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

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

Offset: 1

Author

Alex Ratushnyak, Jul 12 2013

Keywords

• nonn

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^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

Examples

			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.
		

Links

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

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

Programs

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 *)

PARI

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

Formula

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