OFFSET
1,1
COMMENTS
For 1<n<258, we have a(n)=11 (mod 12). Is this true for all n>1? It also appears that for each n there are an infinite number of primes p such that sigma(x)=sigma(p) has exactly n solutions.
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..1199
MATHEMATICA
s=DivisorSigma[1, Range[100000]]; t=Table[Length[Position[s, Prime[n]+1]], {n, PrimePi[Length[s]]}]; u=Union[t]; nLast=First[Complement[Range[u[[ -1]]], u]]-1; Flatten[Table[Prime[Position[t, n, 1, 1]], {n, nLast}]]
PROG
(PARI) sigv(n) = select(i->sigma(i) == n, vector(n, i, i));
a(n) = {p = 2; while (#(sigv(p+1))! = n, p = nextprime(p+1)); p; } \\ Michel Marcus, May 01 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Jan 21 2006
STATUS
approved