[seqfan] Re: Checking the numerical integration for A355594 (should be A355954)

[Rainer Rosenthal]

Dear Hugo,

maybe there would have been thousands or at least dozens of helpful
answers, if you had written A355954 instead of A355594.
As an old fan of xkcd 356 I am fascinated by the problem of the
resistance distance in the infinite grid.
Thanks for your https://oeis.org/A355565, which opened my eyes for
understanding the square grid.
It is fun to see, how the simple progression of resistances along the
diagonal(*) grows by 1/(2n+1) and that the rest is governed by
symmetries and Laplace in a recursive manner.
It's a sure bet that there is some similar recurrence lurking in the
case of the triangular grid(**). The numerical integration you were
asking for could be helpful in the search for that recurrence. So I'll
cross my fingers that you will receive n really helpful answers (n > 0).

Cheers,
Rainer

(*) The corresponding comment in https://oeis.org/A355566 was approved
just yesterday by njas himself (thanks!)
(**) https://oeis.org/A355585


Am 29.07.2022 um 21:04 schrieb Hugo Pfoertner:
> Dear Sequence Fanatics,
>
> In the sequence https://oeis.org/A355954, the correctness of the calculated
> constant depends entirely on the function intnum for numerical integration
> in PARI being able to calculate an integral with a very high level of
> accuracy if the number of digits for the floating-point calculation is set
> to 1000 or even higher, for example.
>
> It's about the integral with which the electrical resistance between two
> nodes of an infinite triangular grid of one-ohm resistors can be
> calculated. It is given, for example, in the comment of
> https://oeis.org/A355589 . In the appendix of the article by Atkinson and
> Steenwijk linked there, it is also specified as a Mathematica function
> Rtri[n_, p_], and it is claimed that Mma can solve this integral for small
> values of n and p (or j and k in my case) in closed form, such that the
> rational representations given in https://oeis.org/A355585 can be
> calculated directly (potentially also using Simplify). That was supposedly
> already possible with Mathematica before the year 1999.
>
> My request to those of you who are good at using Mathematica and can also
> carry out somewhat more expensive calculations (time, memory requirements)
> is as follows:
>
> Calling Mma with the function Rtri(n,p) as given in the attached article on
> page 491. Is a current Mma implementation able to directly deliver the
> results for small n and p that are in the overview table in A355585?
>
> Does Mma reproduce the specified digits (29 in the current draft) of the
> constant A in A355954 when calculating e.g. A = Rtri(20*10^6,0) -
> log(20*10^6)/(Pi*sqrt(3))? The calculation in PARI was executed  with
> \p1500, i.e. 1500 decimal places (Run time some hours), after checking that
> the number of stable digits of A increases monotonically when increasing
> the call parameter in a geometric progression with a growth factor of 1.2.
>
> Is it possible to calculate even more digits of the constant with Mma, e.g.
> by calculating with 5*10^7, 10^8, ... as arguments instead of 2*10^7? PARI
> then fails because the stack size overflows, although I can run it on a
> computer with 256 GB of main memory and use almost the entire memory as a
> stack. A crazy task would be the calculation of Rtri(118805048562,
> 33636581266), for which I expect a result of about 5.00..+ 5.6*10^-23 (from
> the asymptotic approximation). With PARI I have zero chance of verifying
> this result.
>
> Of course, it would be much better than these attempts with brute force if
> someone could transform the integration similar to the method shown in
> (Cserti 2000) for the square lattice https://oeis.org/A355953 in such a way
> that a closed representation of the constant would be found.
>
> I thank you in advance. Have fun with our wonderful database!
>
> Hugo Pfoertner
>
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