A251128 T(n,k)=Number of (n+1)X(k+1) 0..1 arrays with no 2X2 subblock having the sum of its diagonal elements greater than the maximum of its antidiagonal elements.
10, 21, 21, 40, 40, 40, 72, 69, 69, 72, 125, 117, 108, 117, 125, 212, 193, 173, 173, 193, 212, 354, 315, 272, 266, 272, 315, 354, 585, 510, 430, 401, 401, 430, 510, 585, 960, 823, 680, 612, 580, 612, 680, 823, 960, 1568, 1326, 1080, 938, 854, 854, 938, 1080, 1326
Offset: 1
Examples
Some solutions for n=4 k=4 ..0..1..1..1..1....0..0..0..1..1....0..0..1..0..1....1..0..0..0..1 ..0..0..0..0..0....0..0..0..1..0....1..0..1..0..1....1..0..0..0..1 ..1..1..1..1..1....0..0..0..1..0....1..0..1..0..1....1..0..0..0..0 ..0..0..0..0..0....0..0..0..1..0....1..0..1..0..1....1..0..0..0..0 ..0..0..0..0..0....1..1..0..1..0....1..0..1..0..0....1..0..0..0..0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..1104
Crossrefs
Column 1 is A001891(n+2)
Formula
Empirical for column k:
k=1: a(n) = 3*a(n-1) -2*a(n-2) -a(n-3) +a(n-4)
k=2: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5)
k=3: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5) for n>6
k=4: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5) for n>6
k=5: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5) for n>6
k=6: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5) for n>6
k=7: a(n) = 4*a(n-1) -5*a(n-2) +a(n-3) +2*a(n-4) -a(n-5) for n>6
Comments