A052844 -id:A052844 - cp's OEIS frontend

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

Paul Barry, Apr 26 2007

Satisfies the equation e^[x/(1-x),x] = e*[e^(x/(1-x)),x].
Row sums are A052844.
Antidiagonal sums are A129653.

			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;
  ...

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.

Cf. A000262 (column 0), A052844 (row sums).
T(2n,n) gives A350461.
Cf. A052845, A111884, A133314.

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

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 *)

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

A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 15, 9, 1, 1, 64, 66, 16, 1, 1, 325, 490, 190, 25, 1, 1, 1956, 3915, 2120, 435, 36, 1, 1, 13699, 34251, 23975, 6755, 861, 49, 1, 1, 109600, 328804, 283136, 101990, 17696, 1540, 64, 1, 1, 986409, 3452436, 3534636, 1554966, 342846, 40404, 2556, 81, 1

Offset: 0

Peter Luschny, Apr 14 2016

This is the Sheffer (aka exponential Riordan) matrix T = P*L = A007318*A271703 = (exp(x), x/(1-x)). Note that P = A007318 is Sheffer (exp(t), t) (of the Appell type). The Sheffer a-sequence is [1,1,repeat(0)] and the z-sequence has e.g.f. (x/(1+x))*(1 - exp(-x/(1+x)) given in A288869 / A000027. Because the column k=0 has only entries 1, the z-sequence gives fractional representations of 1. See A288869. - Wolfdieter Lang, Jun 20 2017

			Triangle starts:
  1;
  1,    1;
  1,    4,    1;
  1,   15,    9,    1;
  1,   64,   66,   16,   1;
  1,  325,  490,  190,  25,  1;
  1, 1956, 3915, 2120, 435, 36, 1;
  ...
Recurrence: T(3, 2) = (3/2)*4 + 3*1 = 9. - _Wolfdieter Lang_, Jun 20 2017

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Marin Knežević, Vedran Krčadinac, and Lucija Relić, Matrix products of binomial coefficients and unsigned Stirling numbers, arXiv:2012.15307 [math.CO], 2020.

Cf. A000290 (diag n, n-1), A062392 (diag n, n-2).
Cf. A007526 (col. 1), A134432 (col. 2).
Cf. A052844 (row sums), A059110 (matrix inverse).
Cf. A007318, A271703, A288869.

B:=Binomial;
A271705:= func< n,k | k eq 0 select 1 else (&+[B(n, j+k)*B(j+k, k)*B(j+k-1, k-1)*Factorial(j): j in [0..n-k]]) >;
[A271705(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 09 2022

L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
T := (n,k) -> add(L(j,k)*binomial(-j-1,-n-1)*(-1)^(n-j), j=0..n):
seq(seq(T(n,k), k=0..n), n=0..9);

T[n_, k_]:= If[k==0, 1, Sum[((k*j!)/(j+k))*Binomial[n, j+k]*Binomial[j+k, k]^2, {j,0,n-k}]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 09 2022 *)

b=binomial
def A271705(n,k): return 1 if (k==0) else sum(factorial(j-k)*b(n, j)*b(j, k)*b(j-1, k-1) for j in (k..n))
flatten([[A271705(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jan 09 2022

From Wolfdieter Lang, Jun 20 2017: (Start)
T(n, k) = Sum_{m=k..n} A007318(n, m)*A271703(m, k), n >= k >= 0, and 0 for k < m. See also the name.
E.g.f. of column k: exp(x)*(x/(1-x))^k/k! (Sheffer property), k >= 0.
E.g.f. of triangle (or row polynomials in x): exp(z)*exp(x*z/(1-z)).
Recurrence for T(n, k), k >= 1, with T(n, 0) = 1, T(n, k) = 0 if n < k: T(n, k) = (n/k)*T(n-1, k-1) + n*T(n-1, k), n >= 1, k = 1..n. (From the a-sequence with column k=0 as input.) (End)
T(n, k) = Sum_{j=0..n-k} j!*binomial(n, j+k)*binomial(j+k, k)*binomial(j+k-1, k-1) with T(n, 0) = 1. - G. C. Greubel, Jan 09 2022
From Natalia L. Skirrow, Jun 11 2025: (Start)
T(n, k) = C(n, k)*hypergeom([k-n, k], [], -1), which equals C(n, k)*A143409(n-k, k-1) for k>0.
By the saddle point method upon the e.g.f., n-th row polynomial converges with n (for all y) to n^n*exp(2*sqrt(n*y) - n - y/2 + 1)/sqrt(2*sqrt(n/y)); as such, the n-th row's expectation is ~ sqrt(n)-1/4 and the n-th row's variance is ~ (sqrt(n)-1)/2. (End)

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

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