A160892 a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 4.
15, 105, 195, 420, 465, 1365, 855, 1680, 1755, 3255, 1995, 5460, 2745, 5985, 6045, 6720, 4605, 12285, 5715, 13020, 11115, 13965, 8295, 21840, 11625, 19215, 15795, 23940, 13065, 42315, 14895, 26880, 25935, 32235, 26505, 49140, 21105, 40005, 35685, 52080, 25845
Offset: 1
Keywords
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Jin Ho Kwak and Jaeun Lee, Enumeration of graph coverings, surface branched coverings and related group theory, in Combinatorial and Computational Mathematics (Pohang, 2000), ed. S. Hong et al., World Scientific, Singapore 2001, pp. 97-161. See p. 134.
Programs
-
Mathematica
f[p_, e_] := (p^2 + p + 1)*p^(2*e - 2); a[1] = 15; a[n_] := 15*Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* Amiram Eldar, Nov 08 2022 *)
-
PARI
a(n) = {my(f = factor(n)); 15 * prod(i = 1, #f~, (f[i,1]^2 + f[i,1] + 1)*f[i,1]^(2*f[i,2] - 2));} \\ Amiram Eldar, Nov 08 2022
Formula
a(n) = 15*A160889(n). - R. J. Mathar, Feb 07 2011
From Amiram Eldar, Nov 08 2022: (Start)
Sum_{k=1..n} a(k) ~ c * n^3, where c = 5 * Product_{p prime} (1 + 1/p^2 + 1/p^3) = 5 * A330595 = 8.7446649892... .
Sum_{k>=1} 1/a(k) = (zeta(2)*zeta(3)/15) * Product_{p prime} (1 - 2/p^3 + 1/p^5) = 0.09339604419... . (End)