A111999 T(n, k) = [x^k] (-1)^n*Sum_{k=0..n} E2(n, n-k)*(1+x)^(n-k) where E2(n, k) are the second-order Eulerian numbers. Triangle read by rows, T(n, k) for n >= 1 and 0 <= k <= n.

-1, 3, 2, -15, -20, -6, 105, 210, 130, 24, -945, -2520, -2380, -924, -120, 10395, 34650, 44100, 26432, 7308, 720, -135135, -540540, -866250, -705320, -303660, -64224, -5040, 2027025, 9459450, 18288270, 18858840, 11098780, 3678840, 623376, 40320, -34459425, -183783600, -416215800

Offset: 1

Author

Wolfdieter Lang, Sep 12 2005

Keywords

• sign
• easy
• tabl

Comments

Previous name was: A triangle that converts certain binomials into triangle A008276 (diagonals of signed Stirling1 triangle A008275).

Stirling1(n,n-m) = A008275(n,n-m) = Sum_{k=0..m-1}a(m,k)*binomial(n,2*m-k).

The unsigned column sequences start with A001147, A000906 = 2*A000457, 2*|A112000|, 4*|A112001|.

The general results on the convolution of the refined partition polynomials of A133932, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these unsigned polynomials. - Tom Copeland, Sep 20 2016

Examples

			Triangle starts:
  [1]      -1;
  [2]       3,       2;
  [3]     -15,     -20,       -6;
  [4]     105,     210,      130,       24;
  [5]    -945,   -2520,    -2380,     -924,     -120;
  [6]   10395,   34650,    44100,    26432,     7308,     720;
  [7] -135135, -540540,  -866250,  -705320,  -303660,  -64224,  -5040;
  [8] 2027025, 9459450, 18288270, 18858840, 11098780, 3678840, 623376, 40320.
		

References

• Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 152. Table C_{m, nu}.

Links

• Peter Bala, Diagonals of triangles with generating function exp(t*F(x)).
• Steve Butler and Pavel Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, page 4.
• Tom Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms
• Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 6.
• EqWorld, Integral Transforms
• David J. Jeffrey, G. A. Kalugin, N. Murdoch, Lagrange inversion and Lambert W, 17th Int'l Symp. Symb. Numer. Algor. Sci. Comp. (SYNASC 2015).
• Wolfdieter Lang, First 10 rows.
• Lajos Takács, On the number of distinct forests, SIAM J. Discrete Math., 3 (1990), 574-581. Table 3 gives an unsigned version of the triangle.

Crossrefs

Row sums give A032188(m+1)*(-1)^m, m>=1. Unsigned row sums give A032188(m+1), m>=1.

Cf. A008517 (second-order Eulerian triangle) for a similar formula for |Stirling1(n, n-m)|.

Cf. A000457, A000906, A001147, A008275, A008276, A008306, A112001, A133932,

Programs

Maple

CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
E2 := (n, k) -> combinat[eulerian2](n, k):
poly := n -> (-1)^n*add(E2(n, n-k)*(1+x)^(n-k), k = 0..n):
seq(CoeffList(poly(n)), n = 1..8); # Peter Luschny, Feb 05 2021

Mathematica

a[m_, k_] := a[m, k] = Which[m < k + 1, 0, And[m == 1, k == 0], -1, k == -1, 0, True, -(2 m - k - 1)*(a[m - 1, k] + a[m - 1, k - 1])]; Table[a[m, k], {m, 9}, {k, 0, m - 1}] // Flatten (* Michael De Vlieger, Sep 23 2016 *)

Formula

a(m, k)=0 if m

From Tom Copeland, May 05 2010 (updated Sep 12 2011): (Start)

The integral from 0 to infinity w.r.t. w of

exp[-w(u+1)] (1+u*z*w)^(1/z) gives a power series, f(u,z), in z for reversed row polynomials in u of A111999, related to an Euler transform of diagonals of A008275.

Let g(u,x) be obtained from f(u,z) by replacing z^n with x^(n+1)/(n+1)!;

g(u,x)= x - u^2 x^2/2! + (2 u^3 + 3 u^4) x^3/3! - (6 u^4 + 20 u^5 + 15 u^6) x^4/4! + ... , an e.g.f. associated to f(u,z).

Then g^(-1)(u,x)=(1+u)*x - log(1+u*x) is the comp. inverse of g(u,x) in x, and, consequently, A133932 is a refinement of A111999.

With h(u,x)= 1/(dg^(-1)/dx)= (1+u*x)/(1+(1+u)*u*x),

g(u,x)=exp[x*h(u,t)d/dt] t, evaluated at t=0. Also, dg(u,x)/dx = h(u,g(u,x)). (End)

From Tom Copeland, May 06 2010: (Start)

For m,k>0, a(m,k) = Sum(j=2 to 2m-k+1): (-1)^(2m-k+1+j) C(2m-k+1,j) St1d(j,m),

where C(n,j) is the binomial coefficient and St1d(j,m) is the (j-m)-th element of the m-th subdiagonal of A008275 for (j-m)>0 and is 0 otherwise,

e.g., St1d(1,1) = 0, St1d(2,1) = -1, St1d(3,1) = -3, St1d(4,1) = -6. (End)

From Tom Copeland, Sep 03 2011 (updated Sep 12 2011): (Start)

The integral from 0 to infinity w.r.t. w of

exp[-w*(u+1)/u] (1+u*z*w)^(1/(u^2*z)) gives a power series, F(u,z), in z for the row polynomials in u of A111999.

Let G(u,x) be obtained from F(u,z) by replacing z^n with x^(n+1)/(n+1)!;

G(u,x) = x - x^2/2! + (3 + 2 u) x^3/3! - (15 + 20 u + 6 u^2) x^4/4! + ... , an e.g.f. for A111999 associated to F(u,z).

G^(-1)(u,x) = ((1+u)*u*x - log(1+u*x))/u^2 is the comp. inverse of G(u,x) in x.

With H(u,x) = 1/(dG^(-1)/dx) = (1+u*x)/(1+(1+u)*x),

G(u,x) = exp[x*H(u,t)d/dt] t, evaluated at t=0. Also, dG(u,x)/dx = H(u,G(u,x)). (End)

From Tom Copeland, Sep 16 2011: (Start)

f(u,z) and F(u,z) are expressible in terms of the incomplete gamma function Γ(v,p)(see Laplace Transforms for Power-law Functions at EqWorld):

With K(p,s) = p^(-s-1) exp(p) Γ(s+1,p),

f(u,z) = K(p,s)/(u*z) with p=(u+1)/(u*z) and s=1/z , and

F(u,z) = K(p,s)/(u*z) with p=(u+1)/(u^2*z) and s=1/(u^2*z). (End)

Diagonals of A008306 are reversed rows of A111999 (see P. Bala). - Tom Copeland, May 08 2012

Extensions

New name from Peter Luschny, Feb 05 2021

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.