[seqfan] Re: Generalized Erdos-Straus conjecture

[Allan Wechsler]

The sequence I am proposing has 1 <= x <= y <= z. A Lisp program I just
resurrected from the dead says that a(2) = 1, not 0 as I stupidly wrote
above, since 5/2 = 1/1 + 1/1 + 1/2.

Anyway, here are twenty terms: 0, 1, 2, 4, 3, 4, 4, 7, 12, 10, 3, 17, 6,
21, 21, 12, 6, 26, 13, 28.  Confirmation would be nice. I will add this
sometime soon (I have a few OEIS obligations stacking up).



On Tue, Jul 16, 2013 at 12:12 PM, T. D. Noe <noe at sspectra.com> wrote:

> See A075248 (number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z
> satisfying 0 < x < y < z.)
>
> Best regards,
>
> Tony
>
>
> At 9:43 AM -0400 7/16/13, Allan Wechsler wrote:
> >A192787 tells how many ways there are to partition 4/n into three positive
> >integer reciprocals. The famous Erdos-Straus conjecture states that for
> >n>1, this is always nonzero.
> >
> >Is there a similar sequence for partitioning 5/n into three reciprocals?
> If
> >I counted right, the answer is no, because I am getting 0, 0, 2, 4, 3, 4,
> >4, and OEIS has only one match, and it has a -2 in it. But I could have
> >counted wrong.
> >
> >_______________________________________________
> >
> >Seqfan Mailing list - http://list.seqfan.eu/
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>