A188553 T(n,k) = Number of n X k binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally.

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

Offset: 1

Author

R. H. Hardin, Apr 04 2011

Keywords

• nonn
• tabl

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)

Examples

			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
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..9934

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).

Programs

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]

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)