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A251107
Number of (n+1) X (2+1) 0..2 arrays with no 2 X 2 subblock having the maximum of its diagonal elements greater than the absolute difference of its antidiagonal elements
1
147, 810, 3616, 15281, 67518, 304870, 1369052, 6118942, 27356256, 122402144, 547700666, 2450461705, 10963429165, 49051483677, 219461893981, 981894966609, 4393096675995, 19655164658898, 87939229945039, 393449138915943
OFFSET
1,1
COMMENTS
Column 2 of A251113.
FORMULA
Empirical: a(n) = 11*a(n-1) - 53*a(n-2) + 169*a(n-3) - 400*a(n-4) + 717*a(n-5) - 999*a(n-6) + 1063*a(n-7) - 860*a(n-8) + 543*a(n-9) - 268*a(n-10) + 115*a(n-11) - 40*a(n-12) + 9*a(n-13) - a(n-14).
Empirical formula verified: see link. - Robert Israel, Feb 03 2019
EXAMPLE
Some solutions for n=4:
0 0 2 1 2 2 0 2 2 1 0 1 0 0 1 0 0 0 0 0 2
0 0 2 0 0 2 1 0 2 1 0 1 0 0 1 2 2 2 0 0 2
0 0 0 1 0 2 1 0 1 1 1 0 1 0 0 0 0 0 0 0 0
2 0 0 1 0 0 2 0 0 0 1 1 1 0 0 2 2 2 1 0 0
2 2 2 2 2 2 2 2 1 0 0 0 2 1 0 0 0 0 2 1 0
MAPLE
f:= proc(i, j)
local Li, Lj;
Li:= convert(i+27, base, 3)[1..3];
Lj:= convert(j+27, base, 3)[1..3];
if max(Li[1], Lj[2])<=abs(Li[2]-Lj[1])
and max(Li[2], Lj[3])<=abs(Li[3]-Lj[2])
then 1 else 0
fi
end proc:
T:= Matrix(27, 27, f):
u:= Vector(27, 1):
Tu[0]:= u:
for n from 1 to 30 do Tu[n]:= T . Tu[n-1] od:
seq(u^%T . Tu[n], n=1..30); # Robert Israel, Feb 03 2019
CROSSREFS
Sequence in context: A162701 A063701 A261939 * A349987 A183741 A020328
KEYWORD
nonn
AUTHOR
R. H. Hardin, Nov 30 2014
STATUS
approved