A118076 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.

1, 84, 435708, 986076, 1441188, 6066396, 18735444, 78863148

Offset: 1

Author

T. D. Noe, Apr 11 2006

Keywords

• nonn

Comments

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

Examples

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

Crossrefs

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

Programs

Mathematica

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