Three term recurrence (from the conversion property mentioned above, with [d,a] =[4,1]): T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m. T(n, m) = 4*T(n-1, m-1) + (1 + 4*m)*T(n-1, m) for n >= 1, m = 0..n.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
E.g.f. of the row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)*exp(x*(exp(4*z) - 1)). This is the e.g.f. of the triangle.
E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
O.g.f. for sequence of column m: (4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x) (by Laplace transform of the e.g.f.).
T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in
A048993.
A nontrivial recurrence for the column m=0 entries T(n, 0) = 4^n from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)* Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((1 + 4*x) + 4*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 4*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m,m) = 4^m. See a comment and references in
A282629. An example is given below. -
Wolfdieter Lang, Aug 11 2017
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