[seqfan] Re: n X n binary array quasi-packing problem
Ron Hardin
rhhardin at att.net
Thu Feb 9 21:35:03 CET 2017
My sort of non-mathematical intuition about recurrences is the recurrence gets necessary information from the elements it uses as to how many of what kind of interface it faces in adding a new layer to the existing layer, and if the problem itself shields deeply old rows and columns from affecting the addition of new rows and columns, the empirical recurrence is probably good.
In this particular problem, there's not any shielding. A long-period arrangement with an irregular boundary may suddenly come up that now fits the dimension and the straight boundary.
I don't think it doesn't, but it's a different sort of conjecture.
If you would like to add the sequence, incidentally, please do so. You have many more terms than I do.
rhhardin at mindspring.com rhhardin at att.net (either)
From: Rob Pratt <Rob.Pratt at sas.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Thursday, February 9, 2017 3:21 PM
Subject: [seqfan] Re: n X n binary array quasi-packing problem
The conjecture is true at least up through n = 50.
-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Ron Hardin
Sent: Thursday, February 09, 2017 11:19 AM
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: n X n binary array quasi-packing problem
Probably true but the counterspeculation would be that there's some larger-period arrangement that fits better for big n, if it's below the upper bound.
I'll email when I put the sequence up.
rhhardin at mindspring.com rhhardin at att.net (either)
From: Richard J. Mathar <mathar at mpia-hd.mpg.de>
To: seqfan at seqfan.eu
Sent: Thursday, February 9, 2017 7:27 AM
Subject: [seqfan] Re: n X n binary array quasi-packing problem
On behalf of http://list.seqfan.eu/pipermail/seqfan/2017-February/017253.html :
This leads to the immediate conjecture that the case with b=8 is given by
G.f.: -x^2*(2+5*x+3*x^2+3*x^3+x^4)/(1+x+x^2)^2/(x-1)^3 .
<a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,2,-2,0,-1,1).
a(n) = +a(n-1) +2*a(n-3) -2*a(n-4) -a(n-6) +a(n-7).
27*a(n) = -4-9*n+21*n^2 +4*c(n) -12*c(n-2)-c(n-3).
where c(n) = (-1)^n*A099254(n)
and A033582 is a trisection.
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