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a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 4.
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%I #13 Nov 09 2022 07:56:41

%S 15,105,195,420,465,1365,855,1680,1755,3255,1995,5460,2745,5985,6045,

%T 6720,4605,12285,5715,13020,11115,13965,8295,21840,11625,19215,15795,

%U 23940,13065,42315,14895,26880,25935,32235,26505,49140,21105,40005,35685,52080,25845

%N a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 4.

%H Amiram Eldar, <a href="/A160892/b160892.txt">Table of n, a(n) for n = 1..10000</a>

%H Jin Ho Kwak and Jaeun Lee, <a href="https://doi.org/10.1142/9789812799890_0005">Enumeration of graph coverings, surface branched coverings and related group theory</a>, in Combinatorial and Computational Mathematics (Pohang, 2000), ed. S. Hong et al., World Scientific, Singapore 2001, pp. 97-161. See p. 134.

%F a(n) = 15*A160889(n). - _R. J. Mathar_, Feb 07 2011

%F From _Amiram Eldar_, Nov 08 2022: (Start)

%F 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... .

%F Sum_{k>=1} 1/a(k) = (zeta(2)*zeta(3)/15) * Product_{p prime} (1 - 2/p^3 + 1/p^5) = 0.09339604419... . (End)

%t 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 *)

%o (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

%Y Cf. A000010, A002117, A013661, A160889, A330595.

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Nov 19 2009