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Triangle read by rows, T(n,k) for 0<=k<=n, generalizing A098742.
1

%I #15 Mar 17 2020 14:21:25

%S 1,1,1,3,3,1,9,12,6,1,33,51,34,10,1,135,237,193,79,15,1,609,1188,1132,

%T 584,160,21,1,2985,6381,6920,4268,1510,293,28,1,15747,36507,44213,

%U 31542,13576,3464,497,36,1,88761,221400,295314,238261,120206,37839,7231,794

%N Triangle read by rows, T(n,k) for 0<=k<=n, generalizing A098742.

%C Full concordance with A098742 would require two zero rows at the top of the triangle which we omitted for simplicity.

%C Matrix inverse is A137338. - _Peter Luschny_, Sep 21 2012

%F Recurrence: T(0,0)=1, T(0,k)=0 for k>0 and for n>=1 T(n,k) = T(n-1,k-1) + (k+1)*T(n-1,k) + (k+2)*T(n-1,k+1).

%e [0] [1]

%e [1] [1, 1]

%e [2] [3, 3, 1]

%e [3] [9, 12, 6, 1]

%e [4] [33, 51, 34, 10, 1]

%e [5] [135, 237, 193, 79, 15, 1]

%e [6] [609, 1188, 1132, 584, 160, 21, 1]

%e [7] [2985, 6381, 6920, 4268, 1510, 293, 28, 1]

%e [8] [15747, 36507, 44213, 31542, 13576, 3464, 497, 36, 1]

%o (Sage)

%o def A216916_triangle(dim):

%o T = matrix(ZZ,dim,dim)

%o for n in range(dim): T[n,n] = 1

%o for n in (1..dim-1):

%o for k in (0..n-1):

%o T[n,k] = T[n-1,k-1]+(k+1)*T[n-1,k]+(k+2)*T[n-1,k+1]

%o return T

%o A216916_triangle(9)

%K nonn,tabl

%O 0,4

%A _Peter Luschny_, Sep 20 2012