a property of multisets: unit fractions with a twist

[Max Alekseyev]

On Dec 11, 2007 4:50 PM, Max Alekseyev <maxale at gmail.com> wrote:

> > I suspect a bit more work would yield a general approach to finding the
> > number of solutions for (1+1/a_1)(1+1/a_2) = x for arbitrary rational x,
> > which would at least bring a(7) within range.
>
> The number of solutions to (1+1/a_1)(1+1/a_2) = p/q where p and q are
> co-prime integers can be extracted from the prime factorization of N =
> p^2 + p*q - q^2. Namely, the number of solutions equals to the number
> of divisors of N that are congruent to p modulo q-p.
> If we also require that a_1 <= a_2 then we need to restrict our
> counting to divisors of N that are <= sqrt(N).

Ops. I've done my arithmetics wrong. The correct value of N is:
N = p*q (which is easier to factor in general)
and we need to look for divisors that are congruent to -q modulo p-q.

The values of a_1 and a_2 for a particular divisor d of N, congruent
to -q modulo p-q, is given by:
a_1 = (q + d) / (p-q).
a_2 = (q + (p*q/d)) / (p-q).

For example, for p=35 and q=32, we need to find divisors of 35*32 that
are congruent to -32 mod (35-32) = 1 mod 3. There are such 5 divisors
<= sqrt(35*32), namely:
d = 1, 4, 7, 16, 28
giving the following solutions:
(a_1,a_2) = (11, 384), (12, 104), (13, 64), (16, 34), (20, 24).

Regards,
Max