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A095366
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Least k > 1 such that k divides 1^n + 2^n +...+ (k-1)^n.
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2
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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; internal format)
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OFFSET
| 1,1
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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
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EXAMPLE
| a(4) = 7 because k divides 1^4 + 2^4 +...+ k^4 for k=7 but no smaller k > 1.
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MATHEMATICA
| Table[k=2; s=0; While[s=s+(k-1)^n; Mod[s, k]>0, k++ ]; k, {n, 100}]
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CROSSREFS
| Sequence in context: A100667 A096438 A118271 * A029604 A079602 A075572
Adjacent sequences: A095363 A095364 A095365 * A095367 A095368 A095369
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KEYWORD
| nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), Jun 03 2004
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