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

[Max Alekseyev]

Oh, there was a bug in the previous version 1.0 of nsg.gp. The updated
version 1.1 is attached.
This bug does not affect the previously computed terms of A061034 -
they all correct.

Regards,
Max

On Dec 3, 2007 7:17 PM, Max Alekseyev <maxale at gmail.com> wrote:
> Maximilian,
>
> Thanks for the reference!
> I've implemented the formula from that paper in PARI/GP. My
> implementation is attached (nsg.gp). With it we can easily get yet
> another way to compute A006116:
>
> { A006116(n) = numsubgrp(2,vector(n,i,1)) }
>
> and A061034 (see below). Thousands of terms of A061034 can be obtained
> in a matter of seconds now. Here is the first hundred:
>
> 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, 374, 4, 4, 4, 30, 2, 4, 4, 32, 2, 8, 2,
> 10, 12, 4, 2, 134, 10, 16, 4, 10, 2, 56, 4, 32, 4, 4, 2, 20, 2, 4, 12,
> 2825, 4, 8, 2, 10, 4, 8, 2, 96, 2, 4, 16, 10, 4, 8, 2, 134, 212, 4, 2,
> 20, 4, 4, 4, 32, 2, 24, 4, 10, 4, 4, 4, 748, 2, 20, 12, 40
>
> Neil, please fill up A061034 with the terms.
>
> Regards,
> Max
>
> P.S. This is implementation of A061034:
>
> { A061034(n) = local(f=factorint(n)); prod(i=1,matsize(f)[1],
> A061034pp(f[i,1],f[i,2]) ) }
>
> \\ for prime power p^k
> { A061034pp(p,k) = res=0; for(i=1, k, aux_part(p, k-i, i, [])); res }
>
> \\ iterate over all partitions
> { aux_part(p, n, m, v) =
>   v = concat(v,m);
>   if(n,
>     for(i=1, min(m,n), aux_part(p, n-i, i, v))
>     ,
>     res=max(res,numsubgrp(p,v));
>   );
>
> }
>
>
> On Dec 3, 2007 5:55 AM, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> > You could add the reference:
> >
> >     On the Subgroups of an Abelian Group
> >         G. A. Miller
> >         The Annals of Mathematics, 2nd Ser., Vol. 6, No. 1. (Oct.,
> > 1904), pp. 1-6.
> >
> >         Stable URL:
> > http://links.jstor.org/sici?sici=0003-486X%28190410%292%3A6%3A1%3C1%3AOTSOAA%3E2.0.CO%3B2-P
> >
> > Paragraph 4 is entitled "total number of subgroups in a group of order p^m"
> >
> > Maximilian
> >
> >
> >
> > On Dec 1, 2007 9:27 PM, 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
> > >
> >
>
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