

A111999


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


13



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
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OFFSET

1,2


COMMENTS

Stirling1(n,nm) = A008275(n,nm) = Sum_{k=0..m1}a(m,k)*binomial(n,2*mk).
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.
P. Bala, Diagonals of triangles with generating function exp(t*F(x)).
S. Butler, P. Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, page 4.
T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms
EqWorld, Integral Transforms
D. J. Jeffrey, G. A. Kalugin, N. Murdoch, Lagrange inversion and Lambert W, Preprint 2015.
W. Lang, First 10 rows.
L. Takacs, On the number of distinct forests, SIAM J. Discrete Math., 3 (1990), 574581. 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*mk1)*(a(m1, k) + a(m1, k1)) 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 2mk+1): (1)^(2mk+1+j) C(2mk+1,j) St1d(j,m),
where C(n,j) is the binomial coefficient and St1d(j,m) is the (jm)th element of the mth subdiagonal of A008275 for (jm)>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 Powerlaw Functions at EqWorld):
With K(p,s) = p^(s1) 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


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 (secondorder Eulerian triangle) for a similar formula for Stirling1(n, nm).
Cf. A000457, A000906, A001147, A008275, A008276, A008306, A112001, A133932,
Sequence in context: A051917 A223523 A133932 * A286947 A190961 A126323
Adjacent sequences: A111996 A111997 A111998 * A112000 A112001 A112002


KEYWORD

sign,easy,tabl


AUTHOR

Wolfdieter Lang, Sep 12 2005


EXTENSIONS

Typo in an exponent corrected in one of my formulas by Tom Copeland, Dec 19 2015


STATUS

approved



