- --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

Thank you David. A friend told me that your code demonstrates that your g(x) function appears to have my sequence as coefficients. I must confess I am totally ignorant of this aspect of number theory and how you managed to find that g(x) function. My blind spot in this regard is thwarting me in understanding the implications of the form of your g(x) function, and whether it implies that the f(n) function will continue to give the correct values for higher n.

>

>

> --- In primenumbers@yahoogroups.com,

> "Mark" <mark.underwood@> wrote:

>

> > "It is yet to be seen" if the result or a proof is known.

>

> Here is the generating function for Mark's interesting claim:

>

> g(x)=x^4/((1-x)*(1-x^3)*(1-x^8));

>

> f(n)=floor((n^2+4*n+24)/48);

> print(sum(n=0,100,f(n)*x^n)+O(x^101)-g(x));

>

> O(x^101)

>

> David

>

struggling, but amazed as usual at your contributions,

Mark - Actually David gave simply the g.f. (generating function) of the sequence

f(n)=floor((n^2+4*n+24)/48)

In some sense there is no "higher" number theory involved in this

(at least it does not at all refer to primes).

AFAICS David did not explain why

f(prime(n)) = A025428(prime(n)^2)

(which is Mark's A216374).

Maximilian - --- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@...> wrote:

> Actually David gave simply the g.f. (generating function) of the sequence

Indeed. It seemed to me that

> f(n)=floor((n^2+4*n+24)/48)

g(x) = x^4/((1-x)*(1-x^3)*(1-x^8)) = suminf(n=0,f(n)*x^n)

made Mark's conjecture

> f(prime(n)) = A025428(prime(n)^2)

look neater. I made no attempt to prove the conjecture.

David - There is a formula for r_4(n), when you allow zeros and distinguish signs

and order, see http://mathworld.wolfram.com/SumofSquaresFunction.html

r_4(n)=8*sumdiv(n,d,d*(d%4>0)), and this is for n=p^2 (when p>2 prime) is

r_4(p^2)=8*(p^2+p+1).

Essentially you count every 2^4*4! of these terms (the possibility of zero

and two equal terms is small), this gives r_4(p^2)/16/24~p^2/48+O(p), so

your conjecture is in the right order of magnitude.

[Non-text portions of this message have been removed] - I hacked together another formula, one which does away with the floor function and instead uses the mod (%) function.

f2 = ( ( n^2 + 4*n + (19 * (5 * (n % 48) + 2 )^2 ) % 48) - 24) /48

However it doesn't work for the primes n = 3 (and 2).

If anyone knows how to clean up this obscure form, I would be interested!

Mark

--- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:

>

> How many ways can the square of an odd prime n be expressed as the sum of four non zero squares?

>

> The answer surprisingly appears to have the formula:

>

> f = floor((n^2+4*n+24)/48.)

>

> Here's a GP Pari program which counts the number of ways that an odd prime n can be written as the sum of four non zero squares, and compares that count (k) with the formula derivation f = floor((n^2+4*n+24)/48.) .

>

>

> forprime(n=3,100, k=0; for(s1=1,sqrt((n^2)/4),for(s2=s1,sqrt((n^2 - s1^2)/3) ,for(s3=s2,sqrt((n^2-s1^2 - s2^2)/2) ,if(issquare(n^2-s1^2-s2^2-s3^2),k++)))) ; f = floor((n^2+4*n+24)/48.) ; print1([n,k,f]" "))

>

>

>

> [3, 0, 0] [5, 1, 1] [7, 2, 2] [11, 3, 3] [13, 5, 5] [17, 7, 7] [19, 9, 9] [23, 13, 13] [29, 20, 20] [31, 23, 23] [37, 32, 32] [41, 38, 38] [43, 42, 42] [47, 50, 50] [53, 63, 63] [59, 77, 77] [61, 83, 83] [67, 99, 99] [71, 111, 111] [73, 117, 117] [79, 137, 137] [83, 150, 150] [89, 172, 172] [97, 204, 204]

>

>

> Does it always hold? I'm not sure as I have no proof.

>

>

> Early last week I submitted the sequence 0,1,2,3,5,7,9,13,20... to OEIS (but have not heard back).

>

>

> Mark

>