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%I #47 Mar 09 2024 11:50:49
%S 2,3,3,4,5,4,5,8,7,5,6,12,12,9,6,7,17,20,16,11,7,8,23,32,28,20,13,8,9,
%T 30,49,48,36,24,15,9,10,38,72,80,64,44,28,17,10,11,47,102,129,112,80,
%U 52,32,19,11,12,57,140,201,192,144,96,60,36,21,12,13,68,187,303,321,256,176
%N T(n,k) = Number of n X k binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally.
%C From _Miquel A. Fiol_, Feb 06 2024: (Start)
%C Also, T(n,k) is the number of words of length k, x(1)x(2)...x(k), on the alphabet {0,1,...,n}, such that, for i=2,...,k, x(i)=either x(i-1) or x(i)=x(i-1)-1.
%C For the bijection between arrays and sequences, notice that the i-th column consists of 1's and then 0's, and there are x(i)=0 to n of 1's.
%C Such a bijection implies that all the empirical/conjectured formulas in A188554, A188555, A188556, A188557, A188558, and A188559 become correct.
%C (End)
%H R. H. Hardin, <a href="/A188553/b188553.txt">Table of n, a(n) for n = 1..9934</a>
%F Empirical: T(n,k) = (n+1)*2^(k-1) + (1-k)*2^(k-2) for k < n+3, and then the entire row n is a polynomial of degree n in k.
%F From _Miquel A. Fiol_, Feb 06 2024: (Start)
%F The above empirical formula is correct.
%F It can be proved that T(n,k) satisfies the recurrence
%F T(n,k) = Sum_{r=1..n+1} (-1)^(r+1)*binomial(n+1,r)*T(n,k-r)
%F with initial values
%F T(n,k) = Sum_{r=0..k-1} (n+1-r)*binomial(k-1,r) for k = 1..n+1. (End)
%e Table starts
%e ..2..3..4..5...6...7...8...9...10...11...12....13....14....15....16.....17
%e ..3..5..8.12..17..23..30..38...47...57...68....80....93...107...122....138
%e ..4..7.12.20..32..49..72.102..140..187..244...312...392...485...592....714
%e ..5..9.16.28..48..80.129.201..303..443..630...874..1186..1578..2063...2655
%e ..6.11.20.36..64.112.192.321..522..825.1268..1898..2772..3958..5536...7599
%e ..7.13.24.44..80.144.256.448..769.1291.2116..3384..5282..8054.12012..17548
%e ..8.15.28.52..96.176.320.576.1024.1793.3084..5200..8584.13866.21920..33932
%e ..9.17.32.60.112.208.384.704.1280.2304.4097..7181.12381.20965.34831..56751
%e .10.19.36.68.128.240.448.832.1536.2816.5120..9217.16398.28779.49744..84575
%e .11.21.40.76.144.272.512.960.1792.3328.6144.11264.20481.36879.65658.115402
%e Some solutions for 5 X 3:
%e 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1
%e 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1
%e 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1
%e 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1
%e 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0
%e Some solutions for T(5,3): By taking the sums of the columns in the above arrays we get 555, 100, 000, 543, 322, 432, 554. - _Miquel A. Fiol_, Feb 04 2024
%p T:= (n,k)-> `if`(k<=n+1, (2*n+3-k)*2^(k-2), (n+1-k)*binomial(k-1, n) * add(binomial(n, j-1)/(k-j)*T(n, j)*(-1)^(n-j), j=1..n+1)): seq(seq(T(n, 1+d-n), n=1..d), d=1..15); #_Alois P. Heinz_ in the Sequence Fans Mailing List, Apr 04 2011 [We do not permit programs based on conjectures, but this program is now justified by Fiol's comment. - _N. J. A. Sloane_, Mar 09 2024]
%Y Diagonal is A045623.
%Y Column 4 is A086570.
%Y Rows 2..8 are A022856(n+4), A188554, A188555, A188556, A188557, A188558, A188559.
%Y Upper diagonals T(n,n+i) for i=1..8 give: A001792, A001787(n+1), A000337(n+1), A045618, A045889, A034009, A055250, A055251.
%Y Lower diagonals T(n+i,n) for i=1..7 give: A045891(n+1), A034007(n+2), A111297(n+1), A159694(n-1), A159695(n-1), A159696(n-1), A159697(n-1).
%Y Antidiagonal sums give A065220(n+5).
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Apr 04 2011