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.

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

Author

Peter Luschny, Apr 14 2016

Keywords

• nonn
• easy
• tabl

Comments

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

Examples

			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
		

Links

• 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.

Crossrefs

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.

Programs

Magma

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

Maple

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);

Mathematica

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 *)

Sage

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

Formula

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)