[seqfan] Re: A generalization

[Antreas Hatzipolakis]

On Thu, Apr 24, 2014 at 9:26 PM, W. Edwin Clark <wclark at mail.usf.edu> wrote:

> Generalized to n instead of 12 this is the same as the number of 3-element
> subsets S of {1,...,n} where the sum of the element in S has at least 3
> divisors > 1. I get the following sequence starting at n = 1.
> 0, 0, 1, 2, 6, 11, 22, 35, 55, 78, 110, 145, 192, 245, 312, 386, 476, 572,
> 684, 804, 943, 1091, 1261, 1442, 1647, 1864, 2108, 2366, 2651, 2951, 3281,
> 3629, 4010, 4410, 4845, 5299, 5790, 6301, 6850, 7420, 8031, 8665, 9342,
> 10043, 10788, 11559, 12375, 13215, 14101, 15015,
>
> I don't find it in the OEIS. But I'm not sure it should be.  There will be
>



If does not meet the current criteria of OEIS, it is enough it is archived
in
the list's archive !!
Who knows, in the future the criteria may be changed ......

Thank you !

Antreas






> many such sequences. Take the number of k-subsets of {1,...,n} whose sum
> has property X. Perhaps someone has the energy to put some of these in. Not
> me.
>
> --Edwin
>
>
> On Thu, Apr 24, 2014 at 10:25 AM, Antreas Hatzipolakis <
> anopolis72 at gmail.com
> > wrote:
>
> > The circle below has twelve points evenly spaced on its circumference.
> How
> > many triangles can you make by connecting the points with the following
> > property: the sum of the numbers on the vertices of the triangle created
> > has at least 3 factors greater than 1?
> >
> > The answer is: 145
> >
> > Reference:
> > http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=586846
> >
> > How about if the points are n on the circumference?
> > (formula?)
> >
> > Antreas
>