cp's OEIS Frontend

a(n+1) = 1 + Sum_{j=1..n} (-1 + binomial(n+1,j))*a(j). - Jon Perry, Apr 25 2005, corrected by Vaclav Kotesovec, Jan 07 2014

The Hankel transform of this sequence is A121835. - Philippe Deléham, Aug 31 2006

E.g.f. A(x) satisfies A(x) = 1 + Integral_{t=0..x} (A(t)^3 * exp(-t)) dt. - Paul D. Hanna, Jan 24 2008 [Edited by Petros Hadjicostas, May 14 2020]

From Vladimir Kruchinin, May 10 2011: (Start)

a(n) = Sum_{m=1..n} (Sum_{k=m..n} Stirling2(n,k)*k!*binomial(k-1,m-1))*(1/m)*binomial(2*m-2,m-1)*(-1)^(m-1)/2^(m-1), n > 0.

E.g.f. B(x) = Integral_{t = 0..x} A(t) dt satisfies B'(x) = tan(B(x)) + sec(B(x)). (End)

From Peter Bala, Aug 25 2011: (Start)

It follows from Vladimir Kruchinin's formula above that

Sum_{n>=1} a(n-1)*x^n/n! = series reversion (Integral_{t = 0..x} 1/(sec(t)+tan(t)) dt) = series reversion (Integral_{t = 0..x} (sec(t)-tan(t)) dt) = series reversion (x - x^2/2! + x^3/3! - 2*x^4/4! + 5*x^5/5! - 16*x^6/6! + ...) = x + x^2/2! + 2*x^3/3! + 7*x^4/4! + 35*x^5/5! + 226*x^6/6! + ....

Let f(x) = sec(x) + tan(x). Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = (d/dx)(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n) = D^n[f](0). Compare with A190392.

(End)

G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013

a(n) ~ sqrt(2) * n^n / (exp(n) * (log(2))^(n+1/2)). - Vaclav Kotesovec, Jan 07 2014

G.f.: R(0)/(1-x), where R(k) = 1 - x^2*(k+1)*(2*k+1)/(x^2*(k+1)*(2*k+1) - (3*x*k+x-1)*(3*x*k+4*x-1)/R(k+1)); (continued fraction). - Sergei N. Gladkovskii, Jan 30 2014

a(0) = 1 and a(n) = a(n-1) + Sum_{k=1..n-1} binomial(n-1, k-1)*a(k) for n > 0. - Seiichi Manyama, Oct 20 2019

a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n, k)*(2*k-1)!! (see Qi/Ward). - Peter Luschny, Oct 19 2021

a(0) = 1; a(n) = Sum_{k=1..n} (-1)^k * (k/n - 2) * binomial(n,k) * a(n-k). - Seiichi Manyama, Nov 15 2023

Conjecture from Mikhail Kurkov, Jun 24 2025: (Start)

a(n) = R(n,0,2) where

R(0,0,m) = 1,

R(n,0,m) = Sum_{j=0..n-1} R(n-1,j,m),

R(n,k,m) = m*R(n,0,m) - Sum_{j=0..k-1} R(n-1,j,m) for 0 < k <= n.

More generally, R(n,0,m) gives expansion of the e.g.f. (exp(x) / (m - (m-1)*exp(x)))^(1/m) for any m>0. (End)

T(n,k) = C(n+1,2k+1) = Sum_{i=k..n-k} C(i,k) * C(n-i,k).

E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic, Mar 20 2005

G.f.: 1/((1-z)^2-t*z^2). - Emeric Deutsch, Apr 01 2005

T(n,k) = Sum_{j = 0..n} A034839(j,k). - Philippe Deléham, May 18 2005

Pell(n+1) = A000129(n+1) = Sum_{k=0..n} T(n,k) * 2^k = (1/n!) Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007

T(n,k) = A007318(n,2k) + A007318(n,2k+1). - Philippe Deléham, Nov 12 2008

O.g.f for column k, k>=0: (1/(1-x)^2)*(x/(1-x))^(2*k). See the G.f. of this array given above by Emeric Deutsch. - Wolfdieter Lang, Jan 18 2013

T(n,k) = (x^(2*k+1))*((1+x)^n-(1-x)^n)/2. - L. Edson Jeffery, Jan 15 2014

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ (e(2))! * (e(3))! * ... * (e(n))! ].

From Tom Copeland, Sep 06 2011: (Start)

Let h(t) = 1/(df(t)/dt)

= 1/Ev[u./(1-u.t)^2]

= 1/((u_1) + 2*(u_2)*t + 3*(u_3)*t^2 + 4*(u_4)*t^3 + ...),

where Ev denotes umbral evaluation.

Then for the partition polynomials of A133437,

n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,

and the compositional inverse of f(t) is

g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.

Also, dg(t)/dt = h(g(t)). (End)

From Tom Copeland, Oct 20 2011: (Start)

With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators defined by R PS(n,t)=PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are

R = t*h(d/dt) = t* 1/[(u_1) + 2*(u_2)*d/dt + 3*(u_3)*(d/dt)^2 + ...] and

L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2 + (u_3)*(d/dt)^3 + ....

Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)

The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x + u_3 * x^2 + ... + u_n * x^(n-1)]^n evaluated at x=0. - Tom Copeland, Jul 07 2015

From Tom Copeland, Sep 20 2016: (Start)

Let PS(n,u1,u2,...,un) = P(n,t) / t^n, i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.

Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 6 u2 u4 + 3 u3^2, b3 = -21 u2^2 u3, and b4 = 14 u2^4.

The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. A086810) implies that, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = [(n+1)/2] * Sum_{k = 2..n-1} PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).

For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 14 u2^4 - 2 * 21 u2^2 u3 + 1 * 6 u2 u4 + 1 * 3 u3^2 - 0 * u5 = 42 u2^4 - 42 u2^2 u3 + 6 u2 u4 + 3 u3^2 = 3 * [2 * PS(2,1,u2) * PS(4,1,u2,...,u4) + PS(3,1,u2,u3)^2] = 3 * [ 2 * (-u2) (-5 u2^3 + 5 u2 u3 - u4) + (2 u2^2 - u3)^2].

Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.

(End)

From Tom Copeland, Sep 22 2016: (Start)

Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = m!*u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently multiply the diagonals of A132159 by u_m. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.

Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ...)^T * t^n/n! in agreement with A139605. (End)

A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Lah polynomials of A130561 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

The derivative of the partition polynomials of A350499 with respect to a distinguished indeterminate give polynomials proportional to those of this entry. The connection of this derivative relation to the inviscid Burgers-Hopf evolution equation is given in a reference for that entry. - Tom Copeland, Feb 19 2022

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [2!^e(2)*e(2)!*3!^e(3)*e(3)! ... n!^e(n)*e(n)! ].

From Tom Copeland, Sep 05 2011: (Start)

Let h(t) = 1/(df(t)/dt)

= 1/Ev[u.*exp(u.*t)]

= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),

an e.g.f. for the partition polynomials of A133314

(signed A049019) with an index shift.

Then for the partition polynomials of A134685,

n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,

and the compositional inverse of f(t) is

g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.

Also, dg(t)/dt = h(g(t)). (Cf. A000311 and A134991)(End)

From Tom Copeland, Oct 30 2011: (Start)

With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators

defined by R PS(n,t)=PS(n+1,t) and L PS(n,t)= n*PS(n-1,t) are

R = t*h(d/dt) = t * 1/[u_1+(u_2)*d/dt+(u_3)*(d/dt)^2/2!+...], and

L = f(d/dt)=(u_1)*d/dt+(u_2)*(d/dt)^2/2!+(u_3)*(d/dt)^3/3!+....

Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)

The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n-1)/n!]^n evaluated at x=0. - Tom Copeland, Jul 07 2015

Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314. - Tom Copeland, Jul 15 2016

Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016

From Tom Copeland, Sep 20 2016: (Start)

Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.

Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.

The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2,..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).

For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - u3)^2].

Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.

(End)

A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

G.f.: [1-tz-sqrt(1-2tz+t^2*z^2-4z^2)]/(2z^2).

T(n, k) = n!/[k!((n-k)/2)!((n-k)/2-1)! ] = A055151(n, (n-k)/2) if n-k is a nonnegative even number; otherwise T(n, k) = 0.

T(n, k) = C(n, k)*C((n-k)/2)*(1+(-1)^(n-k))/2 if k <= n, 0 otherwise. - Paul Barry, May 18 2005

T(n,k) = A121448(n,k)/2^k. - Philippe Deléham, Aug 17 2006

Sum_{k=0..n} T(n,k)*2^k = A000108(n+1). - Philippe Deléham, Aug 22 2006

Sum_{k=0..n} T(n,k)*3^k = A002212(n+1). - Philippe Deléham, Oct 03 2007

G.f.: 1/(1-x*y-x^2/(1-x*y-x^2/(1-x*y-x^2/.... (continued fraction). - Paul Barry, Dec 15 2008

Sum_{k=0..n} T(n,k)*4^k = A005572(n). - Philippe Deléham, Dec 03 2009

T(n,k) = A007318(n,k)*A126120(n-k). - Philippe Deléham, Dec 12 2009

From Tom Copeland, Jan 23 2016: (Start)

E.g.f.: M(x,t) = e^(xt) AC(t) = e^(xt) I_1(2t)/t = e(xt) * e.g.f.(A126120(t)) = e^(xt) Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp[t P(.,x)].

The e.g.f. of this Appell sequence of polynomials P(n,x) is e^(xt) times the e.g.f. AC(t) of the aerated Catalan numbers A126120. AC(t) = I_1(2t)/t, where I_n(x) = T_n(d/dx) I_0(x) are the modified Bessel functions of the first kind and T_n, the Chebyshev polynomials of the first kind.

P(n,x) has the lowering and raising operators L = d/dx = D and R = d/dD log{M(x,D)} = x + d/dD log{AC(D)} = x + Sum_{n>=0} c(n) D^(2n+1)/(2n+1)! with c(n) = (-1)^n A180874(n+1), i.e., L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).

(P(.,x) + y)^n = P(n,x+y) = Sum_{k=0..n} binomial(n,k) P(k,x) y^(n-k) = (b.+x+y)^n, where (b.)^k = b_k = A126120(k).

Exp(b.D) e^(xt) = exp[(x+b.)t] = exp[P(.,x)t] = e^(b.t) e^(xt) = e^(xt) AC(t).

See p. 12 of the Alexeev et al. link and A055151 for a refinement.

Shifted o.g.f: G(x,t) = [1-tx-sqrt[(1-tx)^2-4x^2]] / 2x = x + t x^2 + (1+t) x^3 + ... has the compositional inverse Ginv(x,t) = x / [1 + tx + x^2] = x - t x^2 +(-1+t^2) x^3 + (2t-t^3) x^4 + (1-3t^2+t^4) x^5 + ..., a shifted o.g.f. for the signed Chebyshev polynomials of the second kind of A049310 (cf. also the Fibonacci polynomials of A011973). Then the inversion formula of A134264, involving non-crossing partitions and free probability with their multitude of interpretations (cf. A125181 also), can be used with h_0 = 1, h_1 = t, and h_2 = 1 to interpret the coefficients of the Motzkin polynomials combinatorially.

(End)

From Tom Copeland, Jan 29 2016: (Start)

Provides coefficients of the inverse of f(x) = x / [1 + h1 x + h2 x^2], a bivariate generating function of A049310 (mod signs).

finv(x) = [(1-h1*x) - sqrt[(1-h1*x)^2-4h2*x^2]]/(2*h2*x) = x + h1 x^2 + (h2 + h1^2) x^3 + (3 h1 h2 + h1^3) x^4 + ... is a bivariate o.g.f. for this entry.

The infinitesimal generator for finv(x) is g(x) d/dx with g(x) = 1 /[df(x)/dx] = x^2 / [(f(x))^2 (1 - h2 x^2)] = (1 + h1 x + h2 x^2)^2 / (1 - h2 x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomials FI(n,h1,h2) of the compositional inverse of f(x), i.e., exp[x g(u)d/du] u |_(u=0) = finv(x) = 1 / [1 -x FI(.,h1,h2)]. Cf. A145271. E.g.,

FI(0,h1,h2) = 0

FI(1,h1,h2) = 1

FI(2,h1,h2) = 1 h1

FI(3,h1,h2) = 1 h2 + 1 h1^2

FI(4,h1,h2) = 3 h2 h1 + 1 h1^3

FI(5,h1,h2) = 2 h2^2 + 6 h2 h1^2 + 1 h1^4

FI(6,h1,h2) = 10 h2^2 h1 + 10 h2 h1^3 + 1 h1^5.

And with D = d/dh1, FI(n+1, h1,h2) = MT(n,h1,h2) = (b.y + h1)^n = Sum_{k=0..n} binomial(n,k) b(k) y^k h1^(n-k) = exp[(b.y D] (h1)^n = AC(y D) (h1)^n, where b(k) = A126120(k), y = sqrt(h2), and AC(t) is defined in my Jan 23 formulas above. Equivalently, AC(y D) e^(x h1) = exp[x MT(.,h1,h2)].

The MT polynomials comprise an Appell sequence in h1 with e.g.f. e^(h1*x) AC(xy) = exp[x MT(.,h1,h2)] with lowering operator L = d/dh1 = D, i.e., L MT(n,h1,h2) = dMT(n,h1,h2)/dh1 = n MT(n-1,h1,h2) and raising operator R = h1 + dlog{AC(y L)}/dL = h1 + Sum_{n>=0} c(n) h2^(n+1) D^(2n+1)/(2n+1)! = h1 + h2 d/dh1 - h2^2 (d/dh1)^3/3! + 5 h2^3 (d/dh1)^5/5! - ... with c(n) = (-1)^n A180874(n+1) (consistent with the raising operator in the Jan 23 formulas).

The compositional inverse finv(x) is also obtained from the non-crossing partitions of A134264 (or A125181) with h_0 = 1, h_1 = h1, h_2 = h2, and h_n = 0 for all other n.

See A238390 for the umbral compositional inverse in h1 of MT(n,h1,h2) and inverse matrix.

(End)

From Tom Copeland, Feb 13 2016: (Start)

z1(x,h1,h2) = finv(x), the bivariate o.g.f. above for this entry, is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,h1,h2),z2(x,h1,h2)) = (z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-h1*x)/(h2*x)] z + 1/h2, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.

The other zero is given by z2(x,h1,h2) = (1 - h1*x)/(h2*x) - z1(x,h1,h2) = [1 - h1*x + sqrt[(1-h1*x)^2 - 4 h2*x^2]] / (2h2*x).

The two are the nontrivial zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2), (see Landweber et al., p. 14, Ellingsrud, and A121448), and the zeros for the Riccati equation z' = (z - z1)(z - z2), associated to soliton solutions of the KdV equation (see Copeland link).

(End)

Comparing the shifted o.g.f. S(x) = x / (1 - h_1 x + h_2 x^2) for the bivariate Chebyshev polynomials S_n(h_1,h_2) of A049310 with the shifted o.g.f. H(x) = x / ((1 - a x)(1 - b x)) for the complete homogeneous symmetric polynomials H_n(a,b) = (a^(n+1)-b^(n+1)) / (a - b) shows that S_n(h_1,h_2) = H_n(a,b) for h_1 = a + b and h_2 = ab and, conversely, a = (h_1 + sqrt(h_1^2 - 4 h_2)) / 2 and b = (h_1 - sqrt(h_1^2 - 4 h_2)) / 2. The compositional inverse about the origin of S(x) gives a bivariate o.g.f. for signed Motzkin polynomials M_n(h_1,h_2) of this entry, and that of H(x) gives one for signed Narayana polynomials N_n(a,b) of A001263, thereby relating the bivariate Motzkin and Narayana polynomials by the indeterminate transformations. E.g., M_2(h_1,h_2) = h_2 + h_1^2 = ab + (a + b)^2 = a^2 + 3 ab + b^2 = N_2(a,b). - Tom Copeland, Jan 27 2024

Sum_{n>=0} Sum_{k=0..n} (-1)^n*T(n, k)*y^(2*k)*x^(2*n+1)/(2*n+1)! = JacobiSN(x, y).

JacobiSN(x, y) = 1*x + (-1/6 - (1/6)*y^2)*x^3 + (1/120 + (7/60)*y^2 + (1/120)*y^4)*x^5 + (-1/5040 - (3/112)*y^4 - (3/112)*y^2 - (1/5040)*y^6)*x^7 + (1/362880 + (307/90720)*y^6 + (913/60480)*y^4 + (307/90720)*y^2 + (1/362880)*y^8)*x^9 + O(x^11).

From Peter Bala, Aug 23 2011: (Start)

Let f(x) = sqrt((1-x^2)*(1-k^2*x^2)).

Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.

Then the coefficient polynomial R(2*n+1,k) of u^(2*n+1)/(2*n+1)! is given by R(2*n+1,k) = D^(2*n)[f](0) - apply [Dominici, Theorem 4.1].

See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x).

(End)

sn(u|k^2) = Sum_{n>=0} a_n(k^2)*u^(2*n+1)/(2*n+1)!. For the recurrence of the row polynomials a_n(k^2) = Sum_{m=0..n} (-1)^n*T(n, m)*k^(2*m), see the Fricke reference. - Wolfdieter Lang, Jul 05 2016

Bergeron et al. give several formulas. Shifts left under "CIJ" (necklace, indistinct, labeled) transform.

E.g.f.: A(x) =

x + (1/2)*x^2 + (1/3)*x^3 + (7/24)*x^4 + (3/10)*x^5 + (49/144)*x^6 + (173/420)*x^7 + (21059/40320)*x^8 + (8887/12960)*x^9 + ...

and satisfies the differential equation A'(x)=log(1/(1-A(x)))+1. - Vladimir Kruchinin, Jan 22 2011

E.g.f. A(x) satisfies: A''(x) = A'(x) * exp(A'(x)-1). - Paul D. Hanna, Apr 17 2015

From Robert Israel, Apr 17 2015 (Start):

E.g.f. A(x) satisfies e*(Ei(1,A'(x)) - Ei(1,1)) = integral(s = 1 .. A'(x), exp(1-s)/s ds) = -x.

a(n) = e^(1-n)*limit(w -> 1, (d^(n-2)/dw^(n-2))(((w-1)/(Ei(1,1)-Ei(1,w)))^(n-1))) for n >= 2. (End)

a(n) = sum(i=1..n-2,binomial(n-2,i)*a(i)*a(n-i))+a(n-1), a(0)=0, a(1)=1. - Vladimir Kruchinin, Jan 24 2011

The following remarks refer to the interpretation of this sequence as counting increasing trees where the nodes of outdegree k come in k+1 colors. Thus we work with the generating function B(x) = A'(x)-1 = x + 2*x^2/2!+7*x^3/3!+36*x^4/4!+.... The degree function phi(x) (see [Bergeron et al.] for definition) for this variety of trees is phi(x) = 1+2*x+3*x^2/2!+4*x^3/3!+5*x^4/4!+... = (1+x)*exp(x). The generating function B(x) satisfies the autonomous differential equation B' = phi(B(x)) with initial condition B(0) = 0. It follows that the inverse function B(x)^(-1) may be expressed as an integral B(x)^(-1) = int {t = 0..x} 1/phi(t) dt = int {t = 0..x} exp(-t)/(1+t) dt. Applying [Dominici, Theorem 4.1] to invert the integral produces the result B(x) = sum {n>=1} D^(n-1)[(1+x)*exp(x)](0)*x^n/n!, where the nested derivative D^n[f](x) of a function f(x) is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. Thus a(n+1) = D^(n-1)[(1+x)*exp(x)](0). - Peter Bala, Aug 30 2011

Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by f_n = (d/dx)^n f(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) f_(n-k). Then R^n = (1, 0, 0, 0,..) [VP * S]^n (1, D, D^2, ..)^T, where S is the shift matrix A129185, representing differentiation in the basis x^n/n!. Cf. A145271. - Tom Copeland, Jul 17 2016

A formula for the coefficients of this matrix is presented in Ihara, obtained from Comtet. - Tom Copeland, Mar 25 2020

Elaborating on my 2011 comments: Let exp[x F(t)] = exp[t p.(x)] be the e.g.f. for the binomial Sheffer sequence of polynomials (p.(x))^n = p_n(x). Then, evaluated at x = t = 0, the coefficient p_(n,k) = (D_x^k/k!) p_n(x) = D_t^n [F(t)]^k/k! = (f(x)D_x)^n x^k/k! = R^n x^k/k!, and so p_(n,k) is the coefficient of D^k of the operator R^n below evaluated at x=0. - Tom Copeland, May 14 2020

Per earlier comments, the action of the differentials of this entry on an exponential is exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)}(x) the compositional inverse of H(x). With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind (see A008275, A048994, and A130534). Consequently, summing the coefficients of this entry over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+4+1) D + (7+4) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. The Connes-Moscovici weights A139002 are a refinement of this entry. - Tom Copeland, Jul 14 2021

T(n,k) = Sum_{j=0..k} (-1)^j*2^(k-j)*binomial(k,j)*binomial(n+k-j-1, 2*k-1) (0 <= k <= n).

G.f.: G(t,x) = (1-x)^2/((1-x)^2 - t*x*(2-x)).

G.f. of column k = x^k*(2-x)^k/(1-x)^{2k} (k>=1) (we have a Riordan array).

Recurrences satisfied by the numbers u_{n,k}=T(n,k) can be found in the Castiglione et al. reference.

Sum_{k=0..n} k*T(n,k) = A181290(n).

T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=0, T(1,1)=2, T(2,0)=0, T(1,1)=3, T(2,2)=4, T(n,k)=0, if k < 0 or if k > n. - Philippe Deléham, Nov 29 2013

JacobiCN(x, y) = 1 - 1/2*x^2 + (1/24 + 1/6*y^2)*x^4 + ( - 1/720 - 11/180*y^2 - 1/45*y^4)*x^6 + (1/40320 + 17/1680*y^2 + 19/840*y^4 + 1/630*y^6)*x^8 + ( - 1/3628800 - 247/56700*y^6 - 461/453600*y^2 - 641/75600*y^4 - 1/14175*y^8)*x^10 + O(x^12).

From Peter Bala, Aug 23 2011: (Start)

The Taylor expansion of the Jacobian elliptic function cn(x,k) begins

cn(x,k) = 1 - x^2/2! + (1+4*k^2)*x^4/4! - (1+44*k^2+16*k^4)*x^6/6! + ....

The coefficient polynomials in this expansion can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1 and Example 4.5]):

Let f(x) = sqrt(k^2-sin^2(x)). Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.

Then the coefficient polynomial R(2*n,k) of x^(2*n)/(2*n)! in the expansion of cn(x,k) is given by R(2*n,k) = D^(2*n)[f](0).

See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x). See A181613 for the expansion of the reciprocal function 1/cn(x,k).

(End)

G.f. 1/(1 - x/(1 - (2*k)^2*x/(1 - 3^2*x/(1 - (4*k)^2*x/(1 - 5^2*x/(1 - ...)))))) = 1 + x + (1 + 4*k^2)*x^2 + (1 + 44*k^2 + 16*k^4)*x^3 + ... (see Wall, 94.19, p. 374). - Peter Bala, Apr 25 2017