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a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 6.
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%I #11 Nov 09 2022 07:56:54

%S 63,1953,7623,31248,49203,236313,176463,499968,617463,1525293,1014615,

%T 3781008,1949283,5470353,5953563,7999488,5590683,19141353,8666343,

%U 24404688,21352023,31453065,18431343,60496128,30751875,60427773,50014503,87525648,46150083,184560453

%N a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 6.

%H Amiram Eldar, <a href="/A160896/b160896.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) = 63*A160893(n). - _R. J. Mathar_, Mar 16 2016

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

%F Sum_{k=1..n} a(k) ~ c * n^5, where c = (63/5) * Product_{p prime} (1 + (p^4-1)/((p-1)*p^5)) = 23.9347523175... .

%F Sum_{k>=1} 1/a(k) = (zeta(4)*zeta(5)/63) * Product_{p prime} (1 - 2/p^5 + 1/p^9) = 0.01658573169... . (End)

%t f[p_, e_] := p^(4*e - 4)*(1 + p + p^2 + p^3 + p^4); a[1] = 63; a[n_] := 63 * Times @@ f @@@ FactorInteger[n]; Array[a, 25] (* _Amiram Eldar_, Nov 08 2022 *)

%o (PARI) a(n) = {my(f = factor(n)); 63 * prod(i = 1, #f~, (f[i,1]^4 + f[i,1]^3 + f[i,1]^2 + f[i,1] + 1)*f[i,1]^(4*f[i,2] - 4));} \\ _Amiram Eldar_, Nov 08 2022

%Y Cf. A000010, A013662, A013663, A160893.

%K nonn

%O 1,1

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