
COMMENTS

"Standard order" here means as produced by Maple's "sort" command.
Row sums are 0 (for n>1). Numbers of terms in rows are partition numbers A000041.
From Tom Copeland, Nov 06 2015: (Start)
With the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... , the partition polynomials of this entry give d[log(f(x))]/dx = L_1(x[1]) + L_2(x[1],x[2]) x + L_3(..) x^2/2! + ... , and the coefficients of the reduced polynomials with x[n] = t are signed A028246.
The raising operator R = x + d[log(f(D)]/dD = x + L_1(x[1]) + L_2[x[1],x[2]) D + L_3(x[1],x[2],x[3]) D^2/2! + ... with D = d/dx generates an Appell sequence of polynomials, given umbrally by P_n(x[1],..,x[n];x) = (x[.] + x)^n = sum[k=0 to n, binomial(n,k) x[k] * x^(nk)] = R^n 1 with the e.g.f. f(t)e^(xt) = exp[t P.(x[1],..,x[.];x)]. P_0 = x[0] = 1.
The umbral compositional inverse Appell sequence is generated by R = x  d[log(f(D))]/dD with e.g.f. e^(xt)/f(t) = exp[t IP.(x[1],..,x[.];x)], so umbrally IP_n(x[1],..,x[n];P.(x[1],..,x[n];x)) = x^n = P_n(x[1],..,x[n];IP.(x[1],..,x[n];x)). An unsigned array for the reduced IP_n(x[1],..,x[n];x) polynomials with IP_0 = x[0] = 1 and x[n] = 1 for n>0 is A154921, for which f(t) = 2  e^t. (End)
From Tom Copeland, Sep 08 2016: (Start)
The Appell formalism allows a matrix representation in the power basis x^n of the raising operator R that incorporates this array's partition polynomials L_n[x[1],..,x[n]):
VP_(n+1) = VP_n * R = VP_n * XPS^(1) * MX * XPS where XPS is the matrix formed from multiplying the nth diagonal of the Pascal matrix PS of A007318 by the indeterminate x[n], with x[0] = 1 for the main diagonal of ones, i.e., XPS[n,k] = PS[n,k] * x[nk]; the matrix MX is A129185; the matrix XPS^(1) is the inverse of XPS, which can be formed by multiplying the diagonals of the Pascal matrix by the partition polynomials IPT(n,x[1],..,x[n]) of A133314, i.e., XPS^(1)[n,k] = PS[n,k] * IPT(nk,x[1],..); and VP_n is the row vector in the power basis representing the Appell polynomial P_n(x) formed from the basic sequence of moments 1,x[1],x[2],..., i.e., umbrally P_n(x) = (x[.] + x)^n = sum_{k = 0,..,n} binomial(n,k) * x[k] * x^(nk).
Then R = XPS^(1) * MX * XPS is the Pascal matrix PS with an additional first superdiagonal of ones and the other lower diagonals multiplied by the partition polynomials of this array, i.e., R[n,k] = PS[n,k] * L_(n+1k)(x[1],..,x[n+1k]) except for the first superdiagonal of ones.
Consistently, VP_n = (1,0,0,..) * R^n = (1,0,0,..) * XPS^(1) * MX^n * XPS = (1,0,0,..) * MX^n * XPS = the nth row vector of XPS, which is the vector representation of P_n(x) = (x[.] + x)^n with x[0] = 1.
See the Copeland link for the umbral representation R = exp[g.*D] * x * exp[h.*D] that reflects the matrix representations.
The Stirling partition polynomials of the first kind St1_n(a[1],a[2],..,a[n]) of A036039, the Stirling partition polynomials of the second kind St2_n(b[1],b[2],..,b[n]) of A036040, and the refined Lah polynomials Lah_n[c[1],c[2],..,c[n]) of A130561 are Appell sequences in the respective distinguished indeterminates a[1], b[1], and c[1]. Comparing the formulas for their raising operators with that in this entry, L_n(x[1],x[2],..,x[n]) evaluates to
A) (n1)! * a[n] for x[n] = St1_n(a[1],a[2],..,a[n])
B) b[n] for x[n] = St2_n(b[1],b[2],..,b[n])
C) n! * c[n] for x[n] = Lah_n(c[1],c[2],..,c[n]).
Conversely, from the respective e.g.f.s (added Sep 12 2016)
D) x[n] = St1[n,L_1(x[1])/0!,..,L_n(x[1],..,x[n])/(n1)!]
E) x[n] = St2[n,L_1(x[1]),..,L_n(x[1],..,x[n])]
F) x[n] = Lah[n,L_1(x[1])/1!,..,L_n(x[1],..,x[n])/n!].
Given only the Appell sequence with no closed form for the e.g.f., the raising operator can be generated using this formalism, as has been partially done for A134264.
(End)
For the Appell sequences above, the raising operator is related to the recursion P_(n+1)(x) = x * P_n(x) + sum_{k 0,..,n} binomial(n,k) * L_(nk+1)(x[1],..,x[n+k1]) * P_k(x). For a derivation and connections to formal cumulants (c_n = L_n(x[1],..)) and moments (m_n = x[n]), see the Copeland link on noncrossing partitions. With x = 0, the recursion reduces to x[n+1] = sum_{k = 0,..,n} binomial(n,k) * L_(nk+1)(x[1],..,x[n+k1]) * x[k] with x[0] = 1. This array is a differently ordered version of A127671.  Tom Copeland, Sep 13 2016
With x[n] = x^(n1), a signed version of A130850 is obtained.  Tom Copeland, Nov 14 2016
