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 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
S. Butler, P. Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, page 4.
EqWorld, Integral Transforms
D. J. Jeffrey, G. A. Kalugin, N. Murdoch, Lagrange inversion and Lambert W, Preprint 2015.
Wolfdieter Lang, First 10 rows.
L. Takacs, 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 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)
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
Cf. A008517 (second-order Eulerian triangle) for a similar formula for |Stirling1(n, n-m)|.
KEYWORD
AUTHOR
Wolfdieter Lang, Sep 12 2005
EXTENSIONS
New name from Peter Luschny, Feb 05 2021
STATUS
approved