|
%I #24 Nov 15 2022 09:17:37
%S 1,64,365,2080,3907,23360,19609,66592,88817,250048,177157,759200,
%T 402235,1254976,1426055,2130976,1508599,5684288,2613661,8126560,
%U 7157285,11338048,6728905,24306080,12210157,25743040,21582653,40786720,21243691,91267520,29583457
%N Number of cyclic subgroups of the group (C_n)^6, where C_n is the cyclic group of order n.
%C Inverse Moebius transform of A160895.
%H Amiram Eldar, <a href="/A344302/b344302.txt">Table of n, a(n) for n = 1..10000</a>
%H László Tóth, <a href="http://arxiv.org/abs/1203.6201">On the number of cyclic subgroups of a finite abelian group</a>, arXiv: 1203.6201 [math.GR], 2012.
%F a(n) = Sum_{x_1|n, x_2|n, ..., x_6|n} phi(x_1)*phi(x_2)* ... *phi(x_6)/phi(lcm(x_1, x_2, ..., x_6)).
%F If p is prime, a(p) = 1 + (p^6 - 1)/(p - 1).
%F From _Amiram Eldar_, Nov 15 2022: (Start)
%F Multiplicative with a(p^e) = 1 + ((p^6 - 1)/(p - 1))*((p^(5*e) - 1)/(p^5 - 1)).
%F Sum_{k=1..n} a(k) ~ c * n^6, where c = (zeta(6)/6) * Product_{p prime} ((1-1/p^5)/(p^2*(1-1/p))) = 0.32592074105... . (End)
%t f[p_, e_] := 1 + ((p^6 - 1)/(p - 1))*((p^(5*e) - 1)/(p^5 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* _Amiram Eldar_, Nov 15 2022 *)
%o (PARI) a160895(n) = sumdiv(n, d, moebius(n/d)*d^6)/eulerphi(n);
%o a(n) = sumdiv(n, d, a160895(d));
%Y Cf. A000010, A013664, A060648, A064969, A160895, A280184, A344219, A344303, A344304, A344305, A344306.
%K nonn,mult
%O 1,2
%A _Seiichi Manyama_, May 14 2021
|
