cp's OEIS Frontend

"Standard order" here means as produced by Maple's "sort" command.

From Petros Hadjicostas, May 27 2020: (Start)

According to the Maple help files for the "sort" command, polynomials in multiple variables are "sorted in total degree with ties broken by lexicographic order (this is called graded lexicographic order)."

Thus for example, x_1^2*x_3 = x_1*x_1*x_3 > x_1*x_2*x_2 = x_1*x_2^2, while x_1^2*x_4 = x_1*x_1*x_4 > x_1*x_2*x_3. (End)

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..n} binomial(n,k) x[k] * x^(n-k) = R^n 1 with the e.g.f. f(t)*e^(x*t) = 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^(x*t)/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 n-th 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[n-k]; 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(n-k, 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^(n-k).

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+1-k}(x[1], ..., x[n+1-k]) 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 n-th 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) (n-1)! * 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])/(n-1)!);

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_(n-k+1)(x[1], ..., x[n+k-1]) * 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_(n-k+1)(x[1], ..., x[n+k-1]) * x[k] with x[0] = 1. This array is a differently ordered version of A127671. - Tom Copeland, Sep 13 2016

With x[n] = x^(n-1), a signed version of A130850 is obtained. - Tom Copeland, Nov 14 2016

See p. 2 of Getzler for a relation to stable graphs called necklaces used in computations for Deligne-Mumford-Knudsen moduli spaces of stable curves of genus 1. - Tom Copeland, Nov 15 2019

For a relation to a combinatorial Faa di Bruno Hopf algebra related to functional composition, as presented by Connes and Moscovici, see Figueroa et al. - Tom Copeland, Jan 17 2020

From Tom Copeland, May 17 2020: (Start)

The e.g.f. of an Appell sequence is f(t) e^(x*t) with f(0) = 1. Given the Laguerre-Polya class function f(t) = e^(-a*t^2 + b*t) Product_m (1 - t/z_m) e^(t/z_m) with a = 0 for simplicity (more generally a >= 0) and b real and where the product runs over all the zeros z_m of f(t) with all zeros real and Sum_m 1/(z_m)^2 convergent, the raising operator of the Appell polynomials is R = x + b - Sum_{k > 0} c_(k+1) D^k with c_k = Sum_m (1/(z_m)^k), i.e., traces of powers of the reciprocals of the zeros. From R in earlier comments, b = L_1(x_1) and otherwise c_k = -L_k(x_1, ..., x_k).

The Laguerre / Turan / de Gua inequalities (Csordas and Williamson and Skovgaard) imply that all the zeros of each Appell polynomial are real and simple and its extrema are local maxima above the x-axis and local minima below and are located above or below the zeros of the next lower degree Appell polynomial. (End)

From Tom Copeland, Oct 15 2020: (Start)

With a_n = n! * b_n = (n-1)! * c_n for n > 0, represent a function with f(0) = a_0 = b_0 = 1 as an

A) exponential generating function (e.g.f), or formal Taylor series: f(x) = e^{a.x} = 1 + Sum_{n > 0} a_n * x^n/n!

B) ordinary generating function (o.g.f.), or formal power series: f(x) = 1/(1-b.x) = 1 + Sum_{n > 0} b_n * x^n

C) logarithmic generating function (l.g.f): f(x) = 1 - log(1 - c.x) = 1 + Sum_{n > 0} c_n * x^n /n.

Expansions of log(f(x)) are given in

I) A127671 and A263634 for the e.g.f: log[ e^{a.*x} ] = e^{L.(a_1,a_2,...)x} = Sum_{n > 0} L_n(a_1,...,a_n) * x^n/n!, the logarithmic polynomials, cumulant expansion polynomials

II) A263916 for the o.g.f.: log[ 1/(1-b.x) ] = log[ 1 - F.(b_1,b_2,...)x ] = -Sum_{n > 0} F_n(b_1,...,b_n) * x^n/n, the Faber polynomials.

Expansions of exp(f(x)-1) are given in

III) A036040 for an e.g.f: exp[ e^{a.x} - 1 ] = e^{BELL.(a_1,...)x}, the Bell/Touchard/exponential partition polynomials, a.k.a. the Stirling partition polynomials of the second kind

IV) A130561 for an o.g.f.: exp[ b.x/(1-b.x) ] = e^{LAH.(b.,...)x}, the Lah partition polynomials

V) A036039 for an l.g.f.: exp[ -log(1-c.x) ] = e^{CIP.(c_1,...)x}, the cycle index polynomials of the symmetric groups S_n, a.k.a. the Stirling partition polynomials of the first kind.

Since exp and log are a compositional inverse pair, one can extract the indeterminates of the log set of partition polynomials from the exp set and vice versa. For a discussion of the relations among these polynomials and the combinatorics of connected and disconnected graphs/maps, see Novak and LaCroix on classical moments and cumulants and the two books on statistical mechanics referenced in A036040. (End)

Ignoring signs, these polynomials appear in Schröder in the set of equations (II) on p. 343 and in Stewart's translation on p. 31. - Tom Copeland, Aug 25 2021

The partitions of number n are grouped by increasing length and in reverse lexical order for partitions of the same length.

This sequence is in the Abramowitz-Stegun ordering, see A036036. - Hartmut F. W. Hoft, Apr 25 2015

In the same manner the unsigned Lah triangle A008297 is obtained from the partition array A130561.

The sequence of row lengths is A000041 (partition numbers) [1, 2, 3, 5, 7, 11, 15, 22, 30, 42,...].

Partition number array M_3(2)= A130561 divided by partition number array M_3 = M_3(1) = A036040.

This triangle is named S2(4)'.

In the same manner the unsigned Lah triangle A008297 is obtained from the partition array A130561.

Inverse of A119274.

Row sums are (-1)^(n+1)*A000321(n+1).

Bell polynomials of the second kind B(n,k)(1,-2). - Vladimir Kruchinin, Mar 25 2011

Also the inverse Bell transform of the quadruple factorial numbers Product_{k=0..n-1} (4*k+2) (A001813) giving unsigned values and adding 1,0,0,0,... as column 0. For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015

This triangle is named S2(3)'.

In the same manner the unsigned Lah triangle A008297 is obtained from the partition array A130561.

This triangle is named S2(5)'.

In the same manner the unsigned Lah triangle A008297 is obtained from the partition array A130561.

This triangle is named S2(6)'.

In the same manner the unsigned Lah triangle A008297 is obtained from the partition array A130561.

The row length is r(n), with r(n) = A008615(n+2) for n >= 2: [1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].

The row sums give A176806(n).

For n = 0 with the trivial [e2, e3] = [0, 0] solution the multinomial is 1 with the row sum A176806(0) = 1. For n = 1 there is no solution (with row sum set to A176806(1) = 0).

This multinomial array for pairs [e2, e3] with 2*2e2 + 3*e3 = n, with nonnegative numbers e2 and e3, is obtained from the multinomial array n!/(e1!*e2!*e3!) with n = e1 + e2 + e3, giving the coefficient x_1^{e1}* x_2^{e2}*x_3^{e3} of (x_1 + x_2 + x_3)^n. Here, in order to find the coefficients of (1 + x^2 + x^3)^n, one sets x_1 = 1, x_2 = x^2 and x_3 = x^3. Hence n = e1 + e2 + e3, and the power of x^n becomes n = 2*e2 + 3*e3. Therefore, e1 = n - (e2 + e3), and the array gives n!/((n-(e2+e3))!*e2!*e3!).