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 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 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), 574-581. Table 3 gives an unsigned version of the triangle. FORMULA a(m, k)=0 if m0, 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 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: 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

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Last modified May 25 19:28 EDT 2017. Contains 287059 sequences.