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A227470 Least k such that n divides sigma(n*k).

%I A227470 #25 May 06 2016 04:48:46
%S A227470 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,
%T A227470 173,4,16,21,27,64,76,22,73,37,6,3,163,10,257,43,6,137,281,11,52,76,
%U A227470 67,45,211,17,109,4,49,173,353,2,169,8,32,93,72,27,401,67
%N A227470 Least k such that n divides sigma(n*k).
%C A227470 Theorem: a(n) always exists.
%C A227470 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
%H A227470 R. J. Mathar, <a href="/A227470/b227470.txt">Table of n, a(n) for n = 1..1000</a>
%F A227470 a(n) = A272349(n)/n. - _R. J. Mathar_, May 06 2016
%e A227470 Least k such that 9 divides sigma(9*k) is k = 10: sigma(90) = 234 = 9*26. So a(9) = 10.
%e A227470 Least k such that 89 divides sigma(89*k) is k = 1024: sigma(89*1024) = 184230 = 89*2070. So a(89) = 1024.
%p A227470 A227470 := proc(n)
%p A227470     local k;
%p A227470     for k from 1 do
%p A227470         if modp(numtheory[sigma](k*n),n) =0 then
%p A227470             return k;
%p A227470         end if;
%p A227470     end do:
%p A227470 end proc: # _R. J. Mathar_, May 06 2016
%t A227470 lknds[n_]:=Module[{k=1},While[!Divisible[DivisorSigma[1,k*n],n],k++];k]; Array[lknds,70] (* _Harvey P. Dale_, Jul 10 2014 *)
%o A227470 (PARI) a227470(n) = {k=1; while(sigma(n*k)%n != 0, k++); k} \\ _Michael B. Porter_, Jul 15 2013
%Y A227470 Indices of 1's: A007691.
%Y A227470 Cf. A000203, A227302, A227303, A097018.
%Y A227470 See A272349 for the sequence [n*a(n)]. - _N. J. A. Sloane_, May 01 2016
%K A227470 nonn
%O A227470 1,2
%A A227470 _Alex Ratushnyak_, Jul 12 2013