cp's OEIS Frontend

a(n,k) = n!/(Product_{j=1..n} e(n,k,j)!) with the exponent e(n,k,j) of j in the k-th partition of n in the A-St ordering of the partitions of n. Exponents 0 can be omitted due to 0!=1.

From Tom Copeland, Sep 18 2011: (Start)

Raising and lowering operators are given for the partition polynomials formed from A130561 in the Copeland link in "Lagrange a la Lah Part I" on pp. 22-23.

An e.g.f. for the partition polynomials is on page 3:

exp[t*:c.*x/(1-c.*x):] = exp[t*(c_1*x + c_2*x^2 + c_3*x^3 + ...)] where :(...): denotes umbral evaluation of the enclosed expression and c. is an umbral coefficient. (End)

From Tom Copeland, Sep 07 2016: (Start)

The row partition polynomials of this array P(n,x_1,x_2,...,x_n), given in the Lang link, are n! * S(n,x_1,x_2,...,x_n), where S(n,x_1,...,x_n) are the elementary Schur polynomials, for which d/d(x_m) S(n,x_1,...,x_n) = S(n-m,x_1,...,x_(n-m)) with S(k,...) = 0 for k < 0, so d/d(x_m) P(n,x_1,...,x_n) = (n!/(n-m)!) P(n-m,x_1,...,x_(n-m)), confirming that the row polynomials form an Appell sequence in the indeterminate x_1 with P(0,...) = 1. See p. 127 of the Ernst paper for more on these Schur polynomials.

With the e.g.f. exp[t * P(.,x_1,x_2,..)] = exp(t*x_1) * exp(x_2 t^2 + x_3 t^3 + ...), the e.g.f. for the partition polynomials that form the umbral compositional inverse sequence U(n,x_1,...,x_n) in the indeterminate x_1 is exp[t * U(.,x_1,x_2,...)] = exp(t*x_1) exp[-(x_2 t^2 + x_3 t^3 + ...)]; therefore, U(n,x_1,x_2,...,x_n) = P(n,x_1,-x_2,.,-x_n), so umbrally P[n,P(.,x_1,-x_2,-x_3,...),x_2,x_3,...,x_n] = (x_1)^n = P[n,P(.,x_1,x_2,...),-x_2,-x_3,...,-x_n]. For example, P(1,x_1) = x_1, P2(x_1,x_2) = 2 x_2 + x_1^2, and P(3,x_1,x_2,x_3) = 6 x_3 + 6 x_2 x_1 + x_1^3, then P[3,P(.,x_1,-x_2,...),x_2,x_3] = 6 x_3 + 6 x_2 P(1,x_1) + P(3,x_1,-x_2,-x_3) = 6 x_3 + 6 x_2 x_1 + 6 (-x_3) + 6 (-x_2) x_1 + x_1^3 = x_1^3.

From the Appell formalism, umbrally [P(.,0,x_2,x_3,...) + y]^n = P(n,y,x_2,x_3,...,x_n).

The indeterminates of the partition polynomials can also be extracted using the Faber polynomials of A263916 with -n * x_n = F(n,S(1,x_1),...,S(n,x_1,...,x_n)) = F(n,P(1,x_1),...,P(n,x_1,...,x_n)/n!). Compare with A263634.

Also P(n,x_1,...,x_n) = ST1(n,x_1,2*x_2,...,n*x_n), where ST1(n,...) are the row partition polynomials of A036039.

(End)

G.f.: Log(1 + Sum_{i >= 1} x_i*t^i/i!) = Sum_{n >= 1} L_n(x_1, x_2, ...)*t^n/n!. [Comtet, p. 140, Eq. [5a]. - corrected by Tom Copeland, Sep 08 2016]

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-2,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = x_k for k > 0. - Mikhail Kurkov, Mar 30 2025

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

From Tom Copeland, Nov 10 2014: (Start)

Terms may be computed symbolically up to order n by using an iterated derivative evaluated at t=0:

with g(t) = 1/{d/dt [t/(1 + 0 t + h_2 t^2 + h_3 t^3 + ... + h_n t^n)]},

evaluate 1/n! * [g(t) d/dt]^n t at t=0, i.e., ask a symbolic math app for the first term in a series expansion of this iterated derivative, to obtain Prt(n-1).

Alternatively, the explicit formula in A134264 for the numerical coefficients of each partition can be used. (End)

From Tom Copeland, Nov 12 2014: (Start)

The first few partitions polynomials formed by taking the reciprocal of the e.g.f. of this entry's e.g.f. (cf. A133314) are

UPrt(0) = 1

UPrt(1;1,0) = 0

UPrt(2;1,0,h_2) = -1 h_2

UPrt(3;1,0,h_2,h_3) = -1 h_3

UPrt(4;1,0,h_2,..,h_4) = -1 h_4 + 4 (h_2)^2

UPrt(5;1,0,h_2,..,h_5) = -1 h_5 + 15 h_2 h_3

UPrt(6;1,0,h_2,..,h_6) = -1 h_6 + 24 h_2 h_4 + 17 (h_3)^2 + -35 (h_2)^3

...

Therefore, umbrally, [Prt(.;1,0,h_2,...) + UPrt(.;1,0,h_2,...)]^n = 0 for n>0 and unity for n=0.

Example of the umbral operation:

(a. + b.)^2 = a.^2 + 2 a.* b. + b.^2 = a_2 + 2 a_1 * b_1 + b_2.

This implies that the umbral compositional inverses (see below) of the partition polynomials of the Lagrange inversion formula (LIF) of A134264 with h_0=1 are given by UPrt(n;1,h_1,h_2,...,h_n) = [UPrt(.;1,0,h_2,...,h_n) + h_1]^n, so that the sequence of polynomials UPrt(n;1,h_1,h_2,...,h_n) is an Appell sequence in the indeterminate h_1. So, if one calculates UPrt(n;1,h_1,...,h_n), the lower order UPrt(n-1;1,h_1,...,h_(n-1)) can be found by taking the derivative w.r.t. h_1 and dividing by n. Same applies for Prt(n;1,h_1,h_2,...,h_n).

This connects the combinatorics of the permutohedra through A133314 and A049019, or their duals, to the noncrossing partitions, Dyck lattice paths, etc. that are isomorphic with the LIF of A134264.

An Appell sequence P(.,x) with the e.g.f. e^(x*t)/w(t) possesses an umbral inverse sequence UP(.,x) with the e.g.f. w(t)e^(x*t), i.e., polynomials such that P(n,UP(.,x))= x^n = UP(n,P(.,x)) through umbral substitution, as in the binomial example. The Bernoulli polynomials with w(t) = t/(e^t - 1) are a good example with the umbral compositional inverse sequence UP(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1) (cf. A074909 and A135278). (End)

a(n) = b(n) - Sum_{k = 1..n-1}((k*binomial(n, k)*b(n-k)*a(k))/n), n >= 1, with b(n) = A000203(n) = sigma(n).

E.g.f. log(1+ Sum_{n >= 1}(b(n)*x^n/n!)), n >= 1, with b(n) = A000203(n) = sigma(n).

a(n) = c(n)/(n-1)! with c(n) = b(n) - Sum_{k=1..n-1}(k*binomial(n, k)*b(n-k)*c(k)), n >= 1 and a(0) = undefined, with b(n) = A001818(n) = ((2*n-1)!!)^2.

a(n) = A000079(n-1) * A005411(n), n >= 1.

O.g.f.: C(x) = 1 + c_1 x + c_2 x^2 + ... = x / (x + m_1 x^2 + m_2 x^3 + m_3 x^4 + ...)^(-1) = x / M^(-1)(x), the shifted reciprocal of the compositional inverse of a power series M(x) = x + m_1 x^2 + m_2 x^3 + ..., the o.g.f. of the free moments m_n in free probability theory.

Row sums: big Schroeder numbers A006318.

Refinement of A060693 and A088617, i.e., by letting m_n = -t and removing all resulting signs, the elements of these two lower triangular matrices are generated.

The coefficients of the highest order terms in m_1^n of the free moment partition polynomials are the signed Catalan numbers A000108.

Taking the derivative with respect to the indeterminate m_1 generates the Lagrange inversion partition polynomials, with shifted indices, of A133437 and A111785 with an overall scale factor. These Lagrange inversion polynomials are the refined Euler characteristic polynomials of the associahedra. E.g.,

D_{m_1} c_5 = 5 (- m_4 + 3 m_2^2 + 6 m_3 m_1 - 21 m_2 m_1^2 + 14 m_1^4). An analogous differential formula that applies to the classical formal cumulants in relation to the permutahedra is stated in my 2012 comment in A127671.

The o.g.f. satisfies the partial differential equations D_{m_1} (x / C(x)) = -(1/3) D_x (x / C(x))^3 and D_{m_1} (C(x) / x) = D_x (x / C(x)), where D_{m_1} and D_x are the partial derivatives with respect to m_1 and x.

More generally, D_{m_n} (x / C(x)) = -(1/(n+2)) D_x (x / C(x))^{n+2), equivalent to D_{m_n} M^(-1)(x) = -(1/(n+2)) D_x (M^(-1)(x))^{n+2). Equations of this type are found in Zhou (see eqn. 44 on p. 11), characterizing the KdV hierarchy. These differential equations can be transformed into the inviscid Burgers-Hopf partial differential equation (see, e.g., A133437, A086810, A001764, A002293, A133932, A134685, and A276850).

The formal free cumulants when identified as the indeterminates of the noncrossing Lagrange inversion partition polynomials NCP_n(c_1,c_2,...,c_n) = m_n of A134264 (as in the example section) satisfy the partial differential equations D_{m_k} NCP_n(c_1, ..., c_n) = d(m_n)/dm_k = delta_{n-k}, where delta_{n} is the Kronecker delta which is zero for all integers n other than n = 0, for which it evaluates as unity. This provides a recursion method for determining the partial derivatives dc_n/dm_k from the partial derivatives dc_p/dm_k and cumulants c_p with k <= p < n. For example, dc_k/dm_j = 0 for j > k and dc_k/dm_k = 1, so dm_3/dm_2 = 0 = D_{m_2} (c_3 + 3 c_2 c_1 + c_1^3) = dc_3/dm_2 + 3 c_1 dc_2/dm_2 = dc_3/dm_2 + 3 c_1 , implying dc_3/dm_2 = -3 c_1 = -3 m_1.

T(n,k) = (n+j-2)!/((n-1)!*Product_{i>=1} s_i!), where (1*s_1 + 2*s_2 + ... = n) is the k-th partition of n and j = s_1 + s_2 + ... is the number of parts. - Andrew Howroyd, Feb 01 2022

Conjecture: free cumulants in terms of the free moments are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = m_k for k > 0. - Mikhail Kurkov, Mar 30 2025