[seqfan] Re: A generalization

W. Edwin Clark wclark at mail.usf.edu
Thu Apr 24 20:26:25 CEST 2014


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
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
>
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