[seqfan] Re: Number of partitions of n into 4 nonzero squares

[Max Alekseyev]

See formulae in http://mathworld.wolfram.com/SumofSquaresFunction.html
In particular,
r_k(0) = 1 for any k>=0.
and for n>0:
r_0(n) = 0,
r_1(n) = 2*issquare(n),
r_2(n) = formula (18),
r_3(n) = formula (35),
r_4(n) = formula (39).

Now,
A025428(n) = \sum_{t=0}^4, (-1)^(k-t) * binomial(k,t) * r_t(n)

I have implementation of this formula in PARI/GP, if anybody is interested.

Regards,
Max

On Fri, Sep 28, 2012 at 3:31 PM, Charles Greathouse
<charles.greathouse at case.edu> wrote:
> I was looking at A025428 recently, and noticed that while new programs
> have been added which are more efficient than the original, they still
> take a long time to run* since they don't take advantage of any number
> theory like Jacobi's theorem. Can anyone give an efficient formula
> here? It should be more "tricky" than "hard", removing the
> double-counting of negatives and taking out the 0s.
>
> Actually my motivation was A216374, for which a formula has been
> proposed which is presumably a special case of the yet-unwritten
> formula for A025428.
>
> * n^(1.5 + o(1)) when it should be n^o(1).
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
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