# Series of Square Coefficients

mlb at fxpt.com mlb at fxpt.com
Sat Apr 26 20:32:13 CEST 2003

```>=Paul D Hanna <pauldhanna at juno.com>
>Consider a power series A(x) with positive integer coefficients
>such that the coefficients of
> A(x)^2 + A(x) - 1
>consist entirely of squares.

>An example of this is found in A083352, which is to be the least case.
>Are there other, more interesting solutions?

>What is somewhat surprising to me is that the terms (after the initial
1)
>of A085332 all seem to be multiples of 3.

Solving

A(x)^2 + A(x) - 1  =  1 + f(x)

for A gives two roots, one with A(0)=-2 and the other with A(0)=1,
whose expansions in powers of f(x) have denominators which are powers of
3.

This 3 simply comes from the particular choice of polynomial in A.  A different
choice gives different denominator powers (eg dropping the 1+ on the right
hand side gives powers of 5 instead).

Anyway, for this case note that the original f's coefficients are powers
of 3.  If we choose  f  such that its coefficients are squares of the multiples
of some  m  we have

1 + f(x)  =  1 + sum((m*n)^2 * x^n)  =  1 - (m^2 x (x+1) / (x-1)^3)

The coefficients of this particular A turn out to be polynomials in m^2,
divided of course by powers of 3.

Not surprisingly, choosing any  m = 3k  cancels those denominators (leaving
as numerators polynomials in k^2 multiplied by 3 k^2).

(Conversely, choosing any m not divisible by 3 seems unlikely to cancel
nicely
to give integers.)

Of course all sorts of other  f  could also be chosen, with approximately
the same qualitative results (eg f(x) with cubes for coefficients).

Perhaps this satisfactorily explains the observed numerology?

```