# [seqfan] Re: Tabl formula?

Richard Mathar mathar at strw.leidenuniv.nl
Fri Apr 1 15:23:20 CEST 2011

```http://list.seqfan.eu/pipermail/seqfan/2011-April/014747.html says

rh> Date: Fri, 1 Apr 2011 04:14:32 -0700 (PDT)
rh> From: Ron Hardin
rh> To: seqfan at list.seqfan.eu
rh> Subject: [seqfan] Tabl formula?
rh>
rh> Is there a better formula formulation?
rh>
rh> Row n is a polynomial of degree n, with enough initial terms from the formula
rh> below to determine the polynomial for each entire row; but the formula itself
rh> explicitly gives only the initial part of each row.
rh>
rh> T(n,k)=Number of nXk binary arrays without the pattern 0 1 diagonally,
rh> vertically, antidiagonally or horizontally

..2..3..4..5...6...7...8...9...10...11...12....13....14....15....16.....17
T(1,k) = 1+k

..3..5..8.12..17..23..30..38...47...57...68....80....93...107...122....138
T(2,k) = k^2/2+3*k/2+3

..4..7.12.20..32..49..72.102..140..187..244...312...392...485...592....714
T(3,k) = (k+2)*(k^2+k+12)/6

..5..9.16.28..48..80.129.201..303..443..630...874..1186..1578..2063...2655
T(4,k) = k^4/24+k^3/12+23*k^2/24+35*k/12+5, gf -3/(x-1)^3-3/(x-1)-1/(x-1)^5-2/(x-1)^4

..6.11.20.36..64.112.192.321..522..825.1268..1898..2772..3958..5536...7599
T(5,k) = (k+2)*(k^4-2*k^3+39*k^2+42*k+360)/120 ,
gf 3/(x-1)^2+1/(x-1)^6+3/(x-1)^5+5/(x-1)^4-3/(x-1)+3/(x-1)^3

..7.13.24.44..80.144.256.448..769.1291.2116..3384..5282..8054.12012..17548
T(6,k) = 1/720*k^6-1/240*k^5+11/144*k^4+3/16*k^3+64/45*k^2+259/60*k+7
gf -4/(x-1)^6-6/(x-1)^3-4/(x-1)-1/(x-1)^7-8/(x-1)^4-8/(x-1)^5

..8.15.28.52..96.176.320.576.1024.1793.3084..5200..8584.13866.21920..33932
T(7,k) = 1/5040*(k+2)*(k^6-9*k^5+109*k^4-183*k^3+2410*k^2+2712*k+20160).
gf -4/(x-1)+14/(x-1)^4+1/(x-1)^8+5/(x-1)^7+4/(x-1)^2+6/(x-1)^3+12/(x-1)^6+16/(x-1)^5

rh> Empirical: T(n,k) = (n+1)*2^(k-1) + (1-k)*2^(k-2) for k<n+3, and then the entire
rh> row n is a polynomial of degree n in k.

No pattern so far. An analytical formula would split the polynomial into
two sums associated with the 2 terms:
1+2^(k-1) = (B_k(3)-B_k)/k with Bell polynomials and Bell numbers B.
1+2^(k-2) = (B_(k-1)(3)-B_(k-1))/(k-1).
1-k+(1-k)2^(k-2) = -(B_(k-1)(3)-B_(k-1)).
then convert to polynomials with eq (13) in
http://mathworld.wolfram.com/BellPolynomial.html .

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