1,2

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 (noe(AT)sspectra.com), Apr 12 2006

n=84 is here because 84 divides each one of sigma_4(n)=53771172, sigma_8(n)=2488859101224132, sigma_16(n)=6144339637187846520573009496452, etc.

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

Cf. A076230 (n divides sigma_2(n) and sigma_4(n)).

Sequence in context: A132052 A097840 A076230 this_sequence A056746 A111194 A096132

Adjacent sequences: A118073 A118074 A118075 this_sequence A118077 A118078 A118079

nonn

T. D. Noe (noe(AT)sspectra.com), Apr 11 2006