A233295 Riordan array ((1+x)/(1-x)^3, 2*x/(1-x)).
1, 4, 2, 9, 10, 4, 16, 28, 24, 8, 25, 60, 80, 56, 16, 36, 110, 200, 216, 128, 32, 49, 182, 420, 616, 560, 288, 64, 64, 280, 784, 1456, 1792, 1408, 640, 128, 81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256, 100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512
Offset: 0
Examples
Triangle begins : 1 4, 2 9, 10, 4 16, 28, 24, 8 25, 60, 80, 56, 16 36, 110, 200, 216, 128, 32 49, 182, 420, 616, 560, 288, 64 64, 280, 784, 1456, 1792, 1408, 640, 128 81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256 100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512
Formula
G.f. for the column k: 2^k*(1+x)/(1-x)^(k+3).
T(n,k) = 2^k*(binomial(n,k)+3*binomial(n,k+1)+2*binomial(n,k+2)), 0<=k<=n.
T(n,0) = 2*T(n-1,0)-T(n-2,0)+2, T(n,k)=2*T(n-1,k)+2*T(n-1,k-1)-2*T(n-2,k-1)-T(n-2,k) for k>=1, T(0,0)=1, T(1,0)=4, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k) = A060188(n+2).
Sum_{k=0..n} T(n,k)*(-1)^k = n+1.
T(n,k) = 2*sum_{j=1..n-k+1} T(n-j,k-1).
T(n,k) = 2^k*A125165(n,k).
T(n,n) = 2^n=A000079(n).
T(n,0) = (n+1)^2=A000290(n+1).
exp(2*x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(2*x)*(16 + 28*x + 24*x^2/2! + 8*x^3/3!) = 16 + 60*x + 200*x^2/2! + 616*x^3/3! + 1792*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), 2*x/(1 - x) ). Cf. A125165. - Peter Bala, Dec 21 2014
Comments