A129652 Exponential Riordan array [e^(x/(1-x)),x].
1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 73, 52, 18, 4, 1, 501, 365, 130, 30, 5, 1, 4051, 3006, 1095, 260, 45, 6, 1, 37633, 28357, 10521, 2555, 455, 63, 7, 1, 394353, 301064, 113428, 28056, 5110, 728, 84, 8, 1, 4596553, 3549177, 1354788, 340284, 63126, 9198, 1092, 108, 9, 1
Offset: 0
Examples
Triangle begins:
1;
1, 1;
3, 2, 1;
13, 9, 3, 1;
73, 52, 18, 4, 1;
501, 365, 130, 30, 5, 1;
4051, 3006, 1095, 260, 45, 6, 1;
...
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- T.-X. He, A symbolic operator approach to power series transformation-expansion formulas, JIS 11 (2008) 08.2.7.
Crossrefs
Programs
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Maple
A129652 := (n, k) -> (-1)^(k-n+1)*binomial(n,k)*KummerU(k-n+1, 2, -1); seq(seq(round(evalf(A129652(n,k),99)),k=0..n),n=0..9); # Peter Luschny, Sep 17 2014 # second Maple program: b:= proc(n) option remember; `if`(n=0, [1$2], add((p-> p+ [0, p[1]*x^j])(b(n-j)*binomial(n-1, j-1)*j!), j=1..n)) end: T:= n-> (p-> seq(coeff(p, x, i)/i!, i=0..n))(b(n)[2]): seq(T(n), n=0..10); # Alois P. Heinz, Feb 21 2022
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Mathematica
T[n_, k_] := If[k==n, 1, n!/k! Sum[Binomial[n-k-1, j]/(j+1)!, {j, 0, n-k-1}]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 14 2019 *)
Formula
Number triangle T(n,k)=(n!/k!)*sum{i=0..n-k, C(n-k-1,i)/(n-k-i)!}
From Peter Bala, May 14 2012 : (Start)
Array is exp(S*(I-S)^(-1)) where S is A132440 the infinitesimal generator for Pascal's triangle.
Column 0 is A000262.
T(n,k) = binomial(n,k)*A000262(n-k).
So T(n,k) gives the number of ways to choose a subset of {1,2,...,n} of size k and then arrange the remaining n-k elements into a set of lists. (End)
T(n,k) = (-1)^(k-n+1)*C(n,k)*KummerU(k-n+1, 2, -1). - Peter Luschny, Sep 17 2014
From Tom Copeland, Mar 11 2016: (Start)
The row polynomials P_n(x) form an Appell sequence with e.g.f. e^(t*P.(x)) = e^[t / (1-t)] e^(x*t), so the lowering and raising operators are L = d/dx = D and the R = x + 1 / (1-D)^2 = x + 1 + 2 D + 3 D^2 + ..., satisfying L P_n(x) = n * P_(n-1)(x) and R P_n(x) = P_(n+1)(x).
(P.(x) + y)^n = Sum_{k=0..n} binomial(n,k) P_k(x) * y^(n-k) = P_n(x+y).
The Appell polynomial umbral compositional inverse sequence has the e.g.f. e^(t*Q.(x)) = e^[-t / (1-t)] e^(x*t) (see A111884 and A133314), so Q_n(P.(x)) = P_n(Q.(x)) = x^n. The lower triangular matrices for the coefficients of these two Appell sequences are a multiplicative inverse pair.
(End)
Sum_{k=0..n} (-1)^k * T(n,k) = A052845(n). - Alois P. Heinz, Feb 21 2022
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