Triangle
Gottfried Helms
Annette.Warlich at t-online.de
Wed Apr 16 15:22:36 CEST 2008
Well -
this was a very fast success - many thanks to the seqfans!
I'll smooth things as much as possible now and shall
post a final result for the description of the
M-matrices.
For more conceptual context you may consider
http://go.helms-net.de/math/tetdocs/ContinuousfunctionalIteration.pdf
a draft handout for the "tetration-forum", where the
M-matrices are described in detail (chap: "fract. iteration
for tetration") and also iterations of other simple functions,
which are defined by powerseries, are discussed in their
basic details.
@Neil: it would be useful to extend the commentary for the
new matrix/sequence, which you already inserted into OEIS.
How far/deep should I go - just describe the iterative
process? It surely would be too much to refer to the whole
context. One or two more examples of K-matrices?
Gottfried
> Am 16.04.2008 10:31 schrieb Gottfried Helms:
>> X1 = // Stirling numbers 2'nd kind, row shifted
>> | 1 . . . . . |
>> | . 1 1 . . . |
>> | . . 1 3 1 . |
>> | . . . 1 7 6 |
>> | . . . . 1 15 |
>> | . . . . . 1 |
>>
>> Then simply
>>
>> H3 = St2 * X1
>>
>> In a matrix-multiplication-scheme
>> X1=
>> | 1 . . . . . |
>> | . 1 1 . . . |
>> | . . 1 3 1 . |
>> | . . . 1 7 6 |
>> | . . . . 1 15 |
>> | . . . . . 1 |
>> St2= H3 =
>> | 1 . . . . . | | 1 . . . . . |
>> | 1 1 . . . . | | 1 1 1 . . . |
>> | 1 3 1 . . . | | 1 3 4 3 1 . |
>> | 1 7 6 1 . . | | 1 7 13 19 13 6 |
>> | 1 15 25 10 1 . | | 1 15 40 85 96 75 |
>> | 1 31 90 65 15 1 | | 1 31 121 335 560 616 |
>>
>> It is nice, that X1 is a simple rowshift of St2.
>> Now it would be good, if also H4 could be computed by a
>> similar simple scheme:
>>
>> H4 =
>> | 1 . . . . . . . . . . . . |
>> | 1 1 1 1 . . . . . . . . . |
>> | 1 3 4 6 4 3 1 . . . . . . |
>> | 1 7 13 26 31 31 25 13 6 1 . . . |
>> | 1 15 40 100 171 220 255 215 156 85 35 10 1 |
>>
>> I fiddled a bit but didn't see it yet ...
>>
>
> Well, that was easy. And seems to give the most simple
> recursive generation scheme.
>
> Row-shifting H3 in the obvious way gives X4
> X4=
> | 1 . . . . . . . . . . . . |
> | . 1 1 1 . . . . . . . . . |
> | . . 1 3 4 3 1 . . . . . . |
> | . . . 1 7 13 19 13 6 1 . . . |
> | . . . . 1 15 40 85 96 75 35 10 1 |
>
> and then again simply
>
> H4 = St2 * X4
>
> H4 =
> | 1 . . . . . . . . . . . . |
> | 1 1 1 1 . . . . . . . . . |
> | 1 3 4 6 4 3 1 . . . . . . |
> | 1 7 13 26 31 31 25 13 6 1 . . . |
> | 1 15 40 100 171 220 255 215 156 85 35 10 1 |
>
> Many thanks for your input!
>
> Gottfried Helms
>
>
>
>>> Proof game for R6R/3Q4/1Q4Q1/4Q3/2 Q4Q/Q4Q2/pp1Q4/kBNN1 KB1:
>> Many thanks! I'm curious who authored the proof game.
Maximilian Hasler:
> I found it on the web, without mention if the poster was the author
> (maybe):
> http://xyqe.com/chess/103-422-re-positions-with-the-most-legal-moves-read.shtml#471269
Thanks. I'm going to guess that Bob (no other info) is indeed the
author of the proof game posted 13 October 2006 but I'm seriously
confused by Bob's posting of the 84-initial-moves-only to that proof
["I have managed to backtrack about 8 moves and have started from the
original position to get the close but have not found a sequence of
moves to arrive at this postion yet."] dated 20 November 2006, because
that post-date is *after* the proof-date.
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