OFFSET
1,2
COMMENTS
This sequence is related to A060594: in fact, for every square root of unity modulo n, there are either one or two such groups of order 2n.
For n >= 3, a(n) is also the number of equivalence classes of automorphisms of order 2 (involutions) of the dihedral group D_n (see the reference below). - Tom Edgar, Jul 03 2013
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
K. K. A. Cunningham, Tom Edgar, A. G. Helminck, B. F. Jones, H. Oh, R. Schwell and J. F. Vasquez, On the Structure of Involutions and Symmetric Spaces of Dihedral Groups, Note di Mat., Volume 34, No. 2, 2014.
Tom Edgar, Generalized Symmetric Spaces of Dihedral Groups, talk slides, (2012).
FORMULA
a(2) = 2, a(2^2) = 4, a(2^k) = 6 for k >= 3.
a(p^k) = 2 for any odd prime number p and k >= 1.
For other values of n, you can find a(n) by using the fact that the sequence is multiplicative.
Dirichlet g.f.: zeta^2(s)*(1+2^s+2^(1-s)-4^(1-s)+6*4^(-s)) / ( zeta(2*s)*(1+2^s) ). - R. J. Mathar, Jun 01 2011
Sum_{k=1..n} a(k) ~ (9*n/Pi^2) * (log(n) - 1 + 2*gamma - 2*log(2)/3 - 12*Zeta'(2) / Pi^2), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 07 2019
From Mats Granvik, Mar 03 2019: (Start)
a(n) = Sum_{m=1..n} Sum_{k=1..n} (k divides n)*(n/k divides m)*(k divides m + 2).
a(n) = Sum_{k=1..n} (k divides n)*Sum_{j=1..2} (gcd(k, n/k) = j)*j. (End)
EXAMPLE
Two such groups that always exist are the cyclic group Z/(2n)Z and the dihedral group Dih_n. If n is prime, these are the only such groups, so a(p)=2 when p is prime.
For even values of n, we also have the direct product Z/nZ x Z/2Z and the dicyclic group Dic_n. If n = 2p with p prime, there are no other groups, so a(2*p)=4 when p is prime.
There exist five groups (up to isomorphism) of order 2*4 = 8. Four of them have Z/4Z as a subgroup, the two abelian groups: Z/8Z and Z/4Z x Z/2Z, also the two nonabelian groups: the Dihedral group Dih_4 and the Quaternion group or Hamiltonian group: Q_8 = H_8 = Dic_2. So, a(4) = 4. The only group which does not have Z/4Z as a subgroup is the abelian group Z/2Z x Z/2Z x Z/2Z = (Z/2Z)^3. - Bernard Schott, Mar 03 2019
MATHEMATICA
a[1] = 1; a[2] = 2; a[4] = 4; a[n_] := a[n] = (f = FactorInteger[n]; np = Length[f]; Which[ np == 1 && f[[1, 1]] == 2 && f[[1, 2]] >= 3, 6, np == 1 && PrimeQ[f[[1, 1]]] && f[[1, 2]] >= 1, 2, np > 1 && f[[1, 1]] != 2, 2^np, np > 1 && f[[1]] == {2, 1}, 2^np, np > 1 && f[[1]] == {2, 2}, 2^(np+1), np > 1 && f[[1, 1]] == 2 && f[[1, 2]] > 1, 3*2^np, True, 0]); Table[a[n], {n, 1, 60}](* Jean-François Alcover, Nov 22 2011 *)
a[n_] := Sum[Sum[If[Mod[n, k] == 0, If[Mod[m, n/k] == 0, 1, 0], 0]*If[Mod[m + 2, k] == 0, 1, 0], {k, 1, n}], {m, 1, n}]; a /@Range[85] (* Dirichlet convolution, Mats Granvik, Mar 03 2019 *)
a[n_] := Sum[If[Mod[n, k] == 0, Sum[If[GCD[k, n/k] == j, j, 0], {j, 2}],
0], {k, n}]; a /@ Range[85] (* GCD sum, Mats Granvik, Mar 03 2019 *)
a[n_] := Sum[j*Count[Divisors[n], d_ /; GCD[d, n/d] == j], {j, 2}];
a/@Range[85] (* After Jean-François Alcover in A034444. - Mats Granvik, Mar 03 2019 *)
f[2, e_] := Which[e == 1, 2, e == 2, 4, e >= 3, 6]; f[p_, e_] := 2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 17 2020 *)
PROG
(PARI) a(n)=my(k=valuation(n, 2)); max(2*min(k, 3), 1)<<omega(n>>k) \\ Charles R Greathouse IV, Nov 22 2011
CROSSREFS
KEYWORD
easy,nice,nonn,mult
AUTHOR
Ilia Smilga (ilia.smilga(AT)ens.fr), Nov 15 2008
EXTENSIONS
Extended comments, references and confirmed "mult" keyword. - Ilia Smilga (ilia.smilga(AT)ens.fr), Nov 17 2008
STATUS
approved