[seqfan] Re: The 4/n problem

[Olivier Gerard]

Hugo,

you might be interested by the following recent article preview by Terence Tao:

http://arxiv.org/abs/1107.1010

on precisely this subject.

Olivier


On Sun, Jul 17, 2011 at 10:31,  <hv at crypt.org> wrote:
> I understand that it remains an open question whether 4/n can be written
> as the sum of 3 unit fractions for all n. It is easy to show that the
> first counterexample, if one exists, would be a prime p of the form
> p == 1 (mod 24); below, I consider only such p.
>
> In that case, if 4/p = 1/a + 1/b + 1/c, with a < b < c [1], we can write
> a = (p + q)/4 to find that 4q / p(p+q) needs to be written as a sum of
> 2 unit fractions (with q == 3 (mod 4)). The first (ie least) q that
> gives a solution can thus be seen in a sense as a measure of how hard
> it is to find a solution for a given p.
>
> When q = 3, for example, there is a solution precisely if (p+3)/4 has
> any prime factor == 2 (mod 3).
>
> Let a(p) represent the least such q, then up to n=10^10 the records
> in a(p) are as follows:
>
>             p => a(p)=q
>            73 => 7
>         1,129 => 11
>         1,201 => 23
>        21,169 => 31
>        67,369 => 35
>        87,481 => 63
>     1,430,641 => 71
>     8,803,369 => 107
>   153,633,769 => 127
>  3,682,770,529 => 131
>  3,853,392,481 => 179
> 10,000,000,000 ------ no further records
>
> This implies that for all p < 10^10, we can find a solution for 4/p with
> at most ceil(179/4) = 45 factorizations.
>
> What sequences revolving around this are interesting enough to submit?
> And what should be the standard values to refer to? Given the motivation,
> it seems clear that we should talk about p rather than about (p - 1)/24;
> it is less obvious whether q is "more fundamental" than (say) (q + 1)/4,
> and though the algebra gets messier, listing (q+1)/4 does highlight that
> it seems unreasonably eager to be composed only of small prime factors:
>  2, 3, 6, 8, 9, 16, 18, 27, 32, 33, 45
>
> (Can anyone see a good reason why that should be so?)
>
> Hugo
>
> [1] Solutions with duplicates among {a, b, c} are in any case not possible.
>
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