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 A188553 T(n,k) = Number of n X k binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally. 8
 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, 30, 49, 48, 36, 24, 15, 9, 10, 38, 72, 80, 64, 44, 28, 17, 10, 11, 47, 102, 129, 112, 80, 52, 32, 19, 11, 12, 57, 140, 201, 192, 144, 96, 60, 36, 21, 12, 13, 68, 187, 303, 321, 256, 176 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS From Miquel A. Fiol, Feb 06 2024: (Start) 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. 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. Such a bijection implies that all the empirical/conjectured formulas in A188554, A188555, A188556, A188557, A188558, and A188559 become correct. (End) LINKS R. H. Hardin, Table of n, a(n) for n = 1..9934 FORMULA 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. From Miquel A. Fiol, Feb 06 2024: (Start) The above empirical formula is correct. It can be proved that T(n,k) satisfies the recurrence T(n,k) = Sum_{r=1..n+1} (-1)^(r+1)*binomial(n+1,r)*T(n,k-r) with initial values T(n,k) = Sum_{r=0..k-1} (n+1-r)*binomial(k-1,r) for k = 1..n+1. (End) EXAMPLE Table starts ..2..3..4..5...6...7...8...9...10...11...12....13....14....15....16.....17 ..3..5..8.12..17..23..30..38...47...57...68....80....93...107...122....138 ..4..7.12.20..32..49..72.102..140..187..244...312...392...485...592....714 ..5..9.16.28..48..80.129.201..303..443..630...874..1186..1578..2063...2655 ..6.11.20.36..64.112.192.321..522..825.1268..1898..2772..3958..5536...7599 ..7.13.24.44..80.144.256.448..769.1291.2116..3384..5282..8054.12012..17548 ..8.15.28.52..96.176.320.576.1024.1793.3084..5200..8584.13866.21920..33932 ..9.17.32.60.112.208.384.704.1280.2304.4097..7181.12381.20965.34831..56751 .10.19.36.68.128.240.448.832.1536.2816.5120..9217.16398.28779.49744..84575 .11.21.40.76.144.272.512.960.1792.3328.6144.11264.20481.36879.65658.115402 Some solutions for 5 X 3: 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 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 MAPLE 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] CROSSREFS Diagonal is A045623. Column 4 is A086570. Rows 2..8 are A022856(n+4), A188554, A188555, A188556, A188557, A188558, A188559. Upper diagonals T(n,n+i) for i=1..8 give: A001792, A001787(n+1), A000337(n+1), A045618, A045889, A034009, A055250, A055251. 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). Antidiagonal sums give A065220(n+5). Sequence in context: A321440 A115729 A115728 * A335680 A026354 A179840 Adjacent sequences: A188550 A188551 A188552 * A188554 A188555 A188556 KEYWORD nonn,tabl AUTHOR R. H. Hardin, Apr 04 2011 STATUS approved

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