[seqfan] Re: Ultra Square Palindromes
hv at crypt.org
hv at crypt.org
Sun Oct 23 16:29:09 CEST 2022
I thought a little about how to generate a larger b-file, and started
by looking at how the possible n-digit endings are distributed over
the last digit (such that row sums give A000993).
I was expecting the shape of this table to be a lot more predictable,
and in particular that at least for 1, 4, 6, 9 each entry would be
either 5 or 10 times the previous entry (with the pattern of 5x and 10x
itself likely predictable), but I see that that's not what happens.
For length n, here are the counts ending in the possible final digits:
n A993(n): 0 1 4 5 6 9
1 6: 1 1 1 1 1 1
2 22: 1 5 5 1 5 5
3 159: 6 25 50 3 50 25
4 1044: 22 250 250 22 250 250
5 9121: 159 2500 1875 212 1875 2500
6 78132: 1044 25000 12500 2088 12500 25000
7 784719: 9121 250000 109375 20848 109375 250000
So for example there are 250 4-digit endings and 1875 5-digit endings
of which the final digit is 4.
(I _think_ this table is a bit too specific to add to OEIS, but I'm happy
to add it if others think it would be of interest.)
Does anyone know of an easy way to look at the n-digit ending of some
square (in decimal) and predict what digits can precede it in (n+1)-digit
endings?
Hugo
Hans Havermann <gladhobo at bell.net> wrote:
:David Griffeath is working on a document dealing with "ultra square palindromes", a draft of which may be read here:
:
:http://chesswanks.com/txt/UltraSquarePalindromes.pdf
:
:Relevant sequences are:
:https://oeis.org/A002778
:https://oeis.org/A002779
:
:My own 1940-term b-files for these were added in 2014 based solely on the 2002 calculations of Feng Yuan. David feels that 20 years later, one should be able to do much better. I've CC'd David herewith. If you can help answer some of the issues presented at the end of his pdf, you may contact him directly. Of course larger b-files are welcome.
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