A066292 -id:A066292 - cp's OEIS frontend

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

T. D. Noe, Apr 11 2006

nonn

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

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

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

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

A118107 Period of the vector sequence d(n)^2^k mod n for k=1,2,3,..., where d(n) is the vector of divisors of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 4, 1, 2, 2, 1, 6, 2, 1, 1, 2, 1, 4, 2, 10, 1, 1, 1, 4, 1, 2, 1, 6, 4, 2, 6, 3, 1, 1, 1, 4, 2, 1, 1, 4, 1, 1, 10, 2, 1, 2, 1, 6, 4, 6, 4, 2, 1, 1, 1, 4, 1, 2, 1, 3, 3, 4, 1, 2, 2, 10, 4, 11, 6, 1, 1, 6, 4, 4

Offset: 1

T. D. Noe, Apr 13 2006

nonn

This sequence is related to the period of sigma_(2^k)(n) mod n, which is important in studying the numbers n dividing sigma_(2^k)(n) for all k>0. See A066292 and A118076. Note that a(n)=1 if n is a power of a prime.

			See A118106 for an example involving d(n)^k.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A118106 (period of the vector sequence d(n)^k mod n).

Table[d=Divisors[n]; k=0; found=False; While[i=0; While[i

A118107(n) = { my(divs=apply(d -> (d%n),divisors(n)), odivs = Vec(divs), vs = Map()); mapput(vs, odivs, 0); for(k=1,oo,divs = vector(#divs,i,(divs[i]*divs[i])%n); if(mapisdefined(vs, divs), return(k-mapget(vs, divs)), mapput(vs, divs, k))); }; \\ Antti Karttunen, Sep 23 2018

A066291 Numbers m such that DivisorSigma(8k-4, m) mod m = 0 holds presumably for all k; that is, (8k-4)-power-sums of divisors of m are divisible by m for all k.

Original entry on oeis.org

1, 34, 492, 5617, 11234, 22468, 67404, 190978, 709937, 763912, 1419874, 2839748, 5073996, 5446841, 7914353, 8519244, 10893682, 11548552, 15828706, 17126233, 21787364, 31657412, 34252466, 43574728, 57928121, 63314824, 65362092, 68504932, 73084632, 94972236

Offset: 1

Labos Elemer, Dec 12 2001

nonn

			Tested for each m with k < 200.
Tested for each m with k < 500. - _Sean A. Irvine_, Oct 07 2023

Cf. A066135, A066284, A066289-A066292.

Table[Union[Table[ IntegerQ[DivisorSigma[8*k-4, Part[t, m]]/Part[t, m]], {k, 1, 200}]], {m, 1, Length[t]}]; where t denotes the table of sequence.

DivisorSigma(8*k-4, m)/m is an integer for k = 1, 2, 3, ..., 200, ...

More terms from Sean A. Irvine, Oct 07 2023

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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.

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A118107 Period of the vector sequence d(n)^2^k mod n for k=1,2,3,..., where d(n) is the vector of divisors of n.

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A066291 Numbers m such that DivisorSigma(8k-4, m) mod m = 0 holds presumably for all k; that is, (8k-4)-power-sums of divisors of m are divisible by m for all k.

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cp's OEIS Frontend

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.

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A118107 Period of the vector sequence d(n)^2^k mod n for k=1,2,3,..., where d(n) is the vector of divisors of n.

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Author

Keywords

Comments

Examples

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Crossrefs

Programs

Mathematica

PARI

A066291 Numbers m such that DivisorSigma(8*k-4, m) mod m = 0 holds presumably for all k; that is, (8*k-4)-power-sums of divisors of m are divisible by m for all k.

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A066291 Numbers m such that DivisorSigma(8k-4, m) mod m = 0 holds presumably for all k; that is, (8k-4)-power-sums of divisors of m are divisible by m for all k.