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%I #40 Nov 29 2022 02:51:41
%S 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,
%T 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,
%U 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
%N Number of subgroups of the group generated by a^n=1, b^3=1 and ab=ba.
%C 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).
%H Reinhard Zumkeller, <a href="/A054584/b054584.txt">Table of n, a(n) for n = 1..10000</a>
%H M. Hampejs, N. Holighaus, L. Tóth and C. Wiesmeyr, <a href="http://arxiv.org/abs/1211.1797">On the subgroups of the group Z_m X Z_n</a>, arXiv preprint arXiv:1211.1797 [math.GR], 2012-2014. - From _N. J. A. Sloane_, Jan 02 2013
%F 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.
%F a(n)/2 is multiplicative with a(3^e)=2e+1 and a(p^e)=e+1 for p<>3.
%F Moebius transform is period 3 sequence [2, 2, 4, ...]. - _Michael Somos_, Sep 20 2005
%F G.f.: Sum_{k>0} x^k(2+2*x^k+4*x^(2k))/(1-x^(3k)).
%F From _Amiram Eldar_, Nov 29 2022: (Start)
%F Dirichlet g.f.: 2 * zeta(s)^2 * (1 + 1/3^s).
%F 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)
%p 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:
%t 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_ *)
%t 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 *)
%o (PARI) a(n)=if(n<1, 0, sumdiv(n,d, (d%3==0)*2+2)) /* _Michael Somos_, Sep 20 2005 */
%o (Haskell)
%o a054584 n = a000005 n + 3 * a079978 n * a000005 (a051176 n) + a035191 n
%o -- _Reinhard Zumkeller_, Aug 27 2012
%Y Cf. A060710, A060724, A062011, A060648, A035191, A000005.
%Y Cf. A001620, A079978, A051176.
%Y A row of A216624.
%K nonn,easy,nice
%O 1,1
%A _John W. Layman_, Apr 12 2000
%E Additional comments from _Vladeta Jovovic_, Oct 25 2001
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