A238096 a(n) = Sum_{k=2..n} floor(n/k)*floor((tau(k)+1)/2), where tau = A000005.
0, 1, 2, 5, 6, 10, 11, 16, 19, 23, 24, 33, 34, 38, 42, 50, 51, 60, 61, 70, 74, 78, 79, 94, 97, 101, 106, 115, 116, 129, 130, 141, 145, 149, 153, 172, 173, 177, 181, 196, 197, 210, 211, 220, 229, 233, 234, 257, 260, 269, 273, 282, 283, 298, 302, 317, 321, 325, 326, 353, 354, 358, 367, 382, 386, 399, 400, 409, 413, 426
Offset: 1
Keywords
Links
- Dorin Andrica and Eugen J. Ionascu, On the number of polynomials with coefficients in [n], An. St. Univ. Ovidius Constanta, Vol. 22(1),2014, 13-23.
Formula
G.f.: Sum_{k>=2} Sum_{d|k} x^(k^2/d)/((1 - x^k)*(1 - x)). - Miles Wilson, Jun 12 2025
Comments