# # solutions to n +a(1) +a(2) +..+a(j) = a(1)a(2)..a(j)

Roland Bacher Roland.Bacher at ujf-grenoble.fr
Tue Sep 23 09:15:35 CEST 2003

The sequence considered below is a kind of special case (diagonal matrices)
of the following more general problem:

Given strictly positive integers d (corresponding to j below), a and b,
how many symmetric positiv definite d times d integral matrices are there with
trace a and determinant b?

(This number is clearly finite since one has $0<m_{i,i}\leq a$ and
$m_{i,i}m_{j,j}-m_{i,j}^2>0$ which bounds all coefficients $m_{i,j}$
of such a matrix).

In fact, the number of symmetric, positive definite d by d matrices with
given trace is already finite as shown by the above proof which does
not involve the determinant (except for its sign).

The corresponding problem where only the determinant is prescribed
is stupid since it leads to infinite sets (see the REMARK below
which leads to finite sets by replacing determinants by permanents
and by considering (not necessarily symmetric) matrices with strictly
positive natural entries).

The problem below arises when one restricts to diagonal matrices
and asks for the number of solutions with a+n=b. However,
the number of solutions in symmetric positive definite integral
d\times d matrices satisfying trace +n=determinant is generally no longer
finite (consider 2 by 2 matrices with diagonal coefficients 2 and \beta
and off-diagonal coefficient \gamma. One has to count solutions of
2+n=\beta-\gamma^2 which has infinitely many solutions with
(\beta,\gamma)\in N^2 for fixed a fixed natural integer n).

REMARK: There is a similar problem with permanents
(the permanent of a matrix $m_{i,j}$ is defined as
$\sum_\sigma \prod_i m_{i,\sigma(i)}$ where the sum is over all permutations
of indices): Count the number of d\times d matrices having strictly positive
natural entries and given prescribed permanent.

I don't know if some sequences associated to the finite sets
considered above are in the EIS. However, they might be quite hard to
compute.

Best whishes    Roland Bacher

> Let n and j be any fixed positive integers.
>
> How many sequences of j positive integers
> {a(1), a(2), ...,a(j)}
>
> satisfy the equation:
>
> n +a(1) +a(2) +a(3) +..+a(j) =
> a(1) *a(2) *a(3) *...*a(j)
>
> ?
>
> Fixing j and finding a closed-form of the number of solutions in-terms of
> n might be interesting.
>
> For example, for j = 2, the number of solutions is
>
> d(n+1), the number of positive divisors of (n+1).
>
> (I think.)
>
>
> There must be a recursive-definition, anyhow, to find the number of
> solutions with j a's (in terms of n) from the numbers of solutions for
> (j-1) a's, as a possibility.
>
> Thanks,
> Leroy Quet