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

[Max Alekseyev]

I apologize, I did not realize that the given formula for *ordered*
representations, not for *unordered* ones counted by A025428.
Btw, functions  r_k(n) are present in the OEIS:

r_0(n) = A000007(n)
r_1(n) = A000122(n)
r_2(n) = A004018(n)
r_3(n) = A005875(n)
r_4(n) = A000118(n)

I'll check how to get unordered representations out of them.

Regards,
Max

On Fri, Sep 28, 2012 at 3:50 PM, Max Alekseyev <maxale at gmail.com> wrote:
> 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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>>>
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