A284082 Smallest positive m such that n divides sigma_m(n) - j where j is some divisor of n, or 0 if no such m exists.

1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 1, 1, 1, 3, 2, 4, 1, 1, 1, 1, 0, 5, 1, 1, 2, 6, 3, 1, 1, 2, 1, 5, 0, 4, 2, 2, 1, 9, 0, 1, 1, 3, 1, 8, 5, 11, 1, 2, 2, 2, 8, 4, 1, 5, 0, 1, 0, 14, 1, 2, 1, 5, 0, 6, 2, 5, 1, 0, 7, 9, 1, 0, 1, 18, 10, 2, 0, 0, 1, 0, 4, 10, 1, 2, 8

Offset: 1

Author

Juri-Stepan Gerasimov, Mar 19 2017

Keywords

• nonn

Comments

If p is a prime, we have a(p) = 1. In general, if n = p^q with p prime, then a(n) <= q. For every prime power p^q < 10^13 it actually holds a(p^q) = q. Is this true for every prime power? - Giovanni Resta, Mar 20 2017

Yes, this is true: sigma[m](p^q) == 1/(1-p^m) (mod p^q); this is never divisible by p, and == 1 (mod p^q) iff m >= q. - Robert Israel, Apr 27 2017

First occurrence of k: 21, 1, 4, 8, 16, 22, 26, 69, 44, 38, 75, 46, 148, 316, 58, 186, ..., . - Robert G. Wilson v, Apr 14 2017

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Integers n such that n divides sigma_k(n) - i, where i is some divisor of n; A205523 (k = 1), A283970 (k = 2).

Cf. A033950.

Programs

Maple

f:= proc(n) local m,mm,F,S,P,D,M0,M1;
      F:= ifactors(n)[2];
      if nops(F) = 1 then return F[1][2] fi;
      P:= map(t -> t[1]^t[2], F);
      S:= mul(add(t[1]^(i*m),i=0..t[2]),t=F);
      D:= subs(n=0,numtheory:-divisors(n));
      for mm from 1 to ilcm(op(map(numtheory:-phi, P)))+max(seq(t[2],t=F)) do
        if member(subs(m=mm,S) mod n, D) then return mm fi;
      od;
      0
end proc:
map(f, [$1..100]); # Robert Israel, Apr 27 2017

Mathematica

a[n_] := Block[{ds, d=Divisors[n], m=0}, While[m <= 2*n, m++; ds = DivisorSigma[m, n]; If[ Select[d, Mod[ds-#, n] == 0 &, 1] != {}, Break[]]]; If[m > 2*n, 0, m]]; Array[a, 85] (* assuming that sigma(m,n) mod n has a period <= 2*n, Giovanni Resta, Mar 20 2017 *)

Extensions

a(56) from Giovanni Resta, Mar 20 2017