A274228 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's with exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.
2, 3, 2, 4, 4, 2, 5, 8, 5, 2, 6, 12, 12, 6, 2, 7, 18, 21, 16, 7, 2, 8, 24, 36, 32, 20, 8, 2, 9, 32, 54, 60, 45, 24, 9, 2, 10, 40, 80, 100, 90, 60, 28, 10, 2, 11, 50, 110, 160, 165, 126, 77, 32, 11, 2, 12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2, 13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2
Offset: 3
Examples
n=3 => 100, 001 -> T(3,0) = 2. n=4 => 0010, 0100, 1001 -> T(4,0) = 3; 0011, 1100 -> T(4,1) = 2. Triangle starts: 2, 3, 2, 4, 4, 2, 5, 8, 5, 2, 6, 12, 12, 6, 2, 7, 18, 21, 16, 7, 2, 8, 24, 36, 32, 20, 8, 2, 9, 32, 54, 60, 45, 24, 9, 2, 10, 40, 80, 100, 90, 60, 28, 10, 2, 11, 50, 110, 160, 165, 126, 77, 32, 11, 2, 12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2, 13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2, ...
Crossrefs
Programs
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Mathematica
Table[(k + 1) (Binomial[Floor[(n + k - 2)/2], k + 1] + Binomial[Floor[(n + k - 3)/2], k + 1]) + 2 Binomial[Floor[(n + k - 3)/2], k], {n, 3, 14}, {k, 0, n - 3}] // Flatten (* Michael De Vlieger, Jun 16 2016 *)
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PARI
T(n,k) = (k+1)*(binomial((n+k-2)\2,k+1)+binomial((n+k-3)\2,k+1))+2*binomial((n+k-3)\2,k); \\ Michel Marcus, Jun 17 2016