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Number of cyclic subgroups of the group (C_n)^n, where C_n is the cyclic group of order n.
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%I #30 Oct 04 2023 06:49:32

%S 1,4,14,136,782,23360,137258,4210816,64576643,2500000768,28531167062,

%T 2229573502976,25239592216022,1852001137606656,54736740117685528,

%U 2305878194659557376,51702516367896047762,6557734713069408616448,109912203092239643840222

%N Number of cyclic subgroups of the group (C_n)^n, where C_n is the cyclic group of order n.

%H Seiichi Manyama, <a href="/A344201/b344201.txt">Table of n, a(n) for n = 1..387</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_n|n} phi(x_1)*phi(x_2)* ... *phi(x_n)/phi(lcm(x_1, x_2, ... , x_n)).

%F a(n) = Sum_{d|n} b(d, n), where b(n, k) = ( Sum_{d|n} mu(n/d) * d^k )/phi(n).

%F If p is prime, a(p) = 1 + (p^p - 1)/(p - 1).

%t b[n_, k_] := DivisorSum[n, MoebiusMu[n/#] * #^k &] / EulerPhi[n]; a[n_] := DivisorSum[n, b[#, n] &]; Array[a, 20] (* _Amiram Eldar_, Oct 04 2023 *)

%o (PARI) b(n, k) = sumdiv(n, d, moebius(n/d)*d^k)/eulerphi(n);

%o a(n) = sumdiv(n, d, b(d, n));

%Y Cf. A060648, A064969, A280184, A344219.

%K nonn

%O 1,2

%A _Seiichi Manyama_, May 12 2021