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A344302
Number of cyclic subgroups of the group (C_n)^6, where C_n is the cyclic group of order n.
8
1, 64, 365, 2080, 3907, 23360, 19609, 66592, 88817, 250048, 177157, 759200, 402235, 1254976, 1426055, 2130976, 1508599, 5684288, 2613661, 8126560, 7157285, 11338048, 6728905, 24306080, 12210157, 25743040, 21582653, 40786720, 21243691, 91267520, 29583457
OFFSET
1,2
COMMENTS
Inverse Moebius transform of A160895.
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_6|n} phi(x_1)*phi(x_2)* ... *phi(x_6)/phi(lcm(x_1, x_2, ..., x_6)).
If p is prime, a(p) = 1 + (p^6 - 1)/(p - 1).
From Amiram Eldar, Nov 15 2022: (Start)
Multiplicative with a(p^e) = 1 + ((p^6 - 1)/(p - 1))*((p^(5*e) - 1)/(p^5 - 1)).
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)
MATHEMATICA
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 *)
PROG
(PARI) a160895(n) = sumdiv(n, d, moebius(n/d)*d^6)/eulerphi(n);
a(n) = sumdiv(n, d, a160895(d));
KEYWORD
nonn,mult
AUTHOR
Seiichi Manyama, May 14 2021
STATUS
approved