cp's OEIS Frontend

Apparently, a raising operator for bivariate polynomials P(n,u,z) having these coefficients is R = (u+z)^2 * z * d/dz with P(0,u,z) = z. E.g., R P(1,u,z) = R^2 P(0,u,z) = R^2 z = u^4 z + 6 u^3 z^2 + 12 u^2 z^3 + 10 u z^4 + 3 z^5 = P(2,u,z). See the Kazarian link. - Tom Copeland, Jun 12 2015

Reverse polynomials seem to be generated by 1 + exp[t*(x+1+z)^2*(1+z)d/dz]z evaluated at z = 0. - Tom Copeland, Jun 13 2015

From Peter Bala, Jun 14 2016: (Start)

T(n,k) = k*T(n,k) + 2*(k - 1)*T(n,k-1) + (k - 2)*T(n,k-2).

n-th diagonal of A008277: Stirling2(N + n - 1,N) = Sum_{k = 1..2*n - 1} T(n,k)*binomial(N - n,k - 1) for N = 1,2,3,....

Row polynomials R(n,z) = Sum_{k >= 1} k^(n+k-1)*( z/(1 + z)*exp(-z/(1 + z)) )^k/k!, n = 1,2,..., follows from the formula given in A008277 for the o.g.f.'s of the diagonals of the Stirling numbers of the second kind.

Consequently, R(n+1,z) = (1 + z)^2*z*d/dz(R(n,z)) for n >= 1 as conjectured above by Copeland.

R(n,z) = (1 + z)^n*P(n,z) where P(n,z) are the row polynomials of A134991.

R(n,z) = (1 + z)^(2*n+1)*B(n,z/(1 + z)), where B(n,z) are the row polynomials of the triangle of second-order Eulerian numbers A008517 (see Barbero et al., Section 6, equation 27). (End)

Based on the comment of Bala the row polynomials have the explicit form R(n, z) = (1+z)^(n+1)*Sum_{k=0..n}(z^k*Sum_{m=0..k}((-1)^(m+k)*binomial(n+k, n+m)* Stirling2(n+m,m))). - Peter Luschny, Jun 15 2016

E.g.f. G(x,y) satisfies G(x,y) = y*(exp(x)*exp(G(x,y)) - G(x,y) - 1). - Andrew Howroyd, Mar 28 2025

E.g.f.(x): T*(1 + 52*T + 328*T^2 + 444*T^3 + 120*T^4)*(1-T)^(-11); where T=T(x) is Euler's tree function (see A000169). - Len Smiley, Nov 17 2001

See A008517 and A134991 for similar e.g.f.s and diagonals of A048993. - Tom Copeland, Oct 03 2011

E.g.f.: d^5/dx^5 {x^5/(T(x)^5*(1-T(x)))}, where T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is the tree function of A000169. - Peter Bala, Aug 05 2012

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

From Peter Bala, Sep 30 2011: (Start)

E.g.f.: A(x,t) = -1-((1+t)/t)*LambertW(-(t/(1+t))*exp((x-t)/(1+t))) = x + (1+t)*x^2/2! + (1+4*t+3*t^2)*x^3/3! + .... A(x,t) is the inverse function of (1+t)*log(1+x)-t*x.

A(x,t) satisfies the partial differential equation (1-x*t)*dA/dx = 1 + A + t*(1+t)*dA/dt. It follows that the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) =(n*t+1)*R(n,t) + t*(1+t)*dR(n,t)/dt. Cf. A054589 and A075856. The polynomials t/(1+t)*R(n,t) are the row polynomials of A134991.

The generating function A(x,t) satisfies the autonomous differential equation dA/dx = (1+A)/(1-t*A). Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees on n+1 vertices where the non-leaf vertices of outdegree k come in t^(k-1)*(1+t) colors. An example is given below. Cf. A006351, which corresponds to the case t = 1. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x) = (1+x)/(1-x*t). Then R(n,t) = (f(x)*d/dx)^n(f(x)) evaluated at x = 0. (End)

Sum_{j=0..n} T(n-j,j) = A000110(n). - Alois P. Heinz, Jun 20 2022

From Mikhail Kurkov, Apr 01 2025: (Start)

E.g.f.: B(y) = -w/(x*(1+w)) where w = LambertW(-x/(1+x)*exp((y-x)/(1+x))) satisfies the first-order ordinary differential equation (1+x)*B'(y) = B(y)*(1+x*B(y))^2, hence row polynomials are P(n,x) = P(n-1,x) + x*Sum_{j=0..n-1} binomial(n, j)*P(j,x)*P(n-j-1,x) for n > 0 with P(0,x) = 1 (see MathOverflow link).

Conjecture: row polynomials are P(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling1(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(j+k)/((n+j-k)!*(i-j)!*k!). (End)

Conjecture: g.f. satisfies 1/(1 - x - x*y/(1 - 2*x - 2*x*y/(1 - 3*x - 3*x*y/(1 - 4*x - 4*x*y/(1 - 5*x - 5*x*y/(1 - ...)))))) (see A383019 for conjectures about combinatorial interpretation and algorithm for efficient computing). - Mikhail Kurkov, Apr 21 2025

a(k, m) = (k+m)*a(k-1, m) + (k+m-1)*a(k-1, m-1) for k >= m >= 0, a(0, 0)=1, a(k, -1):=0, a(k, m)=0 if k < m.

From Tom Copeland, Oct 05 2011: (Start)

With polynomials

P(0,t) = 0

P(1,t) = 1

P(2,t) = -(1 + t)

P(3,t) = 2 + 5 t + 3 t^2

P(4,t) = -( 6 + 26 t + 35 t^2 + 15 t^3)

P(5,t) = 24 + 154 t +340 t^2 + 315 t^3 + 105 t^4

Apparently, P(n,t) = (-1)^(n+1) PW[n,-(1+t)] where PW are the Ward polynomials A134991. If so, an e.g.f. for the polynomials is

A(x,t) = -(x+t+1)/t - LW{-((t+1)/t) exp[-(x+t+1)/t]}, where LW(x) is a suitable branch of the Lambert W Fct. (e.g., see A135338). The comp. inverse in x (about x = 0) is B(x) = x + (t+1) [exp(x) - x - 1]. See A112487 for special case t = 1. These results are a special case of A134685 with u(x) = B(x), i.e., u_1=1 and (u_n)=(1+t) for n>0.

Let h(x,t) = 1/(dB(x)/dx) = 1/[1+(1+t)*(exp(x)-1)], an e.g.f. in x for row polynomials in t of signed A028246 , then P(n,t), is given by

(h(x,t)*d/dx)^n x, evaluated at x=0, i.e., A(x,t)=exp(x*h(u,t)*d/du) u, evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t).

The e.g.f. A(x,t) = -v * Sum_{j>=1} D(j-1,u) (-z)^j / j! where u=-(x+t+1)/t, v=1+u, z=(1+t*v)/(t*v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = -1/[t*(v-A)].(End)

A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=t+1, and (a_n)=t*P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 =0. - Tom Copeland, Oct 08 2011

The row polynomials R(n,x) may be calculated using R(n,x) = 1/x^(n+1)*D^n(x), where D is the operator (x^2+x^3)*d/dx. - Peter Bala, Jul 23 2012

For n>0, Sum_{k=0..n} a(n,k)*(-1/(1+W(t)))^(n+k+1) = (t d/dt)^(n+1) W(t), where W(t) is Lambert W function. For t=-x, this gives Sum_{k>=1} k^(k+n)*x^k/k! = - Sum_{k=0..n} a(n,k)*(-1/(1+W(-x)))^(n+k+1). - Max Alekseyev, Nov 21 2019

Conjecture: row polynomials are R(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling2(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(n+j+k)/((n+j-k)!*(i-j)!*k!). - Mikhail Kurkov, Apr 21 2025

T(n,k) = (-1)^k*FF(n+k,n)*P[n,k](1/(n+1)) where P is the P-transform and FF the falling factorial function. For the definition of the P-transform see the link.

T(n,k) = A268437(n,k)*FF(n+k,n)/(2*n)!.

T(n,k) = (n+k)! [z^{n+k}] (exp(z)-z-1)^k/k!. - Marko Riedel, Apr 14 2016

From Fabián Pereyra, Jan 12 2022: (Start)

T(n,k) = k*T(n-1,k) + (n+k-1)*T(n-1,k-1) for n > 0, T(0,0) = 1, T(n,0) = 0 for n > 0. (See the second Maple program.)

E.g.f.: A(x,t) = 1/((1+t)*(1 + W(-t/(1+t)*exp((x-t)/(1+t))))), where W(x) is the Lambert W-function.

T(n,k) = Sum_{j=0..k} E2(n,j)*binomial(n-j,k-j), where E2(n,k) are the second-order Eulerian numbers A340556.

T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A112486(n,j)*binomial(j,k). (End)

E.g.f.(x): T*(1 +8*T +6*T^2)*(1-T)^(-7); where T=T(x) is Euler's tree function (see A000169). - Len Smiley, Nov 19 2001

See A008517 and A134991 for similar e.g.f.s and diagonals of A048993. - Tom Copeland, Oct 03 2011

E.g.f.: d^3/dx^3 {x^3/(T(x)^3*(1-T(x)))}, where T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is the tree function of A000169. - Peter Bala, Aug 05 2012

a(n) = n*A008788(n). - R. J. Mathar, Oct 31 2015

E.g.f.: T*(1 +22*T +58*T^2 +24*T^3)*(1-T)^(-9); where T is Euler's tree function (see A000169). - Len Smiley, Nov 17 2001

See A008517 and A134991 for similar e.g.f.s and diagonals of A048993. - Tom Copeland, Oct 03 2011

E.g.f.: d^4/dx^4 {x^4/(T(x)^4*(1-T(x)))}, where T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is the tree function of A000169. - Peter Bala, Aug 05 2012

a(n) = Sum_{m=0..n} A112486(n, m), n >= 0.

a(n) = 2*A032188(n+1), n > 0. - Vladeta Jovovic, Jul 11 2007

From Paul D. Hanna, Jun 30 2009: (Start)

E.g.f. A(x) satisfies: A'(x) = A(x)^2 + A(x)^3.

E.g.f. A(x) satisfies: A(x) = exp( Integral[A(x) + A(x)^2]dx ) with A(0)=1. (End)

E.g.f. A(x) satisfies: A(x) = 2*exp(A(x)) - (2+x), where A(x) = Sum_{n>=0} a(n)*x^(n+1)/(n+1)! (the e.g.f. when offset=1). - Paul D. Hanna, Sep 23 2011

From Tom Copeland, Oct 05 2011: (Start)

With c(0)= 0 and c(n+1)= (-1)^n a(n) for n>=0, c(n)=(-1)^(n+1) PW(n,-2) with PW the Ward polynomials A134991. E.g.f. for the c(n) is A(x) = -(x+2)-LW{-2 exp[-(x+2)]}, where LW(x) is a suitable branch of the Lambert W Fct. (see A135338).

The compositional inverse is B(x) = x + 2(exp(x) - x - 1). These results are a special case of A134685 with u(x)=B(x), i.e., u_1=1 and (u_n)=2 for n>0.

Let h(x) = 1/(dB(x)/dx) = 1/[1+2(exp(x)-1)], then c(n) is given by (h(x)*d/dx)^n x, evaluated at x=0, i.e., A(x) = exp(x*h(u)*d/du) u, evaluated at u=0. Also, dA(x)/dx = h(A(x)).

The e.g.f. A(x) = -v * Sum_(j>=1) D(j-1,u) (-z)^j/ j! where u=-(x+2), v=1+u, z=(1+v)/(v^2) and D(j-1,u) are the polynomials of A042977. (End)

a(n) = n!*Sum_{k=0..n} binomial(n+k, n)*Sum_{j=0..k} (-1)^(n+j)*binomial(k, j)*Sum_{l=0..j} binomial(j, l)*(j-l)!*2^(j-l)*(-1)^l*Stirling2(n-l+j, j-l)/(n-l+j)!. - Vladimir Kruchinin, Feb 14 2012

G.f.: 1/Q(0), where Q(k)= 1 + k*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013

a(n) ~ n^n / (exp(n) * (1-log(2))^(n+1/2)). - Vaclav Kotesovec, Aug 14 2017

a(0) = 1; a(n) = n * a(n-1) + Sum_{k=0..n-1} binomial(n,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 02 2020

Row polynomials are P(n,t) = Sum_{j=1..n} C(n,j) * t^j = [ Bell(.,-t) + 2t ]^n, umbrally, where Bell(j,t) are the Touchard/Bell/exponential polynomials described in A008277, with P(0,t) = 1.

E.g.f.: exp{ t * [ -exp(x) + 2x + 1] } and [ P(.,t) + P(.,s) ]^n = P(n,s+t).

The lowering operator gives L[P(n,t)] = n * P(n-1,t) = (D-1)/2 * P(n,t) + Sum_{j>=1} j^(j-1) * 2^(-j) / j! * exp(-j/2) * P(n,t + j/2).

The raising operator gives R[P(n,t)] = P(n+1,t) = 2t * { P(n,t) - Sum_{j>=1} j^(j-1) * 2^(-j) / j! * exp(-j/2) * P(n,t + j/2) } .

Therefore P(n+1,t) = 2t * { [ (1+D)/2 * P(n,t) ] - n * P(n-1,t) }.

P(n,1) = (-1)^n * A074051(n) and P(n,-1) = A126617(n).

See Rota, Roman, Mathworld or Wikipedia on Sheffer sequences and umbral calculus for more formulas, including expansion theorems.

From Tom Copeland, Jan 20 2018: (Start)

Define Q(n,z;w) = [Bell(.,w)+z]^n. Then Q(n,z;w) are a sequence of Appell polynomials with e.g.f. exp[(exp(t)-1+z)*w], lowering operator D = d/dz, and raising operator R = z + w*exp(D), and exp[(exp(D)-1)w] z^n = exp[Bell(.,w)D] z^n = Q(n,z;w) = e^(-w) (w d/dw + z)^n e^w = e^(-w) exp(a.w) = exp[(a. - 1)w] with (a.)^k = a_k = (k + z)^n and (a. - 1)^m = sum{k = 0,..,m} (-1)^k a^(m-k). Then P(n,t) = Q(n,2t;-t).

For example, exp[(a. - 1)w] = (a. - 1)^0 + (a. - 1)^1 w + (a. - 1)^2 w^2/2! + ... = a_0 + (a_1 - a_0) w + (a_2 - 2a_1 + a_0) w^2/2! + ... = z^n + [(1+z)^n - z^n] w + [(2+z)^n - 2(1+z)^n + z^n] w^2/2! + ... . (End)

T(n+1, k) = Sum_{i = 0..n} s(n,k)*binomial(n, i)*T(i, k-1), where s(n,i) = 1 if i = n else -1. - Peter Bala, Feb 13 2025

Given the formal Taylor series or e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ...,

E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 / (D_x f^{(-1)}(x)), where D_x is the derivative w.r.t. x and f^{(-1)}(x) is the (possibly formal) compositional inverse of f(x) about the origin.

E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 f'(f^{(-1)}(x)) by the inverse function theorem, where the prime indicates differentiation w.r.t. the argument of the function f. Note the correspondence to the analytic definitions of the reciprocal tangents [RT] of A356144, consistent with the following algebraic identities.

[E]^{-1} = [P][L] = [P][E][P] = [RT][P], representing, e.g., the substitution of the permutahedra polynomials [P] of A133314 for the indeterminates of the reciprocal tangent polynomials [RT] of A356144. [E] are the refined Eulerian polynomials of A145271, and [L], the classic Lagrange inversion polynomials of A134685.

Since [P]^2 = [L]^2 = [RT]^2 = [I], the substitutional identity, i.e., [P], [L], and [RT] are involutive transformations, many identities follow from the basic ones above, e.g., [L] = [P][E]^{-1} gives an inversion formula for a formal e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ..., and we can identify [E] and [E]^{-1} as a conjugate dual.

With a_n = -x, [E]^{-1} reduces to a signed version of A112493 with an additional initial row, with the row sums of the unsigned coefficients being (1, A006351). A112493 is also given by the diagonals of A124324. See my link above on the reduced polynomials and associated arrays for more detail.

The sequence of row sums of the signed coefficients, i.e., E^{-1}(1,1,...,1), is the sequence (1, 1, 0, 0, 0, 0, ...).

Conjecture: row polynomials are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} binomial(n-1,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = a_k for k > 0. - Mikhail Kurkov, Mar 22 2025