OFFSET
1,1
COMMENTS
Note that a(n) always exists, because sigma(2^(p-2)) = 2^(p-1)-1 is divisible by p for p>2 (by Fermat's little theorem), so there is always a candidate for a(n). Compare A227470, A272349. - N. J. A. Sloane, May 01 2016
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..10000
EXAMPLE
n=11: a(11)=16 is the least number x such that sigma(x) is divisible by the 11th prime, 31.
MATHEMATICA
ln[n_]:=Module[{x=1, p=Prime[n]}, While[!Divisible[DivisorSigma[ 1, x], p], x++]; x]; Array[ln, 60] (* Harvey P. Dale, Sep 07 2014 *)
Module[{nn=5000, ds}, ds=DivisorSigma[1, Range[nn]]; Table[Position[ds, _?(Divisible[#, n]&), 1, 1], {n, Prime[Range[60]]}]]//Flatten (* Much faster than the first program *) (* Harvey P. Dale, May 18 2018 *)
PROG
(PARI) sigma_hunt(x)=local(n=0, g); while(n++, g=sigma(n); if(g%x, , return(n)));
for(x=1, 50, print1(sigma_hunt(prime(x))", ")) /* Phil Carmody, Mar 01 2013 */
(Magma) sol:=[]; p:=PrimesUpTo(10000); for n in [1..53] do k:=2; while Max(PrimeDivisors(SumOfDivisors(k))) ne p[n] do k:=k+1; end while; sol[n]:=k; end for; sol; // Marius A. Burtea, Jun 05 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Aug 23 2004
STATUS
approved