%I #11 May 11 2014 08:30:15

%S 1,3,2,15,8,3,114,56,15,4,1191,568,135,24,5,15993,7536,1710,264,35,6,

%T 263976,123704,27495,4008,455,48,7,5189778,2425320,533565,75696,8050,

%U 720,63,8,118729335,55403008,12121920,1695528,174615,14544,1071,80,9

%N Triangular matrix T, read by rows, that satisfies: T^2 + 2*T = SHIFTUP(T), also T^(n+1) + 2*T^n = SHIFTUP(T^n - D*T^(n-1)) for all n, where D is a diagonal matrix with diagonal(D) = diagonal(T) = {1,2,3,...}.

%C Leftmost column is A082163 (enumerates acyclic automata with 2 inputs). The operation SHIFTUP(T) shifts each column of T up 1 row, dropping the elements that occupied the diagonal of T.

%F G.f. for column k: T(k, k) = k+1 = Sum_{n>=k} T(n, k)*x^(n-k)/(1-2*x)^(n-k) * Product_{j=0..n-k} (1-(j+k+3)*x). Diagonalization: T = P*D*P^-1 where P(r, c) = A103247(r, c)/(r-c)! = (-1)^(r-c)*(c^2+2*c)^(r-c)/(r-c)! for r>=c>=1 and [P^-1](r, c) = A103242(r, c)/(r-c)! and D is a diagonal matrix = {1, 2, 3, ...}.

%e Rows of T begin:

%e [1],

%e [3,2],

%e [15,8,3],

%e [114,56,15,4],

%e [1191,568,135,24,5],

%e [15993,7536,1710,264,35,6],

%e [263976,123704,27495,4008,455,48,7],

%e [5189778,2425320,533565,75696,8050,720,63,8],...

%e Rows of T^2 begin:

%e [1],

%e [9,4],

%e [84,40,9],

%e [963,456,105,16],

%e [13611,6400,1440,216,25],...

%e Rows of T^2+2*T equals SHIFTUP(T):

%e [3],

%e [15,8],

%e [114,56,15],

%e [1191,568,135,24],

%e [15993,7536,1710,264,35],...

%e G.f. for column 0: 1 = (1-3x) + 3*x/(1-2x)*(1-3x)(1-4x) + 15*x^2/(1-2x)^2*(1-3x)(1-4x)(1-5x) + 114*x^3/(1-2x)^3*(1-3x)(1-4x)(1-5x)(1-6x) + ... + T(n,0)*x^n/(1-2*x)^n*(1-3x)(1-4x)*..*(1-(n+3)x) + ...

%e G.f. for column 1: 2 = 2*(1-4x) + 8*x/(1-2x)*(1-4x)(1-5x) + 56*x^2/(1-2x)^2*(1-4x)(1-5x)(1-6x) + 568*x^3/(1-2x)^3*(1-4x)(1-5x)(1-6x)(1-7x) + ... + T(n,1)*x^(n-1)/(1-2*x)^(n-1)*(1-4x)(1-5x)*..*(1-(n+3)x) + ...

%o (PARI) {T(n,k)=if(n<k,0,if(n==k,k+1,polcoeff( k+1-sum(i=k,n-1,T(i,k)*x^i/(1-2*x)^(i-k)* prod(j=0,i-k,1-(j+k+3)*x+x*O(x^n))),n)))}

%Y Cf. A082163, A103247, A103242.

%K nonn,tabl

%O 0,2

%A _Paul D. Hanna_, Jan 31 2005