OFFSET
1,1
COMMENTS
a(n) <= 2p, where p = A002586(n) is the smallest prime factor of (1 + 2^n). (Proof. Since sigma_n(2p) = (1 + 2^n)(1 + p^n) and p is odd, 2p divides sigma_n(2p).) - Jonathan Sondow, Nov 23 2012
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
FORMULA
Sum{d^n} = ka(n), d runs over the divisors of a(n), where k is an integer and a(n) is the smallest suitable number.
MATHEMATICA
Table[m = 2; While[Mod[DivisorSigma[n, m], m] > 0, m++]; m, {n, 100}] (* T. D. Noe, Nov 23 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Dec 06 2001
EXTENSIONS
Definition and formulas corrected by Jonathan Sondow, Nov 23 2012
STATUS
approved