 106, 132, 133, 135, 136, 138, 142, 156, 160, 163, 171, 178, 181, 183, 187, 190, 198, 201, 202, 211, 220, 222, 229, 238, 241, 246, 252

Offset: 1

Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 11 2010

If n is a k-digit number, then we demand that p = (n+2) * 10^k + n - (n+1) is a prime number, obviously of the form p = (n+2) * 10^k - 1, so the decimal representation of p is n+1 followed by k times the digit 9.
The sequence is infinite, proof with Dirichlet's prime number (in arithmetic progressions) theorem.
Note that numbers of the form (n+2)//n + (n+1) are multiples of 3 and do not generate primes.

			2//0 - 1 = 20 - 1 = 19 = prime(8), 0 is first term;
3//1 - 2 = 31 - 2 = 29 = prime(10), 1 is 2nd term;
6//4 - 5 = 64 - 5 = 59 = prime(17), 4 is 3rd term.

Cf. A000040, A065582, A095995, A173976, A177435.

n2ncQ[n_]:=PrimeQ[FromDigits[Join[IntegerDigits[n+2], IntegerDigits[ n]]]- n-1]; Select[Range[0,300],n2ncQ]  (* Harvey P. Dale, Feb 24 2011 *)

cp's OEIS Frontend

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

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