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A344303
Number of cyclic subgroups of the group (C_n)^7, where C_n is the cyclic group of order n.
8
1, 128, 1094, 8256, 19532, 140032, 137258, 528448, 797891, 2500096, 1948718, 9032064, 5229044, 17569024, 21368008, 33820736, 25646168, 102130048, 49659542, 161256192, 150160252, 249435904, 154764794, 578122112, 305191407, 669317632, 581662904, 1133202048
OFFSET
1,2
COMMENTS
Inverse Moebius transform of A160897.
LINKS
László Tóth, On the number of cyclic subgroups of a finite abelian group, arXiv: 1203.6201 [math.GR], 2012.
FORMULA
a(n) = Sum_{x_1|n, x_2|n, ..., x_7|n} phi(x_1)*phi(x_2)* ... *phi(x_7)/phi(lcm(x_1, x_2, ..., x_7)).
If p is prime, a(p) = 1 + (p^7 - 1)/(p - 1).
From Amiram Eldar, Nov 15 2022: (Start)
Multiplicative with a(p^e) = 1 + ((p^7 - 1)/(p - 1))*((p^(6*e) - 1)/(p^6 - 1)).
Sum_{k=1..n} a(k) ~ c * n^7, where c = (zeta(7)/7) * Product_{p prime} ((1-1/p^6)/(p^2*(1-1/p))) = 0.2784611791... . (End)
MATHEMATICA
f[p_, e_] := 1 + ((p^7 - 1)/(p - 1))*((p^(6*e) - 1)/(p^6 - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* Amiram Eldar, Nov 15 2022 *)
PROG
(PARI) a160897(n) = sumdiv(n, d, moebius(n/d)*d^7)/eulerphi(n);
a(n) = sumdiv(n, d, a160897(d));
KEYWORD
nonn,mult
AUTHOR
Seiichi Manyama, May 14 2021
STATUS
approved