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A265649 Triangle of coefficients T(n,k) of polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^k where T(0,0) = 1, and T(n,k) = 0 for k < 0 or k > n, and T(n,k) = T(n-1,k-1) + (2*n-1+k)*T(n-1,k) for n > 0 and 0 <= k <= n.

%I A265649 #40 Mar 11 2024 23:13:21
%S A265649 1,1,1,3,5,1,15,33,12,1,105,279,141,22,1,945,2895,1830,405,35,1,10395,
%T A265649 35685,26685,7500,930,51,1,135135,509985,435960,146685,23310,1848,70,
%U A265649 1,2027025,8294895,7921305,3076290,589575,60270,3318,92,1,34459425,151335135,158799690,69447105,15457365,1915515,136584,5526,117,1
%N A265649 Triangle of coefficients T(n,k) of polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^k where T(0,0) = 1, and T(n,k) = 0 for k < 0 or k > n, and T(n,k) = T(n-1,k-1) + (2*n-1+k)*T(n-1,k) for n > 0 and 0 <= k <= n.
%C A265649 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).
%C A265649 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
%H A265649 Paul Barry, <a href="https://arxiv.org/abs/1702.04011">A Note on d-Hankel Transforms, Continued Fractions, and Riordan Arrays</a>, arXiv:1702.04011 [math.CO], 2017.
%H A265649 Robert S. Maier, <a href="https://arxiv.org/abs/2308.10332">Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers</a>, arXiv:2308.10332 [math.CO], 2023. See pp. 21, 28.
%F A265649 Recurrence: p(0,x) = 1 and p(n+1,x) = (2*n+1+x)*p(n,x) + x*p'(n,x).
%F A265649 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).
%F A265649 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).
%F A265649 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)).
%F A265649 Conjecture: T(n,k) = (-1)^k*Sum_{j=0..n-1} A001497(n-1,j)*A021009(j+1,k).
%F A265649 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
%F A265649 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
%e A265649 The triangle T(n,k) begins:
%e A265649 n\k:        0        1        2        3       4      5     6   7  8
%e A265649   0:        1
%e A265649   1:        1        1
%e A265649   2:        3        5        1
%e A265649   3:       15       33       12        1
%e A265649   4:      105      279      141       22       1
%e A265649   5:      945     2895     1830      405      35      1
%e A265649   6:    10395    35685    26685     7500     930     51     1
%e A265649   7:   135135   509985   435960   146685   23310   1848    70   1
%e A265649   8:  2027025  8294895  7921305  3076290  589575  60270  3318  92  1
%e A265649   etc.
%e A265649 The polynomial corresponding to row 3 is p(3,x) = 15 + 33*x + 12*x^2 + x^3.
%p A265649 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
%t A265649 (* The function RiordanArray is defined in A256893. *)
%t A265649 rows = 10;
%t A265649 R = RiordanArray[1/Sqrt[1 - 2 #]&, 1/Sqrt[1 - 2 #] - 1&, rows, True];
%t A265649 R // Flatten (* _Jean-François Alcover_, Jul 20 2019 *)
%Y A265649 Cf. A000007, A000326, A001147, A001497, A021009, A129890, A035342.
%K A265649 nonn,easy,tabl
%O A265649 0,4
%A A265649 _Werner Schulte_, Dec 11 2015