OFFSET
0,4
COMMENTS
The polynomials p(n,x) satisfy the differential equation: x*y''' + (3*x+1)*y'' + (2*x+2)*y' - 2*n*y = 0 where y' = dy/dx (first derivative).
Appears to be the exponential Riordan array [1/sqrt(1 - 2x), 1/(sqrt(1 - 2x) - 1)]. [Barry, Example 1] - Eric M. Schmidt, Sep 23 2017
LINKS
Paul Barry, A Note on d-Hankel Transforms, Continued Fractions, and Riordan Arrays, arXiv:1702.04011 [math.CO], 2017.
Robert S. Maier, Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers, arXiv:2308.10332 [math.CO], 2023. See pp. 21, 28.
FORMULA
Recurrence: p(0,x) = 1 and p(n+1,x) = (2*n+1+x)*p(n,x) + x*p'(n,x).
T(n,0) = A001147(n), T(n+1,1) = A129890(n), T(n+1,n) = A000326(n+1), and Sum_{k=0..n} (-1)^k*k!*T(n,k) = A000007(n).
Recurrence: k^2*(k+1)*T(n,k+1) = (2*n+2-2*k)*T(n,k-1)-k*(3*k-1)*T(n,k).
Conjecture: T(n,k) = 2^(n-k)*(n-k)!*binomial(n,k)*(Sum_{j=0..n-k} (-1/4)^j* binomial(2*j+k,j)*binomial(n,j+k)).
T(n,k) = (Sum_{i=0..k} (-1)^(k-i) * binomial(k, i)*Product_{j=1..n} (2*j+i-1))/k!. - Werner Schulte, Mar 03 2024
T(n,k) = (2^n/k!)*(Sum_{j=0..k}(-1)^(k-j)*binomial(k,j)*Pochhammer((j + 1)/2, n)). - Peter Luschny, Mar 04 2024
EXAMPLE
The triangle T(n,k) begins:
n\k: 0 1 2 3 4 5 6 7 8
0: 1
1: 1 1
2: 3 5 1
3: 15 33 12 1
4: 105 279 141 22 1
5: 945 2895 1830 405 35 1
6: 10395 35685 26685 7500 930 51 1
7: 135135 509985 435960 146685 23310 1848 70 1
8: 2027025 8294895 7921305 3076290 589575 60270 3318 92 1
etc.
The polynomial corresponding to row 3 is p(3,x) = 15 + 33*x + 12*x^2 + x^3.
MAPLE
T := (n, k) -> local j; 2^n*add((-1)^(k-j)*binomial(k, j)*pochhammer((j+1)/2, n), j=0..k) / k!: for n from 0 to 6 do seq(T(n, k), k=0..n) od; # Peter Luschny, Mar 04 2024
MATHEMATICA
(* The function RiordanArray is defined in A256893. *)
rows = 10;
R = RiordanArray[1/Sqrt[1 - 2 #]&, 1/Sqrt[1 - 2 #] - 1&, rows, True];
R // Flatten (* Jean-François Alcover, Jul 20 2019 *)
CROSSREFS
KEYWORD
AUTHOR
Werner Schulte, Dec 11 2015
STATUS
approved