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A188555 Number of 4 X n binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally. 2
1, 5, 9, 16, 28, 48, 80, 129, 201, 303, 443, 630, 874, 1186, 1578, 2063, 2655, 3369, 4221, 5228, 6408, 7780, 9364, 11181, 13253, 15603, 18255, 21234, 24566, 28278, 32398, 36955, 41979, 47501, 53553, 60168, 67380, 75224, 83736, 92953, 102913, 113655, 125219 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Row 4 of A188553.
From Miquel A. Fiol, Feb 06 2024: (Start)
a(n) is the number of words of length n, x(1)x(2)...x(n), on the alphabet {0,1,...4}, such that, for i=2,...,n, x(i)=either x(i-1) or x(i-1)-1.
For the bijection between arrays and words, notice that the i-th column consists of 1's and then 0's, and there are x(i)=0 to 4 of 1's.
The number of such words satisfy the recurrence given below and, hence, the empirical/conjectured formulas become true. (End)
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..10000 (terms n = 1..200 from R. H. Hardin)
FORMULA
Empirical: a(n) = (1/24)*n^4 - (1/12)*n^3 + (23/24)*n^2 + (13/12)*n + 3.
Conjectures from Colin Barker, Apr 27 2018: (Start)
G.f.: -(2*x^5 - 7*x^4 + 11*x^3 - 6*x^2 + 1)/(x - 1)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)
EXAMPLE
Some solutions for 4 X 3:
1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0
1 1 0 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0
0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0
0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0
For these solutions, the corresponding words are 221, 432, 222, 443, 100, 333, 110. - Miquel A. Fiol, Feb 06 2024
CROSSREFS
Cf. A188553.
Sequence in context: A233184 A356675 A072174 * A020958 A020750 A020713
KEYWORD
nonn,easy
AUTHOR
R. H. Hardin, Apr 04 2011
EXTENSIONS
a(0)=1 prepended by Alois P. Heinz, Feb 10 2024
STATUS
approved

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Last modified May 23 11:57 EDT 2024. Contains 372763 sequences. (Running on oeis4.)