OFFSET
1,3
COMMENTS
For each partition of n into 2 parts, multiply the parts together and find the number of divisors of each product formed. Then add the results to get a(n).
LINKS
MATHEMATICA
Table[Sum[DivisorSigma[0, i (n - i)], {i, 1, Floor[n/2]}], {n, 100}]
PROG
(PARI) a(n) = sum(i=1, n\2, numdiv(i*(n-i))); \\ Michel Marcus, Mar 18 2016
(GAP) List([1..10^4], n->Sum([1..Int(n/2)], i->Tau(i*(n-i)))); # Muniru A Asiru, Feb 04 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, May 01 2015
STATUS
approved