

A227470


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


5



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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^ap^(a1)1) where x is a different prime from p. Similarly. if n = p^a*q^b, then take k = x^(p^ap^(a1)1)*y^(q^bq^(b1)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


LINKS

R. J. Mathar, Table of n, a(n) for n = 1..1000


FORMULA

a(n) = A272349(n)/n.  R. J. Mathar, May 06 2016


EXAMPLE

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.


MAPLE

A227470 := proc(n)
local k;
for k from 1 do
if modp(numtheory[sigma](k*n), n) =0 then
return k;
end if;
end do:
end proc: # R. J. Mathar, May 06 2016


MATHEMATICA

lknds[n_]:=Module[{k=1}, While[!Divisible[DivisorSigma[1, k*n], n], k++]; k]; Array[lknds, 70] (* Harvey P. Dale, Jul 10 2014 *)


PROG

(PARI) a227470(n) = {k=1; while(sigma(n*k)%n != 0, k++); k} \\ Michael B. Porter, Jul 15 2013


CROSSREFS

Indices of 1's: A007691.
Cf. A000203, A227302, A227303, A097018.
See A272349 for the sequence [n*a(n)].  N. J. A. Sloane, May 01 2016
Sequence in context: A225695 A226469 A073341 * A218396 A331926 A070982
Adjacent sequences: A227467 A227468 A227469 * A227471 A227472 A227473


KEYWORD

nonn


AUTHOR

Alex Ratushnyak, Jul 12 2013


STATUS

approved



