A095366 Least k > 1 such that k divides 1^n + 2^n +...+ (k-1)^n.

3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 17, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7, 3, 5, 3, 7, 3, 5, 3, 11, 3, 5, 3, 7

Offset: 1

Author

T. D. Noe, Jun 03 2004

Keywords

• nonn

Comments

This sequence is similar to A094756 but seems to have a simpler periodicity rules:

a(n)=3 when n=1 (mod 2), otherwise

a(n)=5 when n=2 (mod 4), otherwise

a(n)=7 when n=4*m (mod 12) for some m=1,2, otherwise

a(n)=11 when n=12*m (mod 60) for some m=1,2,3,4, otherwise

a(n)=17 when n=60*m (mod 240) for some m=1,2,3, otherwise

a(n)=19 when n=240*m (mod 720) for some m=1,2, otherwise

a(n)=23 when n=720*m (mod 7920) for some m=1,..,10, etc.

Note that only odd primes p given by A095365 seem to appear in this sequence. Given the definition of f(p) in that sequence, let q=A095365(i) and p=A095365(i-1), then the general rule for this sequence seems to be a(n)=q when n=f(p)*m (mod f(q)) for some m=1,...,f(q)/f(p)-1

Examples

			a(4) = 7 because k divides 1^4 + 2^4 +...+ k^4 for k=7 but no smaller k > 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000

Crossrefs

Cf. A094756, A095365.

Programs

Mathematica

Table[k=2; s=0; While[s=s+(k-1)^n; Mod[s, k]>0, k++ ]; k, {n, 100}]

PARI

A095366(n) = { my(k=1,s=0); while(1, k++; s += ((k-1)^n); if(!(s%k), return(k))); }; \\ Antti Karttunen, Dec 19 2018