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?
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