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

1,2

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

References

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

Links

Table of n, a(n) for n=1..39.

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.

Formula

a(m, k)=0 if m<k+1; a(1, 0)=-1; a(m, -1):= 0; a(m, k) = -(2*m-k-1)*(a(m-1, k) + a(m-1, k-1)) else.

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 Gamma(v,p)(see Laplace Transforms for Power-law Functions at EqWorld):

With K(p,s) = p^(-s-1) exp(p) * Gamma(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

Example

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.

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 *)

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,

Sequence in context: A223523 A357613 A133932 * A286947 A392432 A334559

Adjacent sequences: A111996 A111997 A111998 * A112000 A112001 A112002

Keyword

sign,easy,tabl

Author

Wolfdieter Lang, Sep 12 2005

Extensions

New name from Peter Luschny, Feb 05 2021

Status

approved