[seqfan] Re: Simple formulas for Madelung constants?
jonscho at hiwaay.net
jonscho at hiwaay.net
Tue Feb 20 04:02:19 CET 2018
Neil,
Thanks. Today, I've done enough reading about Madelung constants ...
to conclude that the subject is deeper than I want to try to get into
at present. :-/
My thought was, if I had a formula of the form I described that I
could use to generate a sequence of partial sums for, say, A182565,
and -- choosing a simple, straightforward subsequence that did not
diverge -- if I could compute, at sufficiently high precision, a large
enough number of values of that real-valued subsequence and could
identify one or more appropriate convergence acceleration methods
whose application to the selected subsequence would yield a
real-valued output sequence that unmistakably converged to a constant
that not only agreed with all 11 digits currently in the Data for
A182565, but extended several digits beyond that, then that result
might be worth contributing to the OEIS....
A while back, on A259281 (a Madelung-constants-related sequence), I
didn't have to know a lot about crystal lattices to find a way to
extend the existing 44 digits to 400. :-) But I gather that that
sequence, compared to the ones I asked about, was probably a very
simple special case. (For one thing, its partial sums didn't have any
divergent subsequences.) So, as far as attempting to extend any of
the OEIS's remaining Madelung constant sequences that have
keyword:more, I'm willing to leave that to the experts. :-)
(And instead spend more time catching up on neglected necessities,
like housework <sigh> ...) :-(
Thanks again,
-- Jon
Quoting Neil Sloane <njasloane at gmail.com>:
> Jon, The answer is certainly Yes. However, you
> have to be very careful when calculating these Madelung constants. As you
> will
> see if you look at the literature (Finch page 76 is a good start) these
> sums need to
> be interpreted very carefully, since they have subsequences
> which are divergent.
>
> Yes, you can write down expressions like
>
> Sum_{i, j, k not all 0} f(i,j,k)
>
> where f(i,j,k) = (-1)^(i+j+k)/sqrt(i^2+j^2+k^2).
>
> but then you need to say exactly what you mean. We are working here with
> potentially
> divergent sums. My advice would be not to blindly start running some
> computer program.
> Do some research first.
>
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
> On Sun, Feb 18, 2018 at 10:04 PM, <jonscho at hiwaay.net> wrote:
>
>> The Formula section of A085469 gives a formula of the form
>>
>> Sum_{i, j, k not all 0} f(i,j,k)
>>
>> (In this case, f(i,j,k) = (-1)^(i+j+k)/sqrt(i^2+j^2+k^2).)
>>
>> Is there a similar formula (differing only in its function f(i,j,k)) for
>> A181152, A182565, A182566, or A182567?
>>
>> Thanks!
>>
>> -- Jon
>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
> --
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