A131084 A129686 * A007318. Riordan triangle (1+x, x/(1-x)).
1, 1, 1, 0, 2, 1, 0, 2, 3, 1, 0, 2, 5, 4, 1, 0, 2, 7, 9, 5, 1, 0, 2, 9, 16, 14, 6, 1, 0, 2, 11, 25, 30, 20, 7, 1, 0, 2, 13, 36, 55, 50, 27, 8, 1, 0, 2, 15, 49, 91, 105, 77, 35, 9, 1
Offset: 1
Examples
The triangle T(n, k) begins: n\k 0 1 2 3 4 5 6 7 8 9 10 ... 0: 1 1: 1 1 2: 0 2 1 3: 0 2 3 1 4: 0 2 5 4 1 5: 0 2 7 9 5 1 6: 0 2 9 16 14 6 1 7: 0 2 11 25 30 20 7 1 8: 0 2 13 36 55 50 27 8 1 9: 0 2 15 49 91 105 77 35 9 1 10: 0 2 17 64 140 196 182 112 44 10 1 ... Reformatted. - _Wolfdieter Lang_, Jan 06 2015
Formula
A129686(signed): (1,1,1,...) in the main diagonal and (-1,-1,-1, ...) in the subsubdiagonal); * A007318, Pascal's triangle; as infinite lower triangular matrices.
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2*x + 3*x^2/2! + x^3/3!) = 2*x + 7*x^2/2! + 16*x^3/3! + 30*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f. column k: (1+x)*(x/(1-x))^k, k >= 0. (Riordan property). - Wolfdieter Lang, Jan 06 2015
T(n, 0) = 1 if n=0 or n=1 else 0; T(n, k) = binomial(n-1,k-1) + binomial(n-2,k-1)*[n-1 >= k] if n >= k >= 1, where [S] = 1 if S is true, else 0, and T(n, k) = 0 if n < k. - Wolfdieter Lang, Jan 08 2015
Extensions
Edited: Added Riordan property (see Philippe Deléham comment) in name. - Wolfdieter Lang, Jan 06 2015
Comments