cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A134685 Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion.

Original entry on oeis.org

1, -1, 3, -1, -15, 10, -1, 105, -105, 10, 15, -1, -945, 1260, -280, -210, 35, 21, -1, 10395, -17325, 6300, 3150, -280, -1260, -378, 35, 56, 28, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -2100, -1575, -2520, -630, 126, 84, 36, -1
Offset: 1

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Author

Tom Copeland, Jan 26 2008, Sep 13 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[-log(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in reference in A127671 and u_n = (n-1)! * a_n .
If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = Sum_{j>=1} P(j,t) where, with u_n denoted by (n') for brevity,
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 10 (1')^2 (3')^2 + 15 (1')^2 (2') (4') - (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 35 (1')^3 (3')(4') + 21 (1')^3 (2')(5') - (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 17325 (1') (2')^4 (3') + (1')^2 [ 6300 (2')^2 (3')^2 + 3150 (2')^3 (4')] - (1')^3 [280 (3')^3 + 1260 (2')(3')(4') + 378 (2')^2(5')] + (1')^4 [35 (4')^2 + 56 (3')(5') + 28 (2')(6')] - (1')^5 (7') ] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 270270 (1') (2')^5 (3') - (1')^2 [ 138600 (2')^3 (3')^2 + 51975 (2')^4 (4')] + (1')^3 [15400 (2')(3')^3 + 34650 (2')^2(3')(4') + 6930 (2')^3(5')] - (1')^4 [2100 (3')^2(4') + 1575 (2')(4')^2 + 2520 (2')(3')(5') + 630 (2')^2(6') ] + (1')^5 [126 (4')(5') + 84 (3')(6') + 36 (2')(7')] - (1')^6 (8') ] * t^8 / 8!
...
Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 (in the reversed order) and the number of partitions in P(n,t) is given by A000041(n-1).
Combinatorial interpretations are given in the link.
From Tom Copeland, Jul 10 2018: (Start)
Coefficients occurring in prolongation for the special Euclidean group SE(2) and special affine group SA(2) in the Olver presentation on moving frames (MFP) in slides 33 and 42. These are a result of applying an iterated derivative of the form h(x)d/dx = d/dy as in this entry (more generally as g(x) d/dx as discussed in A145271). See also p. 6 of Olver's paper on contact forms, but note that the 12 should be a 15 in the formula for the compositional inverse of S(t).
Change variables in the MFP to obtain connections to the partition polynomials Prt_n = n! * P(n,1) above. Let delta and beta in the formulas for the equi-affine curves in MFP be L and B, respectively, and D_y = (1/(L-B*u_x)) d/dx = (1/w_x) d/dx. Then v_(yy) = (1/B) [-w_(xx)/(w_x)^3] in MFP (there is an overall sign error in MFP for v_(yy) and higher derivatives w.r.t. y), and (d/dy)^n v = v_n = (1/B)* [(1/w_1)*(d/dx)]^(n-2) [-w_2/(w_1)^3] for n > 1, with w_n = (d/dx)^n w. Consequently, in the partition polynomials Prt_n for n > 1 here substitute (n') = -B*u_n = w_n for n > 1 and (1') = L-B*u_1 = w_1, where u_n = (d/dx)^n u, and then divide by B. For example, v_4 = (1/B)*Prt_4 = (1/B)*4!*P(4,1) = (1/B) (L-B*u_n)^(-7) [-15*(-B*u_2)^3 + 10 (L-B*u_1)(-B*u_2)(-B*u_3) - (L-B*u_1)^2 (-B*u_4)], agreeing with v_4 in MFP except for the overall sign.
For the SE(2) transformation formulas in MFP, let w_x = cos(phi) + sin(phi)*u_x, and then the same transformations apply as above with cos(phi) and sin(phi) substituted for L and -B, respectively. (End)

Examples

			Examples and checks:
1) Let u_1 = -1 and u_n = 1 for n>1,
then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1
and g(t) = [e.g.f. of signed A000311];
therefore, the row sums of unsigned [C(j,k)] are A000311 =
(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).
2) Let u_1 = -1 and u_n = (n-1)! for n>1,
then f(t) = -log(1-t)-2t
and g(t) = [e.g.f. of signed (0,A032188)]
with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).
3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then
f(t) = (1+t) log(1+t) - 2t
and g(t) = [e.g.f. of signed (0,A074059)]
with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).
4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,
then f(t) = t(1-t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)
with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).
5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,
then f(t)= t(1-t)
and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).
.
From _Peter Luschny_, Feb 19 2021: (Start)
Triangle starts:
 [1]  1;
 [2] -1;
 [3]  3,     -1;
 [4] -15,     10,    -1;
 [5]  105,   -105,   [10, 15],  -1;
 [6] -945,    1260,  [-280, -210], [35, 21],  -1;
 [7]  10395, -17325, [6300, 3150], [-280, -1260, -378], [35, 56, 28], -1;
 [8] -135135, 270270, [-138600, -51975], [15400, 34650, 6930], [-2100, -1575, -2520, -630], [126, 84, 36], -1
The coefficients can be seen as a refinement of the Ward numbers: Let R(n, k) = Sum T(n, k), where the sum collects adjacent terms with equal sign, as indicated by the square brackets in the table, then R(n+1, k+1) = (-1)^(n-k)*W(n, k), where W(n, k) are the Ward numbers A181996, for n >= 0 and 0 <= k <= n-1.  (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.
  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994, p. 10.
  • J. Riordan, Combinatorial Identities, Robert E. Krieger Pub. Co., 1979, (unsigned partition polynomials in Table 5.2 on p. 181, but may have errors).

Crossrefs

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).
Cf. A181996 (Ward numbers).

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~Module[{h = 1/(1 + Sum[u[k] y^k/k!, {k, n-1}] + O[y]^n), g = y, r}, r = Reap[Do[g = h D[g, y]; Sow[Expand[Normal@g /. {y -> 0}]], {k, n}]][[2, 1, ;;]]; Table[Coefficient[r[[k]], Product[u[t], {t, p}]], {k, 2, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k-1]]}]];
    rows[8] // Flatten (* Andrei Zabolotskii, Feb 19 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [2!^e(2)*e(2)!*3!^e(3)*e(3)! ... n!^e(n)*e(n)! ].
From Tom Copeland, Sep 05 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u.*exp(u.*t)]
= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),
an e.g.f. for the partition polynomials of A133314
(signed A049019) with an index shift.
Then for the partition polynomials of A134685,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (Cf. A000311 and A134991)(End)
From Tom Copeland, Oct 30 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators
defined by R PS(n,t)=PS(n+1,t) and L PS(n,t)= n*PS(n-1,t) are
R = t*h(d/dt) = t * 1/[u_1+(u_2)*d/dt+(u_3)*(d/dt)^2/2!+...], and
L = f(d/dt)=(u_1)*d/dt+(u_2)*(d/dt)^2/2!+(u_3)*(d/dt)^3/3!+....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n-1)/n!]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314. - Tom Copeland, Jul 15 2016
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2,..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Terms in rows 5-8 reordered by Andrei Zabolotskii, Feb 19 2024

A134991 Triangle of Ward numbers T(n,k) read by rows.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 25, 105, 105, 1, 56, 490, 1260, 945, 1, 119, 1918, 9450, 17325, 10395, 1, 246, 6825, 56980, 190575, 270270, 135135, 1, 501, 22935, 302995, 1636635, 4099095, 4729725, 2027025, 1, 1012, 74316, 1487200, 12122110, 47507460, 94594500, 91891800, 34459425
Offset: 1

Views

Author

Tom Copeland, Feb 05 2008

Keywords

Comments

This is the triangle of associated Stirling numbers of the second kind, A008299, read along the diagonals.
This is also a row-reversed version of A181996 (with an additional leading 1) - see the table on p. 92 in the Ward reference. A134685 is a refinement of the Ward table.
The first and second diagonals are A001147 and A000457 and appear in the diagonals of several OEIS entries. The polynomials also appear in Carlitz (p. 85), Drake et al. (p. 8) and Smiley (p. 7).
First few polynomials (with a different offset) are
P(0,t) = 0
P(1,t) = 1
P(2,t) = t
P(3,t) = t + 3*t^2
P(4,t) = t + 10*t^2 + 15*t^3
P(5,t) = t + 25*t^2 + 105*t^3 + 105*t^4
These are the "face" numbers of the tropical Grassmannian G(2,n),related to phylogenetic trees (with offset 0 beginning with P(2,t)). Corresponding h-vectors are A008517. - Tom Copeland, Oct 03 2011
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=-t, and (a_n)=-(1+t) P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 = 0. - Tom Copeland, Oct 08 2011
Beginning with the second column, the rows give the faces of the Whitehouse simplicial complex with the fourth-order complex being three isolated vertices and the fifth-order being the Petersen graph with 10 vertices and 15 edges (cf. Readdy). - Tom Copeland, Oct 03 2014
Stratifications of smooth projective varieties which are fine moduli spaces for stable n-pointed rational curves. Cf. pages 20 and 30 of the Kock and Vainsencher reference and references in A134685. - Tom Copeland, May 18 2017
Named after the American mathematician Morgan Ward (1901-1963). - Amiram Eldar, Jun 26 2021

Examples

			Triangle begins:
  1
  1   3
  1  10   15
  1  25  105  105
  1  56  490 1260   945
  1 119 1918 9450 17325 10395
  ...
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, page 222.

Crossrefs

The same as A269939, with column k = 0 removed.
A reshaped version of the triangle of associated Stirling numbers of the second kind, A008299.
A181996 is the mirror image.
Columns k = 2, 3, 4 are A000247, A000478, A058844.
Diagonal k = n is A001147.
Diagonal k = n - 1 is A000457.
Row sums are A000311.
Alternating row sums are signed factorials (-1)^(n-1)*A000142(n).
Cf. A112493.

Programs

  • Mathematica
    t[n_, k_] := Sum[(-1)^i*Binomial[n, i]*Sum[(-1)^j*(k-i-j)^(n-i)/(j!*(k-i-j)!), {j, 0, k-i}], {i, 0, k}]; row[n_] := Table[t[k, k-n], {k, n+1, 2*n}]; Table[row[n], {n, 1, 9}] // Flatten (* Jean-François Alcover, Apr 23 2014, after A008299 *)

Formula

E.g.f. for the polynomials is A(x,t) = (x-t)/(t+1) + T{ (t/(t+1)) * exp[(x-t)/(t+1)] }, where T(x) is the Tree function, the e.g.f. of A000169. The compositional inverse in x (about x = 0) is B(x) = x + -t * [exp(x) - x - 1]. Special case t = 1 gives e.g.f. for A000311. These results are a special case of A134685 with u(x) = B(x).
From Tom Copeland, Oct 26 2008: (Start)
Umbral-Sheffer formalism gives, for m a positive integer and u = t/(t+1),
[P(.,t)+Q(.,x)]^m = [m Q(m-1,x) - t Q(m,x)]/(t+1) + sum(n>=1) { n^(n-1)[u exp(-u)]^n/n! [n/(t+1)+Q(.,x)]^m }, when the series is convergent for a sequence of functions Q(n,x).
Check: With t=1; Q(n,x)=0^n, for n>=0; and Q(-1,x)=0, then [P(.,1)+Q(.,x)]^m = P(m,1) = A000311(m).
(End)
Let h(x,t) = 1/(dB(x)/dx) = 1/(1-t*(exp(x)-1)), an e.g.f. in x for row polynomials in t of A019538, then the n-th row polynomial in t of the table A134991, P(n,t), is given by ((h(x,t)*d/dx)^n)x evaluated at x=0, i.e., A(x,t) = exp(x*P(.,t)) = exp(x*h(u,t)*d/du) u evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t). - Tom Copeland, Sep 05 2011
The polynomials (1+t)/t*P(n,t) are the row polynomials of A112493. Let f(x) = (1+x)/(1-x*t). Then for n >= 0, P(n+1,t) is given by t/(1+t)*(f(x)*d/dx)^n(f(x)) evaluated at x = 0. - Peter Bala, Sep 30 2011
From Tom Copeland, Oct 04 2011: (Start)
T(n,k) = (k+1)*T(n-1,k) + (n+k+1)*T(n-1,k-1) with starting indices n=0 and k=0 beginning with P(2,t) (as suggested by a formula of David Speyer on MathOverflow).
T(n,k) = k*T(n-1,k) + (n+k-1)*T(n-1,k-1) with starting indices n=1 and k=1 of table (cf. Smiley above and Riordin ref.[10] therein).
P(n,t) = (1/(1+t))^n * Sum_{k>=1} k^(n+k-1)*(u*exp(-u))^k / k! with u=(t/(t+1)) for n>1; therefore, Sum_{k>=1} (-1)^k k^(n+k-1) x^k/k! = [1+LW(x)]^(-n) P{n,-LW(x)/[1+LW(x)]}, with LW(x) the Lambert W-Fct.
T(n,k) = Sum_{i=0..k} ((-1)^i binomial(n+k,i) Sum_{j=0..k-i} (-1)^j (k-i-j)^(n+k-i)/(j!(k-i-j)!)) from relation to A008299. (End)
The e.g.f. A(x,t) = -v * ( Sum_{j=>1} D(j-1,u) (-z)^j / j! ) where u = (x-t)/(1+t), v = 1+u, z = x/((1+t) v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = 1/((1+t)(v-A)) = 1/(1-t*(exp(A)-1)). - Tom Copeland, Oct 06 2011
The general results on the convolution of the refined partition polynomials of A134685, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016
E.g.f.: C(u,t) = (u-t)/(1+t) - W( -((t*exp((u-t)/(1+t)))/(1+t)) ), where W is the principal value of the Lambert W-function. - Cheng Peng, Sep 11 2021
The function C(u,t) in the previous formula by Peng is precisely the function A(u,t) given in the initial 2008 formula of this section and the Oct 06 2011 formula from Copeland. As noted in A000169, Euler's tree function is T(x) = -LambertW(-x), where W(x) is the principal branch of Lambert's function, and T(x) is the e.g.f. of A000169. - Tom Copeland, May 13 2022

Extensions

Reference to A181996 added by N. J. A. Sloane, Apr 05 2012
Further edits by N. J. A. Sloane, Jan 24 2020

A269939 Triangle read by rows, Ward numbers T(n, k) = Sum_{m=0..k} (-1)^(m + k) * binomial(n + k, n + m) * Stirling2(n + m, m), for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 10, 15, 0, 1, 25, 105, 105, 0, 1, 56, 490, 1260, 945, 0, 1, 119, 1918, 9450, 17325, 10395, 0, 1, 246, 6825, 56980, 190575, 270270, 135135, 0, 1, 501, 22935, 302995, 1636635, 4099095, 4729725, 2027025
Offset: 0

Views

Author

Peter Luschny, Mar 26 2016

Keywords

Comments

We propose to call this sequence the 'Ward set numbers' and sequence A269940 the 'Ward cycle numbers'. - Peter Luschny, Nov 25 2022

Examples

			Triangle starts:
  1;
  0, 1;
  0, 1,   3;
  0, 1,  10,   15;
  0, 1,  25,  105,   105;
  0, 1,  56,  490,  1260,    945;
  0, 1, 119, 1918,  9450,  17325,  10395;
  0, 1, 246, 6825, 56980, 190575, 270270, 135135;
		

Crossrefs

Variants: A134991 (main entry for this triangle), A181996.
Row sums are A000311.
Alternating row sums are signed factorials A133942.
Cf. A269940 (Stirling1 counterpart), A268437.

Programs

  • Maple
    # first version
    A269939 := (n,k) -> add((-1)^(m+k)*binomial(n+k,n+m)*Stirling2(n+m, m), m=0..k):
    seq(seq(A269939(n,k), k=0..n), n=0..8);
    # Alternatively:
    T := proc(n,k) option remember;
        `if`(k=0 and n=0, 1,
        `if`(k<=0 or k>n, 0,
        k*T(n-1,k)+(n+k-1)*T(n-1,k-1))) end:
    for n from 0 to 6 do seq(T(n,k),k=0..n) od;
    # simple, third version
    T := (n,k)->  (n+k)!*coeftayl((exp(z)-z-1)^k/k!, z=0, n+k); # Marko Riedel, Apr 14 2016
  • Mathematica
    Table[Sum[(-1)^(m + k) Binomial[n + k, n + m] StirlingS2[n + m, m], {m, 0, k}], {n, 0, 8}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 15 2016 *)
  • PARI
    T(n) = {[Vecrev(Pol(p)) | p<-Vec(serlaplace(1/((1+y)*(1 + lambertw(-y/(1+y)*exp((x-y)/(1+y) + O(x*x^n)))))))]}
    { my(A=T(8)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Jan 14 2022
  • Sage
    T = lambda n,k: sum((-1)^(m+k)*binomial(n+k,n+m)*stirling_number2(n+m,m) for m in (0..k))
    for n in (0..6): print([T(n,k) for k in (0..n)])
    
  • Sage
    # uses[PtransMatrix from A269941]
    PtransMatrix(8, lambda n: 1/(n+1), lambda n, k: (-1)^k*falling_factorial(n+k,n))
    

Formula

T(n,k) = (-1)^k*FF(n+k,n)*P[n,k](1/(n+1)) where P is the P-transform and FF the falling factorial function. For the definition of the P-transform see the link.
T(n,k) = A268437(n,k)*FF(n+k,n)/(2*n)!.
T(n,k) = (n+k)! [z^{n+k}] (exp(z)-z-1)^k/k!. - Marko Riedel, Apr 14 2016
From Fabián Pereyra, Jan 12 2022: (Start)
T(n,k) = k*T(n-1,k) + (n+k-1)*T(n-1,k-1) for n > 0, T(0,0) = 1, T(n,0) = 0 for n > 0. (See the second Maple program.)
E.g.f.: A(x,t) = 1/((1+t)*(1 + W(-t/(1+t)*exp((x-t)/(1+t))))), where W(x) is the Lambert W-function.
T(n,k) = Sum_{j=0..k} E2(n,j)*binomial(n-j,k-j), where E2(n,k) are the second-order Eulerian numbers A340556.
T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A112486(n,j)*binomial(j,k). (End)
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