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

Rainer Rosenthal r.rosenthal at web.de
Tue Aug 2 18:19:55 CEST 2022

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).


(*) 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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