A182822 Exponential Riordan array, defining orthogonal polynomials related to permutations without double falls.
1, 1, 1, 2, 3, 1, 5, 12, 6, 1, 17, 53, 39, 10, 1, 70, 279, 260, 95, 15, 1, 349, 1668, 1914, 880, 195, 21, 1, 2017, 11341, 15330, 8554, 2380, 357, 28, 1, 13358, 86019, 134317, 87626, 29379, 5530, 602, 36, 1, 99377, 722664, 1277604, 954885, 372771, 84231, 11508, 954, 45, 1, 822041, 6655121, 13149441, 11061480, 4924515, 1292445, 211533, 22020, 1440, 55, 1
Offset: 0
Examples
Triangle begins 1; 1, 1; 2, 3, 1; 5, 12, 6, 1; 17, 53, 39, 10, 1; 70, 279, 260, 95, 15, 1; 349, 1668, 1914, 880, 195, 21, 1; 2017, 11341, 15330, 8554, 2380, 357, 28, 1; 13358, 86019, 134317, 87626, 29379, 5530, 602, 36, 1; Production matrix is 1, 1; 1, 2, 1; 0, 4, 3, 1; 0, 0, 9, 4, 1; 0, 0, 0, 16, 5, 1; 0, 0, 0, 0, 25, 6, 1; 0, 0, 0, 0, 0, 36, 7, 1; 0, 0, 0, 0, 0, 0, 49, 8, 1; 0, 0, 0, 0, 0, 0, 0, 64, 9, 1; 0, 0, 0, 0, 0, 0, 0, 0, 81, 10, 1; 0, 0, 0, 0, 0, 0, 0, 0, 0, 100, 11, 1;
Links
- Paul Barry, Constructing Exponential Riordan Arrays from Their A and Z Sequences, Journal of Integer Sequences, 17 (2014), #14.2.6.
Programs
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Mathematica
dim = 11; M[n_, n_] = 1; M[n_ /; 0 <= n <= dim-1, k_ /; 0 <= k <= dim-1] := M[n, k] = M[n-1, k-1] + (k+1)*M[n-1, k] + (k+1)^2*M[n-1, k+1]; M[, ] = 0; Table[M[n, k], {n, 0, dim-1}, {k, 0, n}] (* Jean-François Alcover, Jun 18 2019 *)
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Sage
def A182822_triangle(dim): T = matrix(ZZ,dim,dim) for n in (0..dim-1): T[n,n] = 1 for n in (1..dim-1): for k in (0..n-1): T[n,k] = T[n-1,k-1]+(k+1)*T[n-1,k]+(k+1)^2*T[n-1,k+1] return T A182822_triangle(9) # Peter Luschny, Sep 19 2012
Formula
Exponential Riordan array [exp(x/2)/(cos(sqrt(3)x/2)-sin(sqrt(3)x/2)/sqrt(3)), 2*sin(sqrt(3)x/2)/(sqrt(3)*cos(sqrt(3)x/2)-sin(sqrt(3)x/2))].
From Werner Schulte, Mar 27 2022: (Start)
T(n,k) = T(n-1,k-1) + (k+1) * T(n-1,k) + (k+1)^2 * T(n-1,k+1) for n > 0 with initial values T(0,0) = 1 and T(i,j) = 0 if j < 0 or i < j (see the Sage program below).
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k) * x^k satisfy recurrence equation p(n,x) = (1+x) * (p(n-1,x) + p'(n-1,x)) + x * p"(n-1,x) for n > 0 with initial value p(0,x) = 1 where p' and p" are first and second derivative of p. (End)
Comments