A142198 -id:A142198 - cp's OEIS frontend

A263769 Smallest prime q such that q == -1 (mod prime(n)-1).

2, 3, 3, 5, 19, 11, 31, 17, 43, 83, 29, 71, 79, 41, 137, 103, 173, 59, 131, 139, 71, 233, 163, 263, 191, 199, 101, 211, 107, 223, 251, 389, 271, 137, 443, 149, 311, 647, 331, 859, 1423, 179, 379, 191, 587, 197, 419, 443

Offset: 1

Juri-Stepan Gerasimov, Oct 25 2015

nonn

a(n): A000040(1), A065091(1), A002145(1), A007528(1), A030433(1), A068231(1), A127576(1), A061242(1), A141857(1), A141976(1), A132236(1), A142111(1), A142198(1), A141898(1), A141926(1), A142531(1), A142004(1), A142799(1), A142068(1), A142099(1), ...

Smallest prime q such that (prime(n)^2 + q*prime(n))/(prime(n) + 1) is an integer.

			a(4) = 5 because 5 == -1 (mod prime(4)-1) and is prime.

Robert Israel, Table of n, a(n) for n = 1..6000

Cf. A263730, A263770.

Cf. A000040, A065091, A002145, A007528, A030433, A068231, A127576, A061242, A141857, A141976, A132236, A142111, A142198, A141898, A141926, A142531, A142004, A142799, A142068, A142099.

for n from 1 to 100 do
  k:= ithprime(n)-1;
  q:= 2;
  while (1 + q) mod k <> 0 do
    q:= nextprime(q)
  od;
  A[n]:= q;
od:
seq(A[i],i=1..1000); # Robert Israel, Oct 26 2015

Table[q = 2; z = Prime@ n - 1; While[Mod[q, z] != z - 1, q = NextPrime@ q]; q, {n, 59}] (* Michael De Vlieger, Oct 26 2015 *)

Corrected and edited by Robert Israel, Oct 26 2015,

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

Original entry on oeis.org

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, ...

cp's OEIS Frontend

A263769 Smallest prime q such that q == -1 (mod prime(n)-1).

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