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A284082
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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.
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2
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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
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OFFSET
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1,4
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COMMENTS
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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
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LINKS
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MAPLE
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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:
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MATHEMATICA
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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 *)
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CROSSREFS
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Integers n such that n divides sigma_k(n) - i, where i is some divisor of n; A205523 (k = 1), A283970 (k = 2).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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