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A129652 Exponential Riordan array [e^(x/(1-x)),x].

%I A129652 #40 Sep 20 2024 21:23:29
%S A129652 1,1,1,3,2,1,13,9,3,1,73,52,18,4,1,501,365,130,30,5,1,4051,3006,1095,
%T A129652 260,45,6,1,37633,28357,10521,2555,455,63,7,1,394353,301064,113428,
%U A129652 28056,5110,728,84,8,1,4596553,3549177,1354788,340284,63126,9198,1092,108,9,1
%N A129652 Exponential Riordan array [e^(x/(1-x)),x].
%C A129652 Satisfies the equation e^[x/(1-x),x] = e*[e^(x/(1-x)),x].
%C A129652 Row sums are A052844.
%C A129652 Antidiagonal sums are A129653.
%H A129652 Alois P. Heinz, <a href="/A129652/b129652.txt">Rows n = 0..140, flattened</a>
%H A129652 T.-X. He, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL11/He/he51.html">A symbolic operator approach to power series transformation-expansion formulas</a>, JIS 11 (2008) 08.2.7.
%F A129652 Number triangle T(n,k)=(n!/k!)*sum{i=0..n-k, C(n-k-1,i)/(n-k-i)!}
%F A129652 From _Peter Bala_, May 14 2012 : (Start)
%F A129652 Array is exp(S*(I-S)^(-1)) where S is A132440 the infinitesimal generator for Pascal's triangle.
%F A129652 Column 0 is A000262.
%F A129652 T(n,k) = binomial(n,k)*A000262(n-k).
%F A129652 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)
%F A129652 T(n,k) = (-1)^(k-n+1)*C(n,k)*KummerU(k-n+1, 2, -1). - _Peter Luschny_, Sep 17 2014
%F A129652 From _Tom Copeland_, Mar 11 2016: (Start)
%F A129652 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).
%F A129652 (P.(x) + y)^n = Sum_{k=0..n} binomial(n,k) P_k(x) * y^(n-k) = P_n(x+y).
%F A129652 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.
%F A129652 (End)
%F A129652 Sum_{k=0..n} (-1)^k * T(n,k) = A052845(n). - _Alois P. Heinz_, Feb 21 2022
%e A129652 Triangle begins:
%e A129652      1;
%e A129652      1,    1;
%e A129652      3,    2,    1;
%e A129652     13,    9,    3,   1;
%e A129652     73,   52,   18,   4,  1;
%e A129652    501,  365,  130,  30,  5, 1;
%e A129652   4051, 3006, 1095, 260, 45, 6, 1;
%e A129652   ...
%p A129652 A129652 := (n, k) -> (-1)^(k-n+1)*binomial(n,k)*KummerU(k-n+1, 2, -1);
%p A129652 seq(seq(round(evalf(A129652(n,k),99)),k=0..n),n=0..9); # _Peter Luschny_, Sep 17 2014
%p A129652 # second Maple program:
%p A129652 b:= proc(n) option remember; `if`(n=0, [1$2], add((p-> p+
%p A129652      [0, p[1]*x^j])(b(n-j)*binomial(n-1, j-1)*j!), j=1..n))
%p A129652     end:
%p A129652 T:= n-> (p-> seq(coeff(p, x, i)/i!, i=0..n))(b(n)[2]):
%p A129652 seq(T(n), n=0..10);  # _Alois P. Heinz_, Feb 21 2022
%t A129652 T[n_, k_] := If[k==n, 1, n!/k! Sum[Binomial[n-k-1, j]/(j+1)!, {j, 0, n-k-1}]];
%t A129652 Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* _Jean-François Alcover_, Jun 14 2019 *)
%Y A129652 Cf. A000262 (column 0), A052844 (row sums).
%Y A129652 T(2n,n) gives A350461.
%Y A129652 Cf. A052845, A111884, A133314.
%K A129652 easy,nonn,tabl
%O A129652 0,4
%A A129652 _Paul Barry_, Apr 26 2007