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A227470
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Least k such that n divides sigma(n*k).
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6
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1, 3, 2, 3, 8, 1, 4, 7, 10, 4, 43, 2, 9, 2, 8, 21, 67, 5, 37, 6, 20, 43, 137, 5, 149, 9, 34, 1, 173, 4, 16, 21, 27, 64, 76, 22, 73, 37, 6, 3, 163, 10, 257, 43, 6, 137, 281, 11, 52, 76, 67, 45, 211, 17, 109, 4, 49, 173, 353, 2, 169, 8, 32, 93, 72, 27, 401, 67
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OFFSET
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1,2
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COMMENTS
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Theorem: a(n) always exists.
Proof: If n is a power of a prime, say n = p^a, then, by Euler's generalization of Fermat's little theorem and the multiplicative property of sigma, one can take k = x^(p^a-p^(a-1)-1) where x is a different prime from p. Similarly. if n = p^a*q^b, then take k = x^(p^a-p^(a-1)-1)*y^(q^b-q^(b-1)-1) where {x,y} are primes different from {p,q}. And so on. These k's have the desired property, and so there is always at least one candidate for the minimal k. - N. J. A. Sloane, May 01 2016
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LINKS
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FORMULA
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EXAMPLE
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Least k such that 9 divides sigma(9*k) is k = 10: sigma(90) = 234 = 9*26. So a(9) = 10.
Least k such that 89 divides sigma(89*k) is k = 1024: sigma(89*1024) = 184230 = 89*2070. So a(89) = 1024.
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MAPLE
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local k;
for k from 1 do
if modp(numtheory[sigma](k*n), n) =0 then
return k;
end if;
end do:
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MATHEMATICA
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lknds[n_]:=Module[{k=1}, While[!Divisible[DivisorSigma[1, k*n], n], k++]; k]; Array[lknds, 70] (* Harvey P. Dale, Jul 10 2014 *)
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PROG
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(PARI) a227470(n) = {k=1; while(sigma(n*k)%n != 0, k++); k} \\ Michael B. Porter, Jul 15 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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