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A118076
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Numbers n such that n divides sigma_(2^k)(n), the sum of the 2^k powers of the divisors of n, for all k>0.
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2
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OFFSET
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1,2
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COMMENTS
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Although these numbers have been tested up to k=20, it is conjectured that n divides sigma_(2^k)(n) for all k>0. Intersection of A046762 and A066292.
Let d be the vector of divisors of n. The sequence d^(2^k) mod n has some period p. Thus if n divides sigma_(2^k)(n) for one period, then n divides sigma_(2^k)(n) for all k. For these n, the first period ends for k<14. Hence it is easy to verify divisibility for all k. Intersection of A046762 and A066292. - T. D. Noe, Apr 12 2006
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LINKS
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EXAMPLE
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n=84 is here because 84 divides each one of sigma_4(n)=53771172, sigma_8(n)=2488859101224132, sigma_16(n)=6144339637187846520573009496452, etc.
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MATHEMATICA
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t={}; Do[If[Mod[DivisorSigma[2, n], n]==0, AppendTo[t, n]], {n, 10^8}]; Do[t=Select[t, Mod[DivisorSigma[2^k, # ], # ]==0&], {k, 2, 20}]; t
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CROSSREFS
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Cf. A076230 (n divides sigma_2(n) and sigma_4(n)).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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