# [seqfan] The 4/n problem

hv at crypt.org hv at crypt.org
Sun Jul 17 10:31:46 CEST 2011

```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.

```