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A211402 Triangle T(n,k), 0<=k<=n, given by (0,2,0,4,0,6,0,8,0,10,0,...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9,...) where DELTA is the operator defined in A084938.

%I A211402 #14 Feb 22 2013 14:40:36
%S A211402 1,0,1,0,2,3,0,4,18,15,0,8,84,180,105,0,16,360,1500,2100,945,0,32,
%T A211402 1488,10800,27300,28350,10395,0,64,6048,72240,294000,529200,436590,
%U A211402 135135,0,128,24384,463680,2857680,7938000,11060280,7567560,2027025
%N A211402 Triangle T(n,k), 0<=k<=n, given by (0,2,0,4,0,6,0,8,0,10,0,...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9,...) where DELTA is the operator defined in A084938.
%C A211402 A Galton triangle. Essentially the same as A187075.
%F A211402 T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
%F A211402 T(n,k) = 2^(n-k)*A001147(k)*A048993(n,k).
%F A211402 G.f.: F(x,t) = 1 + x*t + (2*x+3*x^2)*t^2/2! + (4*x+18*x^2+15*x^3)*t^3/3!+ ... = Sum_{n = 0..inf}R(n,x)*t^n/n!.
%F A211402 The row polynomials R(n,x) satisfy the recursion R(n+1,x) = 2*(x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.
%e A211402 Triangle begins :
%e A211402 1
%e A211402 0, 1
%e A211402 0, 2, 3
%e A211402 0, 4, 18, 15
%e A211402 0, 8, 84, 180, 105
%e A211402 0, 16, 360, 1500, 2100, 945
%e A211402 0, 32, 1488, 10800, 27300, 28350, 10395
%Y A211402 Cf. A000079, A001147, A048993.
%K A211402 easy,nonn,tabl
%O A211402 0,5
%A A211402 _Philippe Deléham_, Feb 09 2013