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

[Max Alekseyev]

Correction: in the formula for A025428, we have k=4:
A025428(n) = \sum_{t=0}^4, (-1)^t * binomial(4,t) * r_t(n)

On Fri, Sep 28, 2012 at 3:48 PM, Max Alekseyev <maxale at gmail.com> wrote:
> 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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