cp's OEIS Frontend

E.g.f. A(x) satisfies exp A(x) = 2*A(x) - x + 1.

a(0)=0, a(1)=a(2)=1; for n >= 2, a(n+1) = (n+2)*a(n) + 2*Sum_{k=2..n-1} binomial(n, k)*a(k)*a(n-k+1).

a(1)=1; for n>1, a(n) = -(n-1) * a(n-1) + Sum_{k=1..n-1} binomial(n, k) * a(k) * a(n-k). - Michael Somos, Jun 04 2012

From the umbral operator L in A135494 acting on x^n comes, umbrally, (a(.) + x)^n = (n * x^(n-1) / 2) - (x^n / 2) + Sum_{j>=1} j^(j-1) * (2^(-j) / j!) * exp(-j/2) * (x + j/2)^n giving a(n) = 2^(-n) * Sum_{j>=1} j^(n-1) * ((j/2) * exp(-1/2))^j / j! for n > 1. - Tom Copeland, Feb 11 2008

Let h(x) = 1/(2-exp(x)), an e.g.f. for A000670, then the n-th term of A000311 is given by ((h(x)*d/dx)^n)x evaluated at x=0, i.e., A(x) = exp(x*a(.)) = exp(x*h(u)*d/du) u evaluated at u=0. Also, dA(x)/dx = h(A(x)). - Tom Copeland, Sep 05 2011 (The autonomous differential eqn. here is also on p. 59 of Jones. - Tom Copeland, Dec 16 2019)

A134991 gives (b.+c.)^n = 0^n, for (b_n)=A000311(n+1) and (c_0)=1, (c_1)=-1, and (c_n)=-2* A000311(n) = -A006351(n) otherwise. E.g., umbrally, (b.+c.)^2 = b_2*c_0 + 2 b_1*c_1 + b_0*c_2 =0. - Tom Copeland, Oct 19 2011

a(n) = Sum_{k=1..n-1} (n+k-1)!*Sum_{j=1..k} (1/(k-j)!)*Sum_{i=0..j} 2^i*(-1)^i*Stirling2(n+j-i-1, j-i)/((n+j-i-1)!*i!), n>1, a(0)=0, a(1)=1. - Vladimir Kruchinin, Jan 28 2012

Using L. Comtet's identity and D. Wasserman's explicit formula for the associated Stirling numbers of second kind (A008299) one gets: a(n) = Sum_{m=1..n-1} Sum_{i=0..m} (-1)^i * binomial(n+m-1,i) * Sum_{j=0..m-i} (-1)^j * ((m-i-j)^(n+m-1-i))/(j! * (m-i-j)!). - Peter Regner, Oct 08 2012

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

G.f.: x*Q(0), where Q(k) = 1 - x*(k+1)/(x*(k+1) - (1-k*x)*(1-x-k*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013

a(n) ~ n^(n-1) / (sqrt(2) * exp(n) * (2*log(2)-1)^(n-1/2)). - Vaclav Kotesovec, Jan 05 2014

E.g.f. A(x) satisfies d/dx A(x) = 1 / (1 + x - 2 * A(x)). - Michael Somos, Oct 25 2014

O.g.f.: Sum_{n>=0} x / Product_{k=0..n} (2 - k*x). - Paul D. Hanna, Oct 27 2014

E.g.f.: (x - 1 - 2 LambertW(-exp((x-1)/2) / 2)) / 2. - Vladimir Reshetnikov, Oct 16 2015 (This e.g.f. is given in A135494, the entry alluded to in my 2008 formula, and in A134991 along with its compositional inverse. - Tom Copeland, Jan 24 2018)

a(0) = 0, a(1) = 1; a(n) = n! * [x^n] exp(Sum_{k=1..n-1} a(k)*x^k/k!). - Ilya Gutkovskiy, Oct 17 2017

a(n+1) = Sum_{k=0..n} A269939(n, k) for n >= 1. - Peter Luschny, Feb 15 2021

T(n,k) = 0 if n < k, T(1,1) = 1, T(n,-1) = 0, T(n,k) = k*T(n-1,k) + (2*n-k)*T(n-1,k-1).

a(n,m) = Sum_{k=0..n-m} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-m-k+1, n-m-k+1). - Johannes W. Meijer, Oct 16 2009

From Peter Bala, Sep 29 2011: (Start)

For k = 0,1,2,... put G(k,x,t) := x-(1+2^k*t)*x^2/2+(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 1 and for the Eulerian numbers A008292 when k = 0.

Let v = -t*exp((1-t)^2*x-t) and let B(x,t) = -(1+1/t*LambertW(v))/(1+LambertW(v)). From the e.g.f. given by Copeland above we find B(x,t) = compositional inverse with respect to x of G(1,x,t) = Sum_{n>=1} R(n,t)*x^n/n! = x+(1+2*t)*x^2/2!+(1+8*t+6*t^2)*x^3/3!+.... The function B(x,t) satisfies the differential equation dB/dx = (1+B)*(1+t*B)^2 = 1 + (2*t+1)*B + t*(t+2)*B^2 + t^2*B^3.

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 where each vertex has outdegree <= 3, the vertices of outdegree 1 come in 2*t+1 colors, the vertices of outdegree 2 come in t*(t+2) colors and the vertices of outdegree 3 come in t^2 colors. An example is given below. Cf. A008292. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x,t) = (1+x)*(1+t*x)^2 and let D be the operator f(x,t)*d/dx. Then R(n+1,t) = D^n(f(x,t)) evaluated at x = 0. (End)

From Tom Copeland, Oct 03 2011: (Start)

a(n,k) = Sum_{i=0..k} (-1)^(k-i) binomial(n-i,k-i) A134991(n,i), offsets 0.

P(n+1,t) = (1-t)^(2n+1) Sum_{k>=1} k^(n+k) [t*exp(-t)]^k / k! for n>0; consequently, Sum_{k>=1} (-1)^k k^(n+k) x^k/k!= [1+LW(x)]^(-(2n+1))P[n+1,-LW(x)] where LW(x) is the Lambert W-Function and P(n,t), for n > 0, are the row polynomials as given in Copeland's 2008 comment. (End)

The e.g.f. A(x,t) = -v * { Sum_{j>=1} D(j-1,u) (-z)^j / j! } where u=x*(1-t)^2-t, v=(1+u)/(1-t), z=(t+u)/[(1+u)^2] and D(j-1,u) are the polynomials of A042977. dA(x,t)/dx=(1-t)/[1+u-(1-t)A(x,t)]=(1-t)/{1+LW[-t exp(u)]}, (Copeland's e.g.f. in 2008 comment). - Tom Copeland, Oct 06 2011

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, and (a_n)=-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 compositional inverse (with respect to x) of y = y(t;x) = (x-t*(exp(x)-1)) is 1/(1-t)*y + t/(1-t)^3*y^2/2! + (t+2*t^2)/(1-t)^5*y^3/3! + (t+8*t^2+6*t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed in the Comments section, the rational functions in t are the generating functions for the diagonals of the triangle of Stirling numbers of the second kind (A048993). See the Bala link for a proof. Cf. A112007 and A134991. - Peter Bala, Dec 04 2011

O.g.f. of row n: (1-x)^(2*n+1) * Sum_{k>=0} k^(n+k) * exp(-k*x) * x^k/k!. - Paul D. Hanna, Oct 31 2012

T(n, k) = n!*[x^n][t^k](egf) where egf = (1-t)/(1 + LambertW(-exp(t^2*x - 2*t*x - t + x)*t)) and after expansion W(-exp(-t)t) is substituted by (-t). - Shamil Shakirov, Feb 17 2025

E.g.f. for the polynomials is A(x,t) = (x-t)/(t+1) + T{ (t/(t+1)) * exp[(x-t)/(t+1)] }, where T(x) is the Tree function, the e.g.f. of A000169. The compositional inverse in x (about x = 0) is B(x) = x + -t * [exp(x) - x - 1]. Special case t = 1 gives e.g.f. for A000311. These results are a special case of A134685 with u(x) = B(x).

From Tom Copeland, Oct 26 2008: (Start)

Umbral-Sheffer formalism gives, for m a positive integer and u = t/(t+1),

[P(.,t)+Q(.,x)]^m = [m Q(m-1,x) - t Q(m,x)]/(t+1) + sum(n>=1) { n^(n-1)[u exp(-u)]^n/n! [n/(t+1)+Q(.,x)]^m }, when the series is convergent for a sequence of functions Q(n,x).

Check: With t=1; Q(n,x)=0^n, for n>=0; and Q(-1,x)=0, then [P(.,1)+Q(.,x)]^m = P(m,1) = A000311(m).

(End)

Let h(x,t) = 1/(dB(x)/dx) = 1/(1-t*(exp(x)-1)), an e.g.f. in x for row polynomials in t of A019538, then the n-th row polynomial in t of the table A134991, P(n,t), is given by ((h(x,t)*d/dx)^n)x evaluated at x=0, i.e., A(x,t) = exp(x*P(.,t)) = exp(x*h(u,t)*d/du) u evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t). - Tom Copeland, Sep 05 2011

The polynomials (1+t)/t*P(n,t) are the row polynomials of A112493. Let f(x) = (1+x)/(1-x*t). Then for n >= 0, P(n+1,t) is given by t/(1+t)*(f(x)*d/dx)^n(f(x)) evaluated at x = 0. - Peter Bala, Sep 30 2011

From Tom Copeland, Oct 04 2011: (Start)

T(n,k) = (k+1)*T(n-1,k) + (n+k+1)*T(n-1,k-1) with starting indices n=0 and k=0 beginning with P(2,t) (as suggested by a formula of David Speyer on MathOverflow).

T(n,k) = k*T(n-1,k) + (n+k-1)*T(n-1,k-1) with starting indices n=1 and k=1 of table (cf. Smiley above and Riordin ref.[10] therein).

P(n,t) = (1/(1+t))^n * Sum_{k>=1} k^(n+k-1)*(u*exp(-u))^k / k! with u=(t/(t+1)) for n>1; therefore, Sum_{k>=1} (-1)^k k^(n+k-1) x^k/k! = [1+LW(x)]^(-n) P{n,-LW(x)/[1+LW(x)]}, with LW(x) the Lambert W-Fct.

T(n,k) = Sum_{i=0..k} ((-1)^i binomial(n+k,i) Sum_{j=0..k-i} (-1)^j (k-i-j)^(n+k-i)/(j!(k-i-j)!)) from relation to A008299. (End)

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 = x/((1+t) v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = 1/((1+t)(v-A)) = 1/(1-t*(exp(A)-1)). - Tom Copeland, Oct 06 2011

The general results on the convolution of the refined partition polynomials of A134685, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016

E.g.f.: C(u,t) = (u-t)/(1+t) - W( -((t*exp((u-t)/(1+t)))/(1+t)) ), where W is the principal value of the Lambert W-function. - Cheng Peng, Sep 11 2021

The function C(u,t) in the previous formula by Peng is precisely the function A(u,t) given in the initial 2008 formula of this section and the Oct 06 2011 formula from Copeland. As noted in A000169, Euler's tree function is T(x) = -LambertW(-x), where W(x) is the principal branch of Lambert's function, and T(x) is the e.g.f. of A000169. - Tom Copeland, May 13 2022

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

The polynomials p_n = Sum a[n, k]x^k satisfy p_1=1 and p_(n+1) = x*x*dp_n/dx+n*(1+x)*p_n.

From Peter Bala, Sep 29 2011: (Start)

E.g.f.: series reversion with respect to x of (1-t+(t-1+x*t)*exp(-x)) = x + (1+t)*x^2/2! + (2+4*t+3*t^2)*x^3/3! + ....

The sequence of shifted row polynomials {p_n(1+t)}n>=1 begins [1,2+t,9+10*t+3*t^2,...]. These are the row polynomials of A048160.

(End)

Let f(x) = exp(x)/(1-t*x). The e.g.f. A(x,t) = x + (1+t)*x^2/2! + (2+4*t+3*t^2)*x^3/3! + ... satisfies the autonomous differential equation dA/dx = f(A). The n-th row polynomial (n>=1) equals D^(n-1)(f(x)) evaluated at x = 0, where D is the operator f(x)*d/dx (apply [Dominici, Theorem 4.1]). - Peter Bala, Nov 09 2011

The polynomials (1+t)^(n-1)*p_n(1/(1+t)) are (up to sign) the row polynomials of A042977. - Peter Bala, Jul 23 2012

Let q_n = Sum_{k>=0} a(n,k)*t^(n-k), with q_0 = 1. (So q_1=t, q_2 = t+t^2, and q_3 = 3*t + 4*t^2 + 2*t^3.) Then Sum_{n>=0} q_n*x^n/n! = t - W((t-1-t^2*x)*exp(t-1)), where W is the Lambert function. - Ira M. Gessel, Jan 06 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

E.g.f.: LambertW(exp(1)*(1+4*x)). - Vladeta Jovovic, Nov 19 2003

|a(n)| ~ 4^n * exp(n) * n^(n-1) / (1+exp(2))^(n-1/2). - Vaclav Kotesovec, Jul 09 2013

G.f.: A(x) = B(x)^3*(9-8*B(x)+2*B(x)^2)/(1-B(x))^5, where B(x) is g.f. for rooted trees with n nodes, cf. A000081.

a(n) ~ c * d^n * n^(3/2), where d = A051491 = 2.9557652856519949747148..., c = 0.244665117500618173509... . - Vaclav Kotesovec, Sep 11 2014

G.f.: A(x) = B(x)^4*(64-79*B(x)+36*B(x)^2-6*B(x)^3)/(1-B(x))^7, where B(x) is g.f. for rooted trees with n nodes, cf. A000081.

The row polynomials P(n,t) = Sum_{j=1..n} C(n,j) * t^j satisfy exp[P(.,t) * x] = exp{ -t * [(1+x) * log(1+x) - 2*x] }, with P(0,t) = 1 and [ P(.,x) + P(.,y) ]^n = P(n,x+y). Here, as in the e.g.f., the umbral maneuver P(.,t)^n = P(n,t) is assumed. See Mathworld and Wikipedia on Sheffer sequences and umbral calculus for other general formulas, including expansion theorems.

From Peter Bala, Dec 09 2011: (Start)

E.g.f.: exp(t*(2*x-(1+x)*log(1+x))) = 1 + t*x + (t^2-t)*x^2/2! + (t^3-3*t^2+t)*x^3/3! + ... (Restatement of Copeland's e.g.f. above in umbral notation with P(.,t)^n = P(n,t).).

If a triangular array has an e.g.f. of the form exp(t*F(x)) with F(0) = 0, then the o.g.f.'s for the diagonals of the triangle are rational functions in t (see the Bala link). The rational functions are the coefficients in the compositional inverse (with respect to x) (x-t*F(x))^(-1). In this case (x-t*(2*x-(1+x)*log(1+x)))^(-1) = x/(1-t) - t/(1-t)^3*x^2/2! + (t+2*t^2)/(1-t)^5*x^3/3! - (2*t+6*t^2+7*t^3)/(1-t)^7*x^4/4! + ... . So, for example, the (unsigned) third subdiagonal has o.g.f. (2*t+6*t^2+7*t^3)/(1-t)^7 = 2*t + 20*t^2 + 105*t^3 + 385*t^4 + ... .

(End)