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Triangle T, read by rows, such that diagonal n equals column 0 of T^(n+1), the (n+1)-th matrix power of T.
1

%I #6 Jun 13 2017 22:13:30

%S 1,1,1,1,2,1,1,3,4,1,1,4,9,9,1,1,5,16,32,24,1,1,6,25,78,150,79,1,1,7,

%T 36,155,532,1018,340,1,1,8,49,271,1395,5802,10996,2090,1,1,9,64,434,

%U 3036,21343,116658,212434,20613,1,1,10,81,652,5824,60209,661325,5072504

%N Triangle T, read by rows, such that diagonal n equals column 0 of T^(n+1), the (n+1)-th matrix power of T.

%C Row sums form A098448.

%F T(n, k) = Sum_{i=0..k} T(k, i)*T(n-k+i-1, i), for 0<k<n, else T(0, n)=T(n, n)=1.

%e T(7,3) = T(3,0)*T(3,0) + T(3,1)*T(4,1) + T(3,2)*T(5,2) + T(3,3)*T(6,3)

%e = 1*1 + 3*4 + 4*16 + 1*78 = 155.

%e Rows of T begin:

%e [1],

%e [1,1],

%e [1,2,1],

%e [1,3,4,1],

%e [1,4,9,9,1],

%e [1,5,16,32,24,1],

%e [1,6,25,78,150,79,1],

%e [1,7,36,155,532,1018,340,1],

%e [1,8,49,271,1395,5802,10996,2090,1],

%e [1,9,64,434,3036,21343,116658,212434,20613,1],...

%e Matrix square T^2 begins:

%e [1],

%e [2,1],

%e [4,4,1],

%e [9,14,8,1],

%e [24,53,54,18,1],

%e [79,234,376,280,48,1],

%e [340,1291,2976,4034,2196,158,1],...

%e where column 0 is {1,2,4,9,24,79,340,...} and forms diagonal 1 of T.

%e Matrix cube T^3 begins:

%e [1],

%e [3,1],

%e [9,6,1],

%e [32,33,12,1],

%e [150,219,135,27,1],

%e [1018,2023,1944,744,72,1],...

%e where column 0 is {1,3,9,32,150,1018,...} and forms diagonal 2 of T.

%o (PARI) T(n,k)=if(n<k || k<0,0,if(n==k || k==0,1,sum(i=0,k,T(k,i)*T(n-k+i-1,i));))

%Y Cf. A098448, A098446, A091351.

%K nonn,tabl

%O 0,5

%A _Paul D. Hanna_, Sep 07 2004