[seqfan] Re: A generalization

[M. F. Hasler]

I submitted this ((card,numdiv) = (3,3)) as A2415674
and also some of the closest vN neighbors (2,3), (3,2), (3,4) ...

Maximilian


On Thu, Apr 24, 2014 at 2:26 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
> 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