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 A095366 Least k > 1 such that k divides 1^n + 2^n +...+ (k-1)^n. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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 LINKS Antti Karttunen, Table of n, a(n) for n = 1..20000 Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000 EXAMPLE a(4) = 7 because k divides 1^4 + 2^4 +...+ k^4 for k=7 but no smaller k > 1. MATHEMATICA Table[k=2; s=0; While[s=s+(k-1)^n; Mod[s, k]>0, k++ ]; k, {n, 100}] PROG (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 CROSSREFS Cf. A094756, A095365. Sequence in context: A118271 A260689 A328386 * A029604 A079602 A075572 Adjacent sequences:  A095363 A095364 A095365 * A095367 A095368 A095369 KEYWORD nonn AUTHOR T. D. Noe, Jun 03 2004 STATUS approved

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Last modified September 24 07:17 EDT 2021. Contains 347623 sequences. (Running on oeis4.)