[seqfan] Re: Symmetrize a formula?

[Ron Hardin]

It takes 3 values to determine the formula for a line.  I suppose the formulas for columns 7,8,9 could be used to develop 3 values in any row, and then those 3 values would determine the formula for the row (and also the column by symmetry).  Maybe some math software could then reduce that to a symmetric formula.

 
rhhardin at mindspring.com
rhhardin at att.net (either)


>________________________________
> From: Ron Hardin <rhhardin at att.net>
>To: "seqfan at list.seqfan.eu" <seqfan at list.seqfan.eu> 
>Sent: Monday, December 29, 2014 2:34 PM
>Subject: [seqfan] Symmetrize a formula?
> 
>
>The  a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) recurrence for every row and column of the T(n,k) array for this gives rise to a a*3^n+b*2^n+c formula for each row but I'd expect some symmetric form to result, one of these in n and one of these in k added together with a constant constant for the whole large end of the array.
>
>I don't exactly see how to get there from where I've gotten (see below).
>
>
>/tmp/eur
>T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with every 2X2 subblock diagonal minus antidiagonal sum nondecreasing horizontally and vertically
>
>Table starts
>.....81.....414....1388....3639....8501...19701...48293..126357...346997
>....414....1975....4782....8554...15220...31630...74324..188438...502364
>...1388....4782....7782...11147...19254...39463...89556..218326...561564
>...3639....8554...11147...15730...26831...52912..114717..266911...656997
>...8501...15220...19254...26831...44242...83003..170184..373130...864720
>..19701...31630...39463...52912...83003..147332..285681..590963..1287225
>..48293...74324...89556..114717..170184..285681..526430.1036512..2142374
>.126357..188438..218326..266911..373130..590963.1036512.1956194..3881256
>.346997..502364..561564..656997..864720.1287225.2142374.3881256..7444718
>.982677.1387310.1505134.1694263.2104994.2936843.4611192.7988474.14828736
>
>Empirical for column k:
>k=1: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>10
>k=2: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>k=3: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>k=4: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>k=5: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>k=6: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>k=7: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3) for n>9
>Empirical for column k:
>k=1: a(n) = 400*3^(n-3) + 205*2^(n-1) + 2917 for n>7
>k=2: a(n) = 529*3^(n-3) + 444*2^(n-1) + 3059 for n>6
>k=3: a(n) = 529*3^(n-3) + 673*2^(n-1) + 3635 for n>6
>k=4: a(n) = 529*3^(n-3) + 1039*2^(n-1) + 5372 for n>6
>k=5: a(n) = 529*3^(n-3) + 1832*2^(n-1) + 10087 for n>6
>k=6: a(n) = 529*3^(n-3) + 3431*2^(n-1) + 23248 for n>6
>k=7: a(n) = 529*3^(n-3) + 6631*2^(n-1) + 59197 for n>6
>k=8: a(n) = 529*3^(n-3) + 13031*2^(n-1) + 159679 for n>6
>k=9: a(n) = 529*3^(n-3) + 25831*2^(n-1) + 446341 for n>6
>
>Some.solutions.for.n=4.k=4..
>..0..1..2..2..2....0..2..1..2..1....0..1..1..1..1....1..2..0..0..1..
>..0..0..1..1..1....2..2..1..2..1....2..2..2..2..2....1..2..0..0..1..
>..1..0..1..1..1....2..2..1..2..1....2..1..1..1..1....1..2..0..0..1..
>..1..0..1..1..1....1..1..0..1..0....1..0..0..0..1....1..2..0..0..1..
>..1..0..1..1..2....1..1..0..1..0....1..0..0..0..2....1..2..0..0..1..
>
>
>
>rhhardin at mindspring.com
>rhhardin at att.net (either)
>
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