[seqfan] Re: 1^5+2^5+...+n^5 is a square

[Simon Plouffe]

Hello,

  here is what I have for that sequence :


A031138 11.3246 -(-1-2*x-x^2)/(11*x^2+1-11*x-x^3) ogf
A031138 13.8411 -(x+1)^2/(x-1)/(x^2-10*x+1) ogf
A031138 4.55745 
-13-26*x-13*x^2+(112+55*x+242*x^2-13*x^3)*f(x)+(-1089*x^2-99+1089*x+99*x^3)*f(x)^2 
ogf
A031138 6.55633 
{a(n+3)-11*a(n+2)+11*a(n+1)-a(n),a(0)=1,a(1)=13,a(2)=133} ogf
A031138 7.9513 -(13-33*x+x^3+3*x^2)/(-1+x)/(x+1)/(x^2-10*x+1) lgdogf


The numbers in the second columns is the coefficient of confidence
on the formula based on the length of the original sequence, the
size of the formula and the number of terms in it. The higher the better,

  Best regards,
  Simon Plouffe



Le 22/10/2012 18:15, Ignacio Larrosa Cañestro a écrit :
> El 22/10/2012 16:23, Charles Greathouse escribió:
>> It was recently asked (on MathOverflow, I think) whether the formula 
>> on A031138:
>>
>> a(n) =11*(a(n-1)-a(n-2)) + a(n-3)
>>
>> was proved or merely conjectural. Of course it should be proved to be
>> included as it is, but would someone verify this?
>>
>> This is of course a 6th-degree Diophantine equation:
>>
>> 12m^2 = n^2 (n+1)^2 (2n^2 + 2n - 1)
>>
>> Charles Greathouse
>> Analyst/Programmer
>> Case Western Reserve University
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>
> This is easy , the diofantic is actually quadratic:
>
> In order to
>
> n^2(n+1)^2(2n^2 + 2n - 1)/12
>
> be a square, it is sufficient with that
>
> (2n^2 + 2n - 1)/12 = m^2
>
> And it is certainly a quadratic, and trivial, quadratic diophantic 
> equation.
>