To extend A018216 Maximal number of subgroups in a group with n elements

[Jonathan Post]

Excellent!

We can get another sequence out of this process of seeking the maxima.

a(n) = Sum of number of subgroups of all nonisomorphic groups of order n.

Comment: with multiplicity, i.e. Z_2 X Z_2 has 3 isomorphic subgroups
Z_2, so contributes 3 to the sum for a(4).  Every group has itself and
the trivial group as subgroups, so a(1) = 1 where these are the same,
and n>1 has a(n) => 2.  a(p) = 2 for prime p.

 1, 2, 2, 6, 2, 4, 2, 24, 7, 4, 2, 12, 2, 4, 4

Offset 1,2

examples:
n  a(n)  explanation (not list itself and trivial group)
1  1
2  2
3  2
4  6     Z_4 has Z_2,  Z_2 X Z_2 has 3 of Z_2
5  2
6  4     Z_6 = Z_3 X Z_2 has Z_3, Z_2
7  2
8  24   Z_8 has Z_4 and Z_2;
           Z_4 X Z_2 has Z_2^2 and 2 of Z_4 and 3 of Z_2
           Z_2^3 has 7 of Z_2^2 and 7 of Z_2
9  7     Z_9 has Z_3; Z_3^2 has 4 of Z_3
10 4    Z_10 = Z_5 X Z_2 has Z5, Z_2
11 2
12 12  Z_12 = Z_4 X Z_3 has Z_6, Z_4, Z_3, Z_2;
           Z_6 X Z_2 = Z_3 X Z_2^2 has 6 of Z_6,
           Z_3, 3 of Z_2
13 2
14 4    Z_14 = Z_7 X Z_2 has Z_7, Z_2
15 4    Z_15 = Z_5 X Z_3 has Z_5, Z_3
etc.

This new seq is a row sum of a table that more explicitly embodies the
above explanations. A061034 is the maximum value of each row.

Best,

Jonathan Vos Post

On 12/1/07, Max Alekseyev <maxale at gmail.com> wrote:
> On Nov 30, 2007 4:41 PM, Christian G. Bower <bowerc at usa.net> wrote:
> > I think C2^4 has 67 subgroups (1 trivial, 15 C2, 35 C2^2, 15 C2^3, 1 itself).
> > I would suspect that's the largest case, but I'm not in the mood to check all
> > 14 groups of order 16 (and to dig up a description of the more obsure ones.)
>
> I've checked all abelian groups of order 16 and A006116(4)=67 is
> indeed the largest case:
>
> C16 has 5 subgroups
> C2 x C8 has 11 subgroups
> (C2)^2 x C4 has 27 subgroups
> (C2)^4 has 67 subgroups
> (C4)^2 has 15 subgroups
>
> This gives boost to A061034:
>
> %S A061034 1,2,2,5,2,4,2,16,6,4,2,10,2,4,4,67,2,12,2,10,4,4,2,32,8,4,28,10,2,8,2
>
> Neil, please update A061034 accordingly.
>
> Regards,
> Max
>