A111578 Triangle T(n, m) = T(n-1, m-1) + (4m-3)*T(n-1, m) read by rows 1<=m<=n.

1, 1, 1, 1, 6, 1, 1, 31, 15, 1, 1, 156, 166, 28, 1, 1, 781, 1650, 530, 45, 1, 1, 3906, 15631, 8540, 1295, 66, 1, 1, 19531, 144585, 126651, 30555, 2681, 91, 1, 1, 97656, 1320796, 1791048, 646086, 86856, 4956, 120, 1, 1, 488281, 11984820, 24604420, 12774510

Offset: 1

Author

Gary W. Adamson, Aug 07 2005

Keywords

• nonn
• tabl

Comments

From Peter Bala, Jan 27 2015: (Start)

Working with an offset of 0, this is the exponential Riordan array [exp(z), (exp(4*z) - 1)/4].

This is the triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*4^n * binomial((x - 1)/4,n) }n>=0. An example is given below.

Call this array M and let P denote Pascal's triangle A007318 then P^2 * M = A225469; P^(-1) * M is a shifted version of A075499.

This triangle is the particular case a = 4, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)

Examples

			The triangle starts in row n=1 as:
  1;
  1,1;
  1,6,1;
  1,31,15,1;
  1,156,166,28,1;
Connection constants: Row 4: [1, 31, 15, 1] so
x^3 = 1 + 31*(x - 1) + 15*(x - 1)*(x - 5) + (x - 1)*(x - 5)*(x - 9). - _Peter Bala_, Jan 27 2015
		

Links

• P. Bala, A 3 parameter family of generalized Stirling numbers

Crossrefs

Cf. A111577, A008277, A039755, A016234 (3rd column).

Programs

Mathematica

T[n_, k_] := 1/(4^(k-1)*(k-1)!) * Sum[ (-1)^(k-j-1) * (4*j+1)^(n-1) * Binomial[k-1, j], {j, 0, k-1}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Peter Bala *)

Python

def A096038(n,m):
    if n < 1 or m < 1 or m > n:
        return 0
    elif n <=2:
        return 1
    else:
        return A096038(n-1,m-1)+(4*m-3)*A096038(n-1,m)
print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
# R. J. Mathar, Oct 11 2009

Formula

From Peter Bala, Jan 27 2015: (Start)

The following formulas assume an offset of 0.

T(n,k) = 1/(4^k*k!)*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*(4*j + 1)^n.

T(n,k) = sum {i = 0..n-1} 4^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).

E.g.f.: exp(z)*exp(x/4*(exp(4*z) - 1)) = 1 + (1 + x)*z + (1 + 6*x + x^2)*z^2/2! + ....

O.g.f. for n-th diagonal: exp(-x/4)*sum {k >= 0} (4*k + 1)^(k + n - 1)*((x/4*exp(-x))^k)/k!.

O.g.f. column k: 1/( (1 - x)*(1 - 5*x)*...*(1 - (4*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Oct 11 2009