cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

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%I A111999 #101 Aug 05 2025 14:39:40
%S A111999 -1,3,2,-15,-20,-6,105,210,130,24,-945,-2520,-2380,-924,-120,10395,
%T A111999 34650,44100,26432,7308,720,-135135,-540540,-866250,-705320,-303660,
%U A111999 -64224,-5040,2027025,9459450,18288270,18858840,11098780,3678840,623376,40320,-34459425,-183783600,-416215800
%N 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.
%C A111999 Previous name was: A triangle that converts certain binomials into triangle A008276 (diagonals of signed Stirling1 triangle A008275).
%C A111999 Stirling1(n,n-m) = A008275(n,n-m) = Sum_{k=0..m-1}a(m,k)*binomial(n,2*m-k).
%C A111999 The unsigned column sequences start with A001147, A000906 = 2*A000457, 2*|A112000|, 4*|A112001|.
%C A111999 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
%D A111999 Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 152. Table C_{m, nu}.
%H A111999 Peter Bala, <a href="/A112007/a112007_Bala.txt">Diagonals of triangles with generating function exp(t*F(x)).</a>
%H A111999 Steve Butler and Pavel Karasik, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Butler/butler7.html">A note on nested sums</a>, J. Int. Seq. 13 (2010), 10.4.4, page 4.
%H A111999 Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/12/21/generators-inversion-and-matrix-binomial-and-integral-transforms/">Generators, Inversion, and Matrix, Binomial, and Integral Transforms</a>
%H A111999 Bishal Deb and Alan D. Sokal, <a href="https://arxiv.org/abs/2507.18959">Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness</a>, arXiv:2507.18959 [math.CO], 2025. See p. 6.
%H A111999 EqWorld, <a href="http://eqworld.ipmnet.ru/en/auxiliary/aux-inttrans.htm">Integral Transforms</a>
%H A111999 David J. Jeffrey, G. A. Kalugin, N. Murdoch, <a href="https://doi.org/10.1109/SYNASC.2015.16">Lagrange inversion and Lambert W</a>, 17th Int'l Symp. Symb. Numer. Algor. Sci. Comp. (SYNASC 2015).
%H A111999 Wolfdieter Lang, <a href="/A111999/a111999.txt">First 10 rows</a>.
%H A111999 Lajos Takács, <a href="http://dx.doi.org/10.1137/0403050">On the number of distinct forests</a>, SIAM J. Discrete Math., 3 (1990), 574-581. Table 3 gives an unsigned version of the triangle.
%F A111999 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.
%F A111999 From _Tom Copeland_, May 05 2010 (updated Sep 12 2011): (Start)
%F A111999 The integral from 0 to infinity w.r.t. w of
%F A111999   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.
%F A111999 Let g(u,x) be obtained from f(u,z) by replacing z^n with x^(n+1)/(n+1)!;
%F A111999 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).
%F A111999 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.
%F A111999 With h(u,x)= 1/(dg^(-1)/dx)= (1+u*x)/(1+(1+u)*u*x),
%F A111999 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)
%F A111999 From _Tom Copeland_, May 06 2010: (Start)
%F A111999 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),
%F A111999 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,
%F A111999 e.g., St1d(1,1) = 0, St1d(2,1) = -1, St1d(3,1) = -3, St1d(4,1) = -6. (End)
%F A111999 From _Tom Copeland_, Sep 03 2011 (updated Sep 12 2011): (Start)
%F A111999 The integral from 0 to infinity w.r.t. w of
%F A111999   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.
%F A111999 Let G(u,x) be obtained from F(u,z) by replacing z^n with x^(n+1)/(n+1)!;
%F A111999 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).
%F A111999 G^(-1)(u,x) = ((1+u)*u*x - log(1+u*x))/u^2 is the comp. inverse of G(u,x) in x.
%F A111999 With H(u,x) = 1/(dG^(-1)/dx) = (1+u*x)/(1+(1+u)*x),
%F A111999 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)
%F A111999 From _Tom Copeland_, Sep 16 2011: (Start)
%F A111999 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):
%F A111999 With K(p,s) = p^(-s-1) exp(p) Γ(s+1,p),
%F A111999 f(u,z) = K(p,s)/(u*z) with p=(u+1)/(u*z) and s=1/z , and
%F A111999 F(u,z) = K(p,s)/(u*z) with p=(u+1)/(u^2*z) and s=1/(u^2*z). (End)
%F A111999 Diagonals of A008306 are reversed rows of A111999 (see P. Bala). - _Tom Copeland_, May 08 2012
%e A111999 Triangle starts:
%e A111999   [1]      -1;
%e A111999   [2]       3,       2;
%e A111999   [3]     -15,     -20,       -6;
%e A111999   [4]     105,     210,      130,       24;
%e A111999   [5]    -945,   -2520,    -2380,     -924,     -120;
%e A111999   [6]   10395,   34650,    44100,    26432,     7308,     720;
%e A111999   [7] -135135, -540540,  -866250,  -705320,  -303660,  -64224,  -5040;
%e A111999   [8] 2027025, 9459450, 18288270, 18858840, 11098780, 3678840, 623376, 40320.
%p A111999 CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
%p A111999 E2 := (n, k) -> combinat[eulerian2](n, k):
%p A111999 poly := n -> (-1)^n*add(E2(n, n-k)*(1+x)^(n-k), k = 0..n):
%p A111999 seq(CoeffList(poly(n)), n = 1..8); # _Peter Luschny_, Feb 05 2021
%t A111999 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 *)
%Y A111999 Row sums give A032188(m+1)*(-1)^m, m>=1. Unsigned row sums give A032188(m+1), m>=1.
%Y A111999 Cf. A008517 (second-order Eulerian triangle) for a similar formula for |Stirling1(n, n-m)|.
%Y A111999 Cf. A000457, A000906, A001147, A008275, A008276, A008306, A112001, A133932,
%K A111999 sign,easy,tabl
%O A111999 1,2
%A A111999 _Wolfdieter Lang_, Sep 12 2005
%E A111999 New name from _Peter Luschny_, Feb 05 2021