A054584 Number of subgroups of the group generated by a^n=1, b^3=1 and ab=ba.

2, 4, 6, 6, 4, 12, 4, 8, 10, 8, 4, 18, 4, 8, 12, 10, 4, 20, 4, 12, 12, 8, 4, 24, 6, 8, 14, 12, 4, 24, 4, 12, 12, 8, 8, 30, 4, 8, 12, 16, 4, 24, 4, 12, 20, 8, 4, 30, 6, 12, 12, 12, 4, 28, 8, 16, 12, 8, 4, 36, 4, 8, 20, 14, 8, 24, 4, 12, 12, 16, 4, 40, 4, 8, 18, 12, 8, 24, 4, 20, 18, 8, 4

Offset: 1

Author

John W. Layman, Apr 12 2000

Keywords

• nonn
• easy
• nice

Comments

Also the number of subgroups of the group C_n X C_3 (where C_n is the cyclic group of order n). Number of subgroups of the group C_n X C_m is Sum_{i|n,j|m} gcd(i,j).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• M. Hampejs, N. Holighaus, L. Tóth and C. Wiesmeyr, On the subgroups of the group Z_m X Z_n, arXiv preprint arXiv:1211.1797 [math.GR], 2012-2014. - From _N. J. A. Sloane_, Jan 02 2013

Crossrefs

Cf. A060710, A060724, A062011, A060648, A035191, A000005.

Cf. A001620, A079978, A051176.

A row of A216624.

Programs

Haskell

a054584 n = a000005 n + 3 * a079978 n * a000005 (a051176 n) + a035191 n
-- Reinhard Zumkeller, Aug 27 2012

Maple

for n from 1 to 500 do a := ifactors(n):s := 1:for k from 1 to nops(a[2]) do p := a[2][k][1]:e := a[2][k][2]: if p=3 then b := 2*e+1:else b := e+1:fi:s := s*b:od:printf(`%d,`,2*s); od:

Mathematica

f[d_ /; Mod[d, 3] == 0] = 4; f[] = 2; a[n] := Total[f /@ Divisors[n]]; Table[a[n], {n, 1, 100}](* Jean-François Alcover, Nov 21 2011, after Michael Somos *)
f[p_, e_] := e + 1; f[3, e_] := 2*e + 1; a[1] = 2; a[n_] := 2*Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 29 2022 *)

PARI

a(n)=if(n<1, 0, sumdiv(n,d, (d%3==0)*2+2)) /* Michael Somos, Sep 20 2005 */

Formula

a(n) = tau(n)+3*tau(n/3)+A035191(n) if n is congruent to 0 mod 3 else tau(n)+A035191(n), where A035191(n) is the number of divisors of n that are not congruent to 0 mod 3.

a(n)/2 is multiplicative with a(3^e)=2e+1 and a(p^e)=e+1 for p<>3.

Moebius transform is period 3 sequence [2, 2, 4, ...]. - Michael Somos, Sep 20 2005

G.f.: Sum_{k>0} x^k(2+2*x^k+4*x^(2k))/(1-x^(3k)).

From Amiram Eldar, Nov 29 2022: (Start)

Dirichlet g.f.: 2 * zeta(s)^2 * (1 + 1/3^s).

Sum_{k=1..n} a(k) ~ 2*(4*n*log(n) + (8*gamma - 4 - log(3))*n)/3, where gamma is Euler's constant (A001620). (End)

Extensions

Additional comments from Vladeta Jovovic, Oct 25 2001