A317358 a(n) is the smallest number k > 1 such that 1^(k-1) + 2^(k-1) + ... + n^(k-1) == n (mod k).

2, 3, 5, 2, 2, 7, 11, 2, 2, 3, 3, 2, 2, 17, 17, 2, 2, 3, 3, 2, 2, 23, 29, 2, 2, 5, 3, 2, 2, 31, 37, 2, 2, 37, 35, 2, 2, 3, 41, 2, 2, 43, 47, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 59, 61, 2, 2, 67, 3, 2, 2, 55, 71, 2, 2, 35, 35, 2, 2, 3, 5, 2, 2, 5, 5, 2, 2

Offset: 1

Author

Thomas Ordowski, Jul 26 2018

Keywords

• nonn

Comments

a(n) = 2 if and only if n == {0, 1} (mod 4).

a(n) <= A151800(n).

A133906(n) <= a(n) <= A133907(n).

The sequence is unbounded.

Numbers n such that a(n-1) = n are 2, 3, 7, 23, 31, 43, 59, 139, 283, ...

By the Agoh-Giuga conjecture, if a(n-1) = n, then n is a prime.

It seems that if a(n) > n, then a(n) is a prime (the next prime after n).

If a(n) = n, then n is in A121707. These numbers are 35, 143, 187, 215, ...

Conjecture: all composite terms of the sequence are A121707.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Seiichi Manyama)
• Wikipedia, Agoh-Giuga conjecture

Crossrefs

Cf. A133906, A133907, A151800, A121707, A317058, A317357.

Programs

Mathematica

a[n_] := Block[{k=2}, While[Mod[Sum[PowerMod[j, k-1, k], {j, n}], k] != Mod[n, k], k++]; k]; Array[a, 81] (* Giovanni Resta, Jul 29 2018 *)

PARI

a(n) = for(k=2,oo, if (sum(j=1,n, Mod(j,k)^(k-1)) == n, return (k));); \\ Michel Marcus, Jul 26 2018

Python

def g(n,p,q): # compute (-n + sum_{k=1,n} k^p)  mod q
    c = (-n) % q
    for k in range(1,n+1):
        c = (c+pow(k,p,q)) % q
    return c
def A317358(n):
    k = 2
    while g(n,k-1,k):
        k += 1
    return k # Chai Wah Wu, Jul 30 2018

Extensions

More terms from Michel Marcus, Jul 26 2018