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The e.g.f. T(x) = Sum_{n>=1} n^(n-1)*x^n/n! satisfies T(x) = x*exp(T(x)), so T(x) is the functional inverse (series reversion) of x*exp(-x).

Also T(x) = -LambertW(-x) where W(x) is the principal branch of Lambert's function.

T(x) is sometimes called Euler's tree function.

a(n) = A000312(n-1)*A128434(n,1)/A128433(n,1). - Reinhard Zumkeller, Mar 03 2007

E.g.f.: LambertW(x)=x*G(0); G(k) = 1 - x*((2*k+2)^(2*k))/(((2*k+1)^(2*k)) - x*((2*k+1)^(2*k))*((2*k+3)^(2*k+1))/(x*((2*k+3)^(2*k+1)) - ((2*k+2)^(2*k+1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 30 2011

a(n) = Sum_{i=1..n} binomial(n-1,i-1)*i^(i-2)*(n-i)^(n-i). - Dmitry Kruchinin, Oct 28 2013

Limit_{n->oo} a(n)/A000312(n-1) = e. - Daniel Suteu, Jul 23 2016

From Amiram Eldar, Nov 20 2020: (Start)

Sum_{n>=1} 1/a(n) = A098686.

Sum_{n>=1} (-1)^(n+1)/a(n) = A262974. (End)

a(n) = Sum_{k=0..n-1} (-1)^(n+k-1)*Pochhammer(n, k)*Stirling2(n-1, k). - Mélika Tebni, May 07 2023

In terms of Eulerian numbers A340556(n,k) of the second order Sum_{m>=1} m^(m+n) z^m/m! = 1/(1-T(z))^(2n+1) * Sum_{k=0..n} A2(n,k) T(z)^k. - Marko Riedel, Jan 10 2024

E.g.f.: A(x, y)=(1-x+x*y)*B(x, y)-B(x, y)^2/2. B(x, y): e.g.f. of A055302.

T(n, k) = binomial(n+1, k)*Sum( binomial(n+1-k, i)*(-1)^(n+1-k-i)*i^(n-1), i=0..n+1-k).

T(n, k) = (n!/k!)*Stirling2(n-2, n-k). - Vladeta Jovovic, Jan 28 2004

E.g.f.: x/(1-T), where T=T(x) is Euler's tree function (see A000169).

a(n) = Sum_{k=1..n} A055302(n, k)*k.

a(n) = the n-th term of the (n-1)-th binomial transform of {1, 1, 4, 18, 96, ..., (n-1)*(n-1)!, ...} (cf. A001563). - Paul D. Hanna, Nov 17 2003

a(n) = (n-1)^(n-1) + Sum_{i=2..n} (n-1)^(n-i)*binomial(n-1, i-1)*(i-1) *(i-1)!. - Paul D. Hanna, Nov 17 2003

a(n) = [x^(n-1)] 1/(1 - (n-1)*x)^2. - Paul D. Hanna, Dec 27 2012

a(n) ~ exp(-1) * n^n. - Vaclav Kotesovec, Nov 14 2014

Column-wise binomial transform of A327377.

E.g.f.: exp(x + U(x,y) + B(x*(1-y) + R(x,y))), where R(x,y) is the e.g.f. of A055302, U(x,y) is the e.g.f. of A055314 and B(x) + x is the e.g.f. of A059167. - Andrew Howroyd, Oct 05 2019

N(J(n,r)) = C(n,r)*S(n,r+1)*r! where S(n, r + 1) is a Stirling number of the second kind (given by A048993 with zeros removed); generating function = (x+1)^(n-1).

From Peter Bala, Oct 22 2008: (Start)

Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).

Let f(x) = 1 + a*x + a*x^2/2! + a*x^3/3! + ... . Then the e.g.f. for this table is I(f(x)) = 1 + a*x +(a + 2*a^2)*x^2/2! + (a + 9*a^2 + 6*a^3)*x^3/3! + (a + 28*a^2 + 72*a^3 + 24*a^4)*x^4/4! + ... . Note, if we take f(x) = 1 + a*x + a*x^2 + a*x^3 + ... then I(f(x)) is the o.g.f. of the Narayana triangle A001263. (End)

A generator for this array is given by the inverse, g(x,t), of f(x,t)= x/(1 + t * (e^x-1)). Then A248927 gives h(x,t)= x / f(x,t) = 1 + t*(e^x-1)= 1 + t * (x + x^2/2! + x^3/3! + ...) and g(x,t)= x * (1 + t * x + (t + 2 t^2) * x^2/2! + (t + 9 t^2 + 6 t^3) * x^3/3! + ...), so by Bala's arguments A248927 is a refinement of A141618 with row sums A000272. The connection to Narayana numbers is reflected in the relation between A248927 and A134264. See A145271 for more relations that g(x,t) and f(x,t) must satisfy. - Tom Copeland, Oct 17 2014

T(n,k) = C(n,k-1) * A028246(n,k) = C(n,k-1) * A019538(n,k)/k = A055302(n+1,n+1-k) / (n+1). - Tom Copeland, Oct 25 2014

E.g.f. is the series reversion of log(1 + x)/(1 + t*x) with respect to x. Cf. A198204. - Peter Bala, Oct 21 2015

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.

If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.

T(j,k) is (j-1)! P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).

For example, from g(t) above, T(6,3) = 5! * [6!/(3!*2!)]/(2!)^2 = 1800 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = 5! * [6!/4!]/(2!*3!) = 300.

If the initial factorial and final denominator of T(n,k) are removed and the expression divided by j and the partitions reversed in order, then A134264 is obtained, a refinement of the Narayana numbers.

For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums are A000169(n) = n^(n-1) = n* A000272(n).

Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/[d{x/[h_0+h_1*x+ ...]/dx]. Then the partition polynomials above are given by (W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)=exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.

With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(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 * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).

Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)

Following the notes connected to the Lagrange reversion theorem in A248927, a generator for the n-th partition polynomial P_n of this entry is (d/dx)^(n-1) (h (x))^n, and -log(1-t*P.) = (t*Q.) / (1 - t*Q.), umbrally, where (Q.)^n = Q_n is the n-th partition polynomial of A248927. - Tom Copeland, Nov 25 2016

With h_0 = 1, the n-th partition polynomial is obtained as the n-th element (with initial index 0) of the first column of M^{n+1}, where M is the matrix with M_{i,j}= binomial(i,j) h_{i-j}, i.e., the lower triangular Pascal matrix with its n-th diagonal multiplied by h_n. This follows from the Lagrange inversion theorem and the relation between powers of matrices such as M and powers of formal Taylor series discussed in A133314. This is equivalent to repeated binomial convolution of the coefficients of the Taylor series with itself. - Tom Copeland, Nov 13 2019

T(n,k) = n*A248927(n,k). - Andrew Howroyd, Feb 02 2022

Column-wise inverse binomial transform of A327369.

E.g.f.: exp(-x)*exp(x + U(x,y) + B(x*(1-y) + R(x,y))), where R(x,y) is the e.g.f. of A055302, U(x,y) is the e.g.f. of A055314 and B(x) + x is the e.g.f. of A059167. - Andrew Howroyd, Oct 05 2019

E.g.f.: x^3/(2*(1-x)^3).

a(n) = (n-2)!*t(n-2)*t(n-1) = (n-2)!*(n-2)*(n-1)^2*n/4 = n!*(n-2)*(n-1)/4 = n!*t(n-2)/2 where t = A000217. - Jon Perry, Feb 22 2004

D-finite with recurrence: (n-3)*a(n) - (n^2 - n)*a(n-1) = 0. - Georg Fischer, Aug 17 2021

a(n) = 3 * A001754(n). - Alois P. Heinz, Aug 17 2021

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.

If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.

T(j,k) is [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).

For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.

If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.

For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).

Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (1/n)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)= exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.

With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(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 * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).

Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)

As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - Tom Copeland, Nov 25 2016

T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 02 2022

T(n,k) = n!/k! * Stirling2(n,n-k).

T(n,0) = n!.

T(n,k) = A055302(n,k)*(n-k) + A055302(n,k+1)*(k+1). The first term (on rhs of this equation) is the number of such functions in which the preimage of f(n) contains more than one element. The second term is the number of such functions in which the preimage of f(n) contains exactly one element.

T(n,k) = binomial(n,k) Sum_{j=0..n-k}(-1)^j*binomial(n-k,j)*(n-k-j)^n. - Geoffrey Critzer, Aug 20 2013

E.g.f.: 1/(1 - (A(x,y) - y*x + x)) where A(x,y) is the e.g.f. for A055302. - Geoffrey Critzer, Jan 24 2022

From Alois P. Heinz, Jan 24 2022: (Start)

Sum_{k=0..n} k * T(n,k) = A209290(n).

Sum_{k=0..n} (-1)^k * T(n,k) = A344053(n). (End)