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

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.

Links

R. H. Hardin, Table of n, a(n) for n = 1..210

Robert Israel, Maple-assisted proof of empirical recurrence

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

Adjacent sequences: A251104 A251105 A251106 * A251108 A251109 A251110

Keyword

nonn

Author

R. H. Hardin, Nov 30 2014

Status

approved