OFFSET
1,2
COMMENTS
a(n) is the number of lattices L in Z^4 such that the quotient group Z^4 / L is C_nm x (C_m)^3 (and also (C_nm)^3 x C_m), for every m>=1. - Álvar Ibeas, Oct 30 2015
LINKS
Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000
Jin Ho Kwak and Jaeun Lee, Enumeration of graph coverings, surface branched coverings and related group theory, in Combinatorial and Computational Mathematics (Pohang, 2000), ed. S. Hong et al., World Scientific, Singapore 2001, pp. 97-161. See p. 134.
FORMULA
a(n) = J_4(n)/J_1(n) = J_4(n)/phi(n) = A059377(n)/A000010(n), where J_k is the k-th Jordan totient function. - Enrique Pérez Herrero, Oct 19 2010
Multiplicative with a(p^e) = p^(3e-3)*(1+p+p^2+p^3). - R. J. Mathar, Jul 10 2011
For squarefree n, a(n) = A000203(n^3). - Álvar Ibeas, Oct 30 2015
Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^3/((p^3-1)*(p^3+p^2+p+1))) = 1.115923965261131974852254388404911045036763705978837384729819264463715993... - Vaclav Kotesovec, Sep 20 2020
Sum_{k=1..n} a(k) ~ c * n^4, where c = (1/4) * Product_{p prime} (1 + 1/p^2 + 1/p^3 + 1/p^4) = 0.4629765396... . - Amiram Eldar, Nov 08 2022
EXAMPLE
There are 1395 = A160870(8,4) lattices of volume 8 in Z^4. Among them, a(8) = 960 give the quotient group C_8 and a(2) = 15 give C_2 x C_2 x C_2.
Among the lattices of volume 64 in Z^4, there are a(4) = 120 such that the quotient group is C_4 x C_4 x C_4 and other 120 with quotient group C_8 x (C_2)^3.
MAPLE
A160891 := proc(n) a := 1 ; for f in ifactors(n)[2] do p := op(1, f) ; e := op(2, f) ; a := a*p^(3*e-3)*(1+p+p^2+p^3) ; end do; a; end proc:
seq(A160891(n), n=1..20) ; # R. J. Mathar, Jul 10 2011
MATHEMATICA
A160891[n_]:=DivisorSum[n, MoebiusMu[n/#]*#^(5-1)/EulerPhi[n]&] (* Enrique Pérez Herrero, Oct 19 2010 *)
f[p_, e_] := p^(3 e - 3)*(1 + p + p^2 + p^3); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* Amiram Eldar, Nov 08 2022 *)
PROG
(PARI) vector(50, n, sumdiv(n^3, d, if(ispower(d, 4), moebius(sqrtnint(d, 4))*sigma(n^3/d), 0))) \\ Altug Alkan, Oct 30 2014
(PARI) a(n) = {f = factor(n); for (i=1, #f~, p = f[i, 1]; f[i, 1] = p^(3*f[i, 2]-3)*(1+p+p^2+p^3); f[i, 2] = 1; ); factorback(f); } \\ Michel Marcus, Nov 12 2015
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
N. J. A. Sloane, Nov 19 2009
EXTENSIONS
Definition corrected by Enrique Pérez Herrero, Aug 22 2010
STATUS
approved