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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A133314 Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.).

Original entry on oeis.org

1, -1, -1, 2, -1, 6, -6, -1, 8, 6, -36, 24, -1, 10, 20, -60, -90, 240, -120, -1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720, -1, 14, 42, -126, 70, -630, 840, -420, -630, 5040, -4200, 2520, -12600, 15120, -5040, -1, 16, 56, -168, 112, -1008, 1344, 70
Offset: 0

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Author

Tom Copeland, Oct 18 2007, Oct 29 2007, Nov 16 2007

Keywords

Comments

The list partition transform of a sequence a(n) for which a(0)=1 is illustrated by:
b_0 = 1
b_1 = -a_1
b_2 = -a_2 + 2 a_1^2
b_3 = -a_3 + 6 a_2 a_1 - 6 a_1^3
b_4 = -a_4 + 8 a_3 a_1 + 6 a_2^2 - 36 a_2 a_1^2 + 24 a_1^4
... .
The unsigned coefficients are A049019 with a leading 1. The sign is dependent on the partition as evident from inspection (replace a_n's by -1).
Expressed umbrally, i.e., with the umbral operation (a.)^n := a_n,
exp(a.x) exp(b.x) = exp[(a.+b.)x] = 1; i.e., (a.+b.)^n = 1 for n=0 and 0 for all other values of n.
Expressed recursively,
b_0 = 1, b_n = -Sum_{j=1..n} binomial(n,j) a_j b_{n-j}; which is conditionally self-inverse, i.e., the roles of a_k and b_k may be reversed with a_0 = b_0 = 1.
Expressed in matrix form, b_n form the first column of B = matrix inverse of A .
A = Pascal matrix diagonally multiplied by a_n, i.e., A_{n,k} = binomial(n,k)* a_{n-k}.
Some examples of reciprocal pairs of sequences under these operations are:
1) A084358 and -A000262 with the first term set to 1.
2) (1,-1,0,0,...) and (0!,1!,2!,3!,...) with the unsigned associated matrices A128229 and A094587.
3) (1,-1,-1,-1,...) and A000670.
5) (1,-2,-2,0,0,0,...) and (0! c_1,1! c_2,2! c_3,3! c_4,...) where c_n = A000129(n) with the associated matrices A110327 and A110330.
6) (1,-2,2,0,0,0,...) and (1!,2!,3!,4!,...).
7) Sequences of rising and signed lowering factorials form reciprocal pairs where a_n = (-1)^n m!/(m-n)! and b_n = (m-1+n)!/(m-1)! for m=0,1,2,... .
Denote the action of the list partition transform on the sequence a. or an invertible matrix M by LPT(a.) = b. or LPT(M)= M^(-1).
If the matrix equation M = exp(T) also holds, then exp[a.*T]*exp[b.*T] = exp[(a.+b.)*T] = I, the identity matrix, because (a.+b.)^n = delta_n, the Kronecker delta with delta_n = 1 and delta_n = 0 otherwise, i.e., (0)^n = delta_n.
Therefore, [exp(a.*T)]^(-1) = exp[b.*T] = exp[LPT(a.)*T] = LPT[exp(a.*T)].
The fundamental Pascal (A007318), unsigned Lah (A105278) and associated Laguerre matrices can be generated by exponentiation of special infinitesimal matrices (see A132440, A132710 and A132681) such that finding LPT(a.) amounts to multiplying the k'th diagonal of the fundamental matrices by a_k for every diagonal followed by matrix inversion and then extraction of the b_n factors from the first column (simplest for the Pascal formulas above).
Conversely, the inverses of matrices formed by diagonally multiplying the three fundamental matrices by a_k are given by diagonally multiplying the fundamental matrices by b_k.
If LPT(M) is defined differently as application of the top formula to a_n = M^n, then b_n = (-M)^n and the formalism could even be applied to more general sequences of matrices M., providing the reciprocal of exp[t*M.].
The group of fundamental lower triangular matrices M = exp(T) such that LPT[exp(a.*T)] = exp[LPT(a.)*T] = [exp[a.*T]]^(-1) are obtained by infinitesimal generator matrices of the form T =
0;
t(0), 0;
0, t(1), 0;
0, 0, t(2), 0;
0, 0, 0, t(3), 0;
... .
T^m has trivially vanishing terms except along the m'th subdiagonal, which is a sequence of generalized factorials:
[ t(0)*t(1)...t(m-2)*t(m-1), t(1)*t(2)...t(m-1)*t(m), t(2)*t(3)...t(m)*t(m+1), ... ].
Therefore the principal submatrices of T (given by setting t(j) = 0 for j > n-1) are nilpotent with at least [Tsub_n]^(n+1) = 0.
The general group of matrices GM[a.] = exp[a.*T] can also be obtained through diagonal multiplication of M = exp(T) by the sequence a_n, as in the Pascal matrix example above and their inverses by diagonal multiplication by b. = LPT(a.).
Weighted-mappings interpretation for the top partition equation:
Given n pre-nodes (Pre) and k post-nodes (Post), each Pre is connected to only one Post and each Post has at least one Pre connected to it (surjections or onto functions/maps). Weight each Post by -a_m where m is the number of connections to the Post.
Weight each map by the product of the Post weights and multiply by the number of maps that share the same connectivity. Sum over the possible mappings for n Pre. The result is b_n.
E.g., b_3 = [ 3 Pre to 1 Post ] + [ 3 Pre to 2 Post ] + [ 3 Pre to 3 Post ]
= [1 map with 1 Post with 3 connections] + [ 6 maps with 1 Post with 2 connections and 1 Post with 1 connection] + [6 maps with 3 Post with 1 connection each]
= -a_3 + 6 * [-a_2*(-a_1)] + 6 * [-a_1*(-a_1)*(-a_1)].
See A263633 for the complementary formulation for the reciprocal of o.g.f.s rather than e.g.f.s and computations of these partition polynomials as Gram determinants. - Tom Copeland, Dec 04 2016
The coefficients of the partition polynomials enumerate the faces of the convex, bounded polytopes called permutohedra, and the absolute value of the sum of the coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is unity. In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With the fundamental matrix chosen to be the lower triangular Pascal matrix M, the matrix MA whose n-th diagonals are multiplied by a_n (i.e., MA_{i,j} = PM_{i,j} * a_{i-j}) gives a matrix representation of the e.g.f. associated to the Appell polynomial sequence defined by e^{a.t}e^{xt}= e^{(a.+x)t} = e^{A.(x)t} where umbrally (A.(x))^n = A_n(x) = (a. + x)^n = sum_{k=0..n} binomial(n,k) a_k x^{n-k} are the associated Appell polynomials. Left multiplication of the column vector (1,x,x^2,..) by MA gives the Appell polynomial sequence, and multiplication of the two e.g.f.s e^{a.t} and e^{b.t} corresponds to multiplication of their respective matrix representations MA and MB. Forming the reciprocal of an e.g.f. corresponds to taking the matrix inverse of its matrix representation as noted above. A263634 gives an associated modified Pascal matrix representation of the raising operator for the Appell sequence. - Tom Copeland, Nov 13 2019
The diagonal of MA consists of all ones. Let MAN be the truncated square submatrix of MA containing the coefficients of the first N Appell polynomials A_k=(a.+x)^k = Sum(j=0 to k) MAN(k,j) x^j. Then by the Cayley-Hamilton theorem (I-MAN)^N = 0; therefore, MAN^(-1) = Sum(k=1 to N) binomial(N,k) (-MAN)^{k-1} = MBN, the inverse of MAN, containing the coefficients of the first N rows of the Appell polynomials B_k(x) = (b. + x)^k = Sum(j=0 to k) MBN(k,j) x^j, which are the umbral compositional inverses of the Appell row polynomials A_k(x) of MAN; that is, A_k(B.(x)) = x^k = B_k(A.(x)), where, e.g., (A.(x))^k = A_k(x). - Tom Copeland, May 13 2020
The use of the term 'list partition transform' resulted from one of my first uses of these partition polynomials in relating A000262 to A084358 with their simple e.g.f.s. Other appropriate names would be the permutohedra polynomials since they are refined Euler characteristics of the permutohedra or the reciprocal polynomials since they give the multiplicative inverses of e.g.f.s with a constant of 1. - Tom Copeland, Oct 09 2022

Examples

			Table starts:
[0] [ 1]
[1] [-1]
[2] [-1,  2]
[3] [-1,  6, -6]
[4] [-1,  8,  6, -36,  24]
[5] [-1, 10, 20, -60, -90,  240, -120]
[6] [-1, 12, 30, -90,  20, -360,  480, -90, 1080, -1800, 720]
		

Crossrefs

Programs

  • Mathematica
    b[0] = 1; b[n_] := b[n] = -Sum[Binomial[n, j]*a[j]*b[n-j], {j, 1, n}];
    row[0] = {1}; row[n_] := Coefficient[b[n], #]& /@ (Times @@ (a /@ #)&) /@ IntegerPartitions[n];
    Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 23 2014 *)
  • Sage
    def A133314_row(n): return [(-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in Partitions(n)]
    for n in (0..10): print(A133314_row(n)) # Peter Luschny, Sep 18 2015

Formula

b_{n-1} = (1/n)(d/da(1))p_n[a_1, a_2, ..., a_n] where p_n are the row partition polynomials of the cumulant generator A127671. - Tom Copeland, Oct 13 2012
(E.g.f. of matrix B) = (e.g.f. of b)·exp(xt) = exp(b.t)·exp(xt) = exp(xt)/exp(a.t) = (e.g.f. of A^(-1)) and (e.g.f. of matrix A) = exp(a.t)·exp(xt) = exp(xt)/exp(b.t) = (e.g.f. of B^(-1)), where the umbral evaluation of exp(b.t) = Sum{n >= 0} (b.t)^n / n! = Sum_{n >= 0} b_n t^n / n! is understood in the denominator. These e.g.f.s define Appell sequences of polynomials. - Tom Copeland, Mar 22 2014
Sum of the n-th row is (-1)^n. - Peter Luschny, Sep 18 2015
The unsigned coefficients for the partitions a_2*a_1^n for n >= 0 are the Lah numbers A001286. - Tom Copeland, Aug 06 2016
G.f.: 1 / (1 + Sum_{n > 0} a_n x^n/n!) = 1 / exp(a.x). - Tom Copeland, Oct 18 2016
Let a_1 = 1 + x + B_1 = x + 1/2 and a_n = B_n = (B.)^n, where B_n are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted signed polynomials of A019538: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... , p_n(x) = n * b_{n-1}. - Tom Copeland, Oct 18 2016
With a_n = 1/(n+1), b_n = B_n, the Bernoulli numbers. - Tom Copeland, Nov 08 2016
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of A145271; [E]^(-1), A356145, the inverse set to [E]; [P], the permutohedra polynomials of this entry; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

More terms from Jean-François Alcover, Apr 23 2014

A134264 Coefficients T(j, k) of a partition transform for Lagrange compositional inversion of a function or generating series in terms of the coefficients of the power series for its reciprocal. Enumeration of noncrossing partitions and primitive parking functions. T(n,k) for n >= 1 and 1 <= k <= A000041(n-1), an irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1, 1, 7, 7, 7, 21, 42, 21, 21, 35, 105, 35, 35, 70, 21, 1, 1, 8, 8, 8, 4, 28, 56, 56, 28, 28, 56, 168, 84, 168, 14, 70, 280, 140, 56, 140, 28, 1, 1, 9, 9, 9, 9, 36, 72
Offset: 1

Views

Author

Tom Copeland, Jan 14 2008

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal to 0, form h(t) = t / f(t) and let h_n denote the coefficient of t^n in h(t).
Lagrange inversion gives the compositional inverse about t=0 as g(t) = Sum_{j>=1} ( t^j * (1/j) * Sum_{permutations s with s(1) + s(2) + ... + s(j) = j - 1} h_s(1) * h_s(2) * ... * h_s(j) ) = t * T(1,1) * h_0 + Sum_{j>=2} ( t^j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) ), where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.
Denoting h_n by (n') for brevity, to 8th order in t,
g(t) = t * (0')
+ t^2 * [ (0') (1') ]
+ t^3 * [ (0')^2 (2') + (0') (1')^2 ]
+ t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ]
+ t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ]
+ t^6 * [ (0')^5 (5') + 5 (0')^4 (1') (4') + 5 (0')^4 (2') (3') + 10 (0')^3 (1')^2 (3') + 10 (0')^3 (1') (2')^2 + 10 (0')^2 (1')^3 (2') + (0') (1')^5 ]
+ t^7 * [ (0')^6 (6') + 6 (0')^5 (1') (5') + 6 (0')^5 (2') (4') + 3 (0')^5 (3')^2 + 15 (0')^4 (1')^2 (4') + 30 (0')^4 (1') (2') (3') + 5 (0')^4 (2')^3 + 20 (0')^3 (1')^3 (3') + 30 (0')^3 (1')^2 (2')^2 + 15 (0')^2 (1')^4 (2') + (0') (1')^6]
+ t^8 * [ (0')^7 (7') + 7 (0')^6 (1') (6') + 7 (0')^6 (2') (5') + 7 (0')^6 (3') (4') + 21 (0')^5 (1')^2* (5') + 42 (0')^5 (1') (2') (4') + 21 (0')^5 (1') (3')^2 + 21 (0')^5 (2')^2 (3') + 35 (0')^4 (1')^3 (4') + 105 (0)^4 (1')^2 (2') (3') + 35 (0')^4 (1') (2')^3 + 35 (0')^3 (1')^4 (3') + 70 (0')^3 (1')^3 (2')^2 + 21 (0')^2 (1')^5 (2') + (0') (1')^7 ]
+ ..., where from the formula section, for example, T(8,1',2',...,7') = 7! / ((8 - (1'+ 2' + ... + 7'))! * 1'! * 2'! * ... * 7'!) are the coefficients of the integer partitions (1')^1' (2')^2' ... (7')^7' in the t^8 term.
A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.
If the coefficients of partitions with the same exponent for h_0 are summed within rows, A001263 is obtained, omitting the leading 1.
From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014
From Tom Copeland, Oct 28-29 2014: (Start)
Operating with d/d(1') = d/d(h_1) on the n-th partition polynomial Prt(n;h_0,h_1,..,h_n) in square brackets above associated with t^(n+1) generates n * Prt(n-1;h_0,h_1,..,h_(n-1)); therefore, the polynomials are an Appell sequence of polynomials in the indeterminate h_1 when h_0=1 (a special type of Sheffer sequence).
Consequently, umbrally, [Prt(.;1,x,h_2,..) + y]^n = Prt(n;1,x+y,h_2,..); that is, Sum_{k=0..n} binomial(n,k) * Prt(k;1,x,h_2,..) * y^(n-k) = Prt(n;1,x+y,h_2,..).
Or, e^(x*z) * exp[Prt(.;1,0,h_2,..) * z] = exp[Prt(.;1,x,h_2,..) * z]. Then with x = h_1 = -(1/2) * d^2[f(t)]/dt^2 evaluated at t=0, the formal Laplace transform from z to 1/t of this expression generates g(t), the comp. inverse of f(t), when h_0 = 1 = df(t)/dt eval. at t=0.
I.e., t / (1 - t*(x + Prt(.;1,0,h_2,..))) = t / (1 - t*Prt(.;1,x,h_2,..)) = g(t), interpreted umbrally, when h_0 = 1.
(End)
Connections to and between arrays associated to the Catalan (A000108 and A007317), Riordan (A005043), Fibonacci (A000045), and Fine (A000957) numbers and to lattice paths, e.g., the Motzkin, Dyck, and Łukasiewicz, can be made explicit by considering the inverse in x of the o.g.f. of A104597(x,-t), i.e., f(x) = P(Cinv(x),t-1) = Cinv(x) / (1 + (t-1)*Cinv(x)) = x*(1-x) / (1 + (t-1)*x*(1-x)) = (x-x^2) / (1 + (t-1)*(x-x^2)), where Cinv(x) = x*(1-x) is the inverse of C(x) = (1 - sqrt(1-4*x)) / 2, a shifted o.g.f. for the Catalan numbers, and P(x,t) = x / (1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x / (1-t*x). Then h(x,t) = x / f(x,t) = x * (1+(t-1)Cinv(x)) / Cinv(x) = 1 + t*x + x^2 + x^3 + ..., i.e., h_1=t and all other coefficients are 1, so the inverse of f(x,t) in x, which is explicitly in closed form finv(x,t) = C(Pinv(x,t-1)), is given by A091867, whose coefficients are sums of the refined Narayana numbers above obtained by setting h_1=(1')=t in the partition polynomials and all other coefficients to one. The group generators C(x) and P(x,t) and their inverses allow associations to be easily made between these classic number arrays. - Tom Copeland, Nov 03 2014
From Tom Copeland, Nov 10 2014: (Start)
Inverting in x with t a parameter, let F(x;t,n) = x - t*x^(n+1). Then h(x) = x / F(x;t,n) = 1 / (1-t*x^n) = 1 + t*x^n + t^2*x^(2n) + t^3*x^(3n) + ..., so h_k vanishes unless k = m*n with m an integer in which case h_k = t^m.
Finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). [Added braces to disambiguate the formula. - N. J. A. Sloane, Oct 20 2015]
This relation reveals properties of the partitions and sums of the coefficients of the array. For n=1, h_k = t^k for all k, implying that the row sums are the Catalan numbers. For n = 2, h_k for k odd vanishes, implying that there are no blocks with only even-indexed h_k on the even-numbered rows and that only the blocks containing only even-sized bins contribute to the odd-row sums giving the Fuss-Catalan numbers for n=2. And so on, for n > 2.
These relations are reflected in any combinatorial structures enumerated by this array and the partitions, such as the noncrossing partitions depicted for a five-element set (a pentagon) in Wikipedia.
(End)
From Tom Copeland, Nov 12 2014: (Start)
An Appell sequence possesses an umbral inverse sequence (cf. A249548). The partition polynomials here, Prt(n;1,h_1,...), are an Appell sequence in the indeterminate h_1=u, so have an e.g.f. exp[Prt(.;1,u,h_2...)*t] = e^(u*t) * exp[Prt(.;1,0,h2,...)*t] with umbral inverses with an e.g.f e^(-u*t) / exp[Prt(.;1,0,h2,...)*t]. This makes contact with the formalism of A133314 (cf. also A049019 and A019538) and the signed, refined face partition polynomials of the permutahedra (or their duals), which determine the reciprocal of exp[Prt(.,0,u,h2...)*t] (cf. A249548) or exp[Prt(.;1,u,h2,...)*t], forming connections among the combinatorics of permutahedra and the noncrossing partitions, Dyck paths and trees (cf. A125181), and many other important structures isomorphic to the partitions of this entry, as well as to formal cumulants through A127671 and algebraic structures of Lie algebras. (Cf. relationship of permutahedra with the Eulerians A008292.)
(End)
From Tom Copeland, Nov 24 2014: (Start)
The n-th row multiplied by n gives the number of terms in the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n+1)]^n under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1^2], and 3 * A134264(3) = 3 *(1,1)= (3,3) the number of summands in the two homogeneous polynomials in the square brackets. For n=3, [a + b + c + d]^3 = [a^3 + b^3 + ...] + 3 [a*b^2 + a*c^2 + ...] + 6 [a*b*c + a*c*d + ...] maps to [4 * h_3] + 3 [12 * h_1 * h_2] + 6 [4 * (h_1)^3], and the number of terms in the brackets is given by 4 * A134264(4) = 4 * (1,3,1) = (4,12,4).
The further reduced expression is 4 h_3 + 36 h_1 h_2 + 24 (h_1)^3 = A248120(4) with h_0 = 1. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
Abramowitz and Stegun give combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link.
(End)
These partition polynomials are dubbed the Voiculescu polynomials on page 11 of the He and Jejjala link. - Tom Copeland, Jan 16 2015
See page 5 of the Josuat-Verges et al. reference for a refinement of these partition polynomials into a noncommutative version composed of nondecreasing parking functions. - Tom Copeland, Oct 05 2016
(Per Copeland's Oct 13 2014 comment.) The number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition, where the ordering of integer partitions is first by total, then by length, then lexicographically by the reversed sequence of parts. - Gus Wiseman, Feb 15 2019
With h_0 = 1 and the other h_n replaced by suitably signed partition polynomials of A263633, the refined face partition polynomials for the associahedra of normalized A133437 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Sep 09 2019
Number of primitive parking functions associated to each partition of n. See Lemma 3.8 on p. 28 of Rattan. - Tom Copeland, Sep 10 2019
With h_n = n + 1, the d_k (A006013) of Table 2, p. 18, of Jong et al. are obtained, counting the n-point correlation functions in a quantum field theory. - Tom Copeland, Dec 25 2019
By inspection of the diagrams on Robert Dickau's website, one can see the relationship between the monomials of this entry and the connectivity of the line segments of the noncrossing partitions. - Tom Copeland, Dec 25 2019
Speicher has examples of the first four inversion partition polynomials on pp. 22 and 23 with his k_n equivalent to h_n = (n') here with h_0 = 1. Identifying z = t, C(z) = t/f(t) = h(t), and M(z) = f^(-1)(t)/t, then statement (3), on p. 43, of Theorem 3.26, C(z M(z)) = M(z), is equivalent to substituting f^(-1)(t) for t in t/f(t), and statement (4), M(z/C(z)) = C(z), to substituting f(t) for t in f^(-1)(t)/t. - Tom Copeland, Dec 08 2021
Given a Laurent series of the form f(z) = 1/z + h_1 + h_2 z + h_3 z^2 + ..., the compositional inverse is f^(-1)(z) = 1/z + Prt(1;1,h_1)/z^2 + Prt(2;1,h_1,h_2)/z^3 + ... = 1/z + h_1/z^2 + (h_1^2 + h_2)/z^3 + (h_1^3 + 3 h_1 h_2 + h_3)/z^4 + (h_1^4 + 6 h_1^2 h_2 + 4 h_1 h_3 + 2 h_2^2 + h_4)/z^5 + ... for which the polynomials in the numerators are the partition polynomials of this entry. For example, this formula applied to the q-expansion of Klein's j-invariant / function with coefficients A000521, related to monstrous moonshine, gives the compositional inverse with the coefficients A091406 (see He and Jejjala). - Tom Copeland, Dec 18 2021
The partition polynomials of A350499 'invert' the polynomials of this entry giving the indeterminates h_n. A multinomial formula for the coefficients of the partition polynomials of this entry, equivalent to the multinomial formula presented in the first four sentences of the formula section below, is presented in the MathOverflow question referenced in A350499. - Tom Copeland, Feb 19 2022

Examples

			1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).
2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).
3) With f(t) = (e^(-a*t)-1) / (-a), h(t) = Sum_{n>=0} Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = Sum_{n>=1} a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, Sum_{permutations s with s(1)+s(2)+...+s(j)=j-1} h_s(1) * h_s(2) * ... * h_s(j) = j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) = a^(j-1). Note, in turn, Sum_{a=1..m} a^(j-1) = (Bernoulli(j,m+1) - Bernoulli(j)) / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x) = t / (x-1+1/(1-t)), then h(t,x) = x-1+1/(1-t), giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = (1-(1-x)*t-sqrt(1-2*(1+x)*t+((x-1)*t)^2)) / 2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.
5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)
6) With f(t,x) = x / ((1-t*x)*(1-(1+t)*x)), an o.g.f. for A074909, the reverse face polynomials of the simplices, h(t,x) = (1-t*x) * (1-(1+t)*x) with h_0=1, h_1=-(1+2*t), and h_2=t*(1+t), giving as the inverse in x about 0 the o.g.f. (1+(1+2*t)*x-sqrt(1+(1+2*t)*2*x+x^2)) / (2*t*(1+t)*x) for signed A033282, the reverse face polynomials of the Stasheff polytopes, or associahedra. Cf. A248727. (Added Jan 21 2015 by author.)
7) With f(x,t) = x / ((1+x)*(1+t*x)), an o.g.f. for the polynomials (-1)^n * (1 + t + ... + t^n), h(t,x) = (1+x) * (1+t*x) with h_0=1, h_1=(1+t), and h_2=t, giving as the inverse in x about 0 the o.g.f. (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2)) / (2*x*t) for the Narayana polynomials A001263. Cf. A046802. (Added Jan 24 2015 by author.)
From _Gus Wiseman_, Feb 15 2019: (Start)
Triangle begins:
   1
   1
   1   1
   1   3   1
   1   4   2   6   1
   1   5   5  10  10  10   1
   1   6   6   3  15  30   5  20  30  15   1
   1   7   7   7  21  42  21  21  35 105  35  35  70  21   1
Row 5 counts the following non-crossing set partitions:
  {{1234}}  {{1}{234}}  {{12}{34}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
            {{123}{4}}  {{14}{23}}  {{1}{23}{4}}
            {{124}{3}}              {{12}{3}{4}}
            {{134}{2}}              {{1}{24}{3}}
                                    {{13}{2}{4}}
                                    {{14}{2}{3}}
(End)
		

References

  • A. Nica and R. Speicher (editors), Lectures on the Combinatorics of Free Probability, London Mathematical Society Lecture Note Series: 335, Cambridge University Press, 2006 (see in particular, Eqn. 9.14 on p. 141, enumerating noncrossing partitions).

Crossrefs

(A001263,A119900) = (reduced array, associated g(x)). See A145271 for meaning and other examples of reduced and associated.
Other orderings are A125181 and A306438.
Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).
Cf. A248927 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A091867 and A125181, for relations to lattice paths and trees.
Cf. A249548 for use of Appell properties to generate the polynomials.
Cf. A133314, A049019, A019538, A127671, and A008292 for relations to permutahedra, Eulerians.
Cf. A006013.

Programs

  • Mathematica
    Table[Binomial[Total[y],Length[y]-1]*(Length[y]-1)!/Product[Count[y,i]!,{i,Max@@y}],{n,7},{y,Sort[Sort/@IntegerPartitions[n]]}] (* Gus Wiseman, Feb 15 2019 *)
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(Vec(p)) | p<-partitions(n-1)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

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 [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.
For example from g(t) above, T(5,4) = (5! / ((5-3)! * 2!)) / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.
From Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.
Then for the partition polynomials of A134264,
Poly[n;h_0,...,h_(n-1)]=(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)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1.
(End)
From Tom Copeland, Oct 20 2011: (Start)
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) = t*((h_0) + (h_1)d/dt + (h_2)(d/dt)^2 + ...)^2 / ((h_0) - (h_2)(d/dt)^2 - 2(h_3)(d/dt)^3 - 3(h_4)(d/dt)^4 + ...), and
L = (d/dt)/h(d/dt) = (d/dt) 1/((h_0) + (h_1)*d/dt + (h_2)*(d/dt)^2 + ...)
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
(End)
Using the formalism of A263634, the raising operator for the partition polynomials of this array with h_0 = 1 begins as R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 + 5 h_2^3 - 7 h_3^2 - 9 h_2 h_4) D^5/5! + (h_7 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... with D = d/d(h_1). - Tom Copeland, Sep 09 2016
Let h(x) = x/f^{-1}(x) = 1/[1-(c_2*x+c_3*x^2+...)], with c_n all greater than zero. Then h_n are all greater than zero and h_0 = 1. Determine P_n(t) from exp[t*f^{-1}(x)] = exp[x*P.(t)] with f^{-1}(x) = x/h(x) expressed in terms of the h_n (cf. A133314 and A263633). Then P_n(b.) = 0 gives a recursion relation for the inversion polynomials of this entry a_n = b_n/n! in terms of the lower order inversion polynomials and P_j(b.)P_k(b.) = P_j(t)P_k(t)|{t^n = b_n} = d{j,k} >= 0 is the coefficient of x^j/j!*y^k/k! in the Taylor series expansion of the formal group law FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)]. - Tom Copeland, Feb 09 2018
A raising operator for the partition polynomials with h_0 = 1 regarded as a Sheffer Appell sequence in h_1 is described in A249548. - Tom Copeland, Jul 03 2018

Extensions

Added explicit t^6, t^7, and t^8 polynomials and extended initial table to include the coefficients of t^8. - Tom Copeland, Sep 14 2016
Title modified by Tom Copeland, May 28 2018
More terms from Gus Wiseman, Feb 15 2019
Title modified by Tom Copeland, Sep 10 2019

A133437 Irregular triangle of coefficients of a partition transform for direct Lagrange inversion of an o.g.f., complementary to A134685 for an e.g.f.; normalized by the factorials, these are signed, refined face polynomials of the associahedra.

Original entry on oeis.org

1, -2, 12, -6, -120, 120, -24, 1680, -2520, 360, 720, -120, -30240, 60480, -20160, -20160, 5040, 5040, -720, 665280, -1663200, 907200, 604800, -60480, -362880, -181440, 20160, 40320, 40320, -5040, -17297280, 51891840, -39916800, -19958400, 6652800, 19958400, 6652800, -1814400, -1814400, -3628800, -1814400, 362880, 362880, 362880, -40320
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Sum_{n>=1} u_n * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -2 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 12 (2')^2 - 6 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -120 (2')^3 + 120 (1')(2')(3') - 24 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 1680 (2')^4 - 2520 (1') (2')^2 (3') + 360 (1')^2 (3')^2 + 720 (1')^2 (2') (4') - 120 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -30240 (2')^5 + 60480 (1') (2')^3 (3') - 20160 (1')^2 (2') (3')^2 - 20160 (1')^2 (2')^2 (4') + 5040 (1')^3 (3')(4') + 5040 (1')^3 (2')(5') - 720 (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 665280 (2')^6 - 1663200 (1')(2')^4(3') + (1')^2 [907200 (2')^2(3')^2 + 604800 (2')^3(4')] - (1')^3 [60480 (3')^3 + 362880 (2')(3')(4') + 181440 (2')^2(5')] + (1')^4 [20160 (4')^2 + 40320 (3')(5') + 40320 (2')(6')] - 5040 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -17297280 (2')^7 + 51891840 (1')(2')^5(3') - (1')^2 [39916800 (2')^3(3')^2 + 19958400 (2')^4(4')] + (1')^3 [6652800 (2')(3')^3 + 19958400 (2')^2(3')(4') + 6652800 (2')^3(5')] - (1')^4 [1814400 (3')^2(4') + 1814400 (2')(4')^2 + 3628800 (2')(3')(5') + 1814400 (2')^2(6')] + (1')^5 [362880 (4')(5') + 362880 (3')(6') + 362880 (2')(7')] - 40320 (1')^6(8')] * t^8 / 8!
...
See A134685 for more information.
A111785 is obtained from A133437 by dividing through the bracketed terms of the P(n,t) by n! and unsigned A111785 is a refinement of A033282 and A126216. - Tom Copeland, Sep 28 2008
For relation to the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see the Loday and McCammond links. E.g., P(5,t) = (1')^(-9) * [ 14 (2')^4 - 21 (1') (2')^2 (3') + 6 (1')^2 (2') (4')+ 3 (1')^2 (3')^2 - 1 (1')^3 (5') ] * t^5 is related to the 3-D associahedron with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Summing over faces of the same dimension gives A033282 or A126216. - Tom Copeland, Sep 29 2008
A relation between this Lagrange inversion for an o.g.f. and partition polynomials formed from the "refined Lah numbers" A130561 is presented in the link "Lagrange a la Lah" along with umbral binary tree representations.
With f(x,t) = x + t*Sum_{n>=2} u_n*x^n, the compositional inverse in x is related to the velocity profile of particles governed by the inviscid Burgers's, or Hopf, eqn. See A001764 and A086810. - Tom Copeland, Feb 15 2014
Newton was aware of this power series expansion for series reversion. See the Ferraro reference p. 75 eqn. 52. - Tom Copeland, Jan 22 2017
The coefficients of the partition polynomials divided by the associated factorial enumerate the faces of the convex, bounded polytopes called the associahedra, and the absolute value of the sum of the renormalized coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is either n! (unnormalized) or unity (normalized). In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With u_1 = 1 and the other u_n replaced by suitably signed partition polynomials of A263633, the partition polynomials enumerating the refined noncrossing partitions of A134264 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Nov 16 2019
Relations between associahedra and oriented n-simplices are presented by Halvorson and by Street. - Tom Copeland, Dec 08 2019
Let f(x;t,n) = x - t * x^(n+1), giving u_1 = (1') = 1 and u_(n+1) = (n+1) = -t. Then inverting in x with t a parameter gives finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). Comparing this with the same result in A134264 gives relations between the faces of associahedra and noncrossing partitions (and other combinatorial constructs related to these inversion formulas and those listed in A145271). - Tom Copeland, Jan 27 2020
From Tom Copeland, Mar 24 2020: (Start)
There is a mapping between the faces of K_n, the associahedron of dimension (n-1), and polygon dissections. The dissecting noncrossing diagonals (i.e., nonintersecting in the interior) form subpolygons. Assign the indeterminate x_k to a subpolygon where k = (number of vertices of the subpolygon) - 1. Multiply the x_k together to form the monomials for the inversion formula.
For the 3-dimensional associahedron K_4, the fundamental polygon is the hexagon, which can be dissected into pentagons, associated to x_4; tetragons, to x_3; and triangles, to x_2; for example, there are six distinguished partitions of the hexagon into one triangle and one pentagon, sharing two vertices, associated to the monomial 6 * x_2 * x_4 since the unshared vertex of the triangle can be moved consecutively from one vertex of the hexagon to the next. This term corresponds to 720 (1')^2 (2') (4') / 5! in P(5,t) above, denumerating the six pentagonal facets of K_4. (End)

References

  • G. Ferraro, The Rise and Development of the Theory of Series up to the Early 1820s, Springer Science and Business Media, 2007.
  • H. Halvorson (editor), Deep Beauty: Understanding the Quantum World Through Innovation, Cambridge Univ. Press, 2011.
  • H. Turnbull (editor), The Correspondence of Isaac Newton Vol. II 1676-1687, Cambridge Univ. Press, 1960, p. 147.

Crossrefs

Cf. A145271, (A086810, A181289) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k, {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

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

Extensions

Missing coefficient in P(6,t) replaced by Tom Copeland, Nov 06 2008
P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Title modified by Tom Copeland, Jan 13 2020
Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 07 2024

A178867 Irregular triangle read by rows: multinomial coefficients, version 3; alternatively, row n gives coefficients of the n-th complete exponential Bell polynomial B_n(x_1, x_2, ...) with monomials sorted into some unknown order.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 4, 3, 1, 1, 10, 10, 15, 5, 10, 1, 1, 15, 20, 45, 15, 60, 6, 15, 15, 10, 1, 1, 21, 35, 105, 35, 210, 21, 105, 105, 70, 7, 105, 35, 21, 1, 1, 28, 56, 210, 70, 560, 56, 420, 420, 280, 28, 840, 280, 168, 8, 280, 210, 105, 56, 35, 28, 1
Offset: 1

Views

Author

Johannes W. Meijer and Manuel Nepveu (Manuel.Nepveu(AT)tno.nl), Jun 21 2010, Jun 24 2010

Keywords

Comments

"Exponential" here means in contrast to "ordinary", cf. A263633 (see Comtet). "Standard order" means as produced by Maple's "sort" command. - N. J. A. Sloane, Oct 28 2015
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.
It appears that the authors' n-th row does give the coefficients of the n-th complete exponential Bell polynomial. Starting with row 6, however, it is unknown in what order the monomials of the n-th complete exponential Bell polynomial follow. It is definitely not the standard order nor any other known order. (End)
This version of the multinomial coefficients was discovered while calculating the probability that two 23 year old chessplayers would play each other on their birthday during a Dutch Chess Championship. This unique encounter took place on Apr 05 2008. Its probability is 1 in 50000 years, see the Meijer-Nepveu article.
The third version of the multinomial coefficients can be constructed with the basic multinomial coefficients A178866; see the formulas below. These multinomial coefficients appear in the columns of the multinomial coefficient matrix MCM[n,m] (n >= 1 and m >= 1).
We observe that the sum of the C(m,n) coefficients follow the A000296(n) sequence. The missing C(m, n=1) corresponds to A000296(1) = 0.
The number of multinomial coefficients in a triangle row leads to the partition numbers A000041. The row sums lead to the Bell numbers A000110.

Examples

			The first few complete exponential Bell polynomials are:
(1) x[1];
(2) x[1]^2 + x[2];
(3) x[1]^3 + 3*x[1]*x[2] + x[3];
(4) x[1]^4 + 6*x[1]^2*x[2] + 4*x[1]*x[3] + 3*x[2]^2 + x[4];
(5) x[1]^5 + 10*x[1]^3*x[2] + 10*x[1]^2*x[3] + 15*x[1]*x[2]^2 + 5*x[1]*x[4] + 10*x[2]*x[3] + x[5];
(6) x[1]^6 + 15*x[1]^4*x[2] + 20*x[1]^3*x[3] + 45*x[1]^2*x[2]^2 + 15*x[1]^2*x[4] + 60*x[1]*x[2]*x[3] + 6*x[1]*x[5] + 15*x[2]^3 + 15*x[2]*x[4] + 10*x[3]^2 + x[6].
(7) x[1]^7 + 21*x[1]^5*x[2] + 35*x[1]^4*x[3] + 105*x[1]^3*x[2]^2 + 35*x[1]^3*x[4] + 210*x[1]^2*x[2]*x[3] + 21*x[1]^2*x[5] + 105*x[1]*x[2]^3 + 105*x[1]*x[2]*x[4] + 70*x[1]*x[3]^2 + 7*x[1]*x[6] + 105*x[2]^2*x[3] + 35*x[3]*x[4] + 21*x[2]*x[5] + x[7].
...
The first few rows of the triangle are
  1;
  1,  1;
  1,  3,  1;
  1,  6,  4,   3,  1;
  1, 10, 10,  15,  5,  10,  1;
  1, 15, 20,  45, 15,  60,  6,  15,  15, 10, 1;
  1, 21, 35, 105, 35, 210, 21, 105, 105, 70, 7, 105, 35, 21, 1;
  ...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 134, 307-310.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, Chapter 2, Section 8 and table on page 49.

Crossrefs

Cf. A036040 (version 1 of multinomial coefficients), A080575 (version 2).

Programs

  • Maple
    with(combinat): nmax:=11; A178866(1):=1: T:=1: for n from 1 to nmax do y(n):=numbpart(n): P(n):=sort(partition(n)): k:=0: for r from 1 to y(n) do if P(n)[r,1]>1 then k:=k+1; B(k):=P(n)[r]: fi; od: A002865(n):=k; for k from 1 to A002865(n) do s:=0: j:=1: while sA002865(n))], `>`): for k from 1 to A002865(n) do M3[n,k]:=a[k] od: for k from 1 to A002865(n) do T:=T+1: A178866(T):= M3[n,k]: od: od:
    mmax:=nmax: n:=1: for m from 1 to mmax do MCM[n,m]:= A178866(n) od: n:=2: r:=1: for i from 2 to nmax do p:=A002865(i): r:=r+1: while p>0 do for m from 1 to mmax do MCM[n,m]:=A178866(n)*binomial(m,r) od: p:=p-1: n:=n+1: od: od: T:=0: for m from 1 to mmax do for n from 1 to numbpart(m) do T:=T+1; A178867(T):= MCM[n,m]; od: od; seq(A178867(n), n=1..T);
    # To produce the complete exponential Bell polynomials in standard order, from N. J. A. Sloane, Oct 28 2015
    M:=12;
    EE:=add(x[i]*t^i/i!,i=1..M);
    t1:=exp(EE);
    t2:=series(t1,t,M);
    Q:=k->sort(expand(k!*coeff(t2,t,k)));
    for k from 1 to 8 do lprint(k,Q(k)); od;
    # To produce the coefficients of the complete exponential Bell polynomials in standard order:
    triangle := proc(numrows) local E, s, Q;
    E := add(x[i]*t^i/i!, i=1..numrows);
    s := series(exp(E), t, numrows+1);
    Q := k -> sort(expand(k!*coeff(s, t, k)));
    seq(print(coeffs(Q(k))), k=1..numrows) end:
    triangle(6); # Peter Luschny, May 27 2020

Formula

G.f.: Exp(Sum_{i >= 1} x_i*t^i/i!) = 1 + Sum_{n >= 1} B_n(x_1, x_2,...)*t^n/n!. [Comtet, p. 134, Eq. [3b]. - N. J. A. Sloane, Oct 28 2015]
For m >= 1, the formulas for the first few matrix columns are:
MCM[1,m] = A178866(1)*C(m,0) = 1*C(m,0);
MCM[2,m] = A178866(2)*C(m,2) = 1*C(m,2);
MCM[3,m] = A178866(3)*C(m,3) = 1*C(m,3);
MCM[4,m] = A178866(4)*C(m,4) = 3*C(m,4) and
MCM[5,m] = A178866(5)*C(m,4) = 1*C(m,4);
MCM[6,m] = A178866(6)*C(m,5) = 10*C(m,5) and
MCM[7,m] = A178866(7)*C(m,5) = 1*C(m,5);
MCM[8,m] = A178866(8)*C(m,6) = 15*C(m,6) and
MCM[9,m] = A178866(9)*C(m,6) = 15*C(m,6) and
MCM[10,m] = A178866(10)*C(m,6) = 10*C(m,6) and
MCM[11,m] = A178866(11)*C(m,6) = 1*C(m,6).

Extensions

Alternative definition as coefficients of complete Bell polynomials added by N. J. A. Sloane, Oct 28 2015
Various sections and name edited by Petros Hadjicostas, May 28 2020

A263634 Irregular triangle read by rows: row n gives coefficients of n-th logarithmic polynomial L_n(x_1, x_2, ...) with monomials sorted into standard order.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -6, 12, -4, -3, 1, 24, -60, 20, 30, -5, -10, 1, -120, 360, -120, -270, 30, 120, 30, -6, -15, -10, 1, 720, -2520, 840, 2520, -210, -1260, -630, 42, 210, 140, 210, -7, -21, -35, 1
Offset: 1

Views

Author

N. J. A. Sloane, Oct 29 2015

Keywords

Comments

"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

Examples

			The first few polynomials are:
(1) x[1].
(2) -x[1]^2 + x[2].
(3) 2*x[1]^3 - 3*x[1]*x[2] + x[3].
(4) -6*x[1]^4 + 12*x[1]^2*x[2] - 4*x[1]*x[3] - 3*x[2]^2 + x[4].
(5) 24*x[1]^5 - 60*x[1]^3*x[2] + 20*x[1]^2*x[3] + 30*x[1]*x[2]^2 - 5*x[1]*x[4] - 10*x[2]*x[3] + x[5].
(6) -120*x[1]^6 + 360*x[1]^4*x[2] - 120*x[1]^3*x[3] - 270*x[1]^2*x[2]^2 + 30*x[1]^2*x[4] + 120*x[1]*x[2]*x[3] + 30*x[2]^3 - 6*x[1]*x[5] - 15*x[2]*x[4] - 10*x[3]^2 + x[6].
...
[1]    1
[2]   -1,    1
[3]    2,   -3,     1
[4]   -6,   12,    -4,    -3,   1
[5]   24,  -60,    20,    30,  -5,  -10,   1
[6] -120,  360,  -120,  -270,  30,  120,  30, -6, -15, -10, 1
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 140, 156, 308.

Crossrefs

Programs

  • Maple
    triangle := proc(numrows) local E, s, Q;
    E := add(x[i]*t^i/i!, i=1..numrows);
    s := series(log(1 + E), t, numrows+1);
    Q := k -> sort(expand(k!*coeff(s, t, k)));
    seq(print(coeffs(Q(k))), k=1..numrows) end:
    triangle(6); # updated by Peter Luschny, May 27 2020

Formula

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

A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.

Original entry on oeis.org

1, -1, 3, -2, -15, 20, -6, 105, -210, 40, 90, -24, -945, 2520, -1120, -1260, 420, 504, -120, 10395, -34650, 25200, 18900, -2240, -15120, -9072, 1260, 2688, 3360, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -50400, -56700, -120960, -75600, 18144, 20160, 25920, -5040
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 210 (1') (2')^2 (3') + 40 (1')^2 (3')^2 + 90 (1')^2 (2') (4') - 24 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 2520 (1') (2')^3 (3') - 1120 (1')^2 (2') (3')^2 - 1260 (1')^2 (2')^2 (4') + 420 (1')^3 (3')(4') + 504 (1')^3 (2')(5') - 120 (1')^4 (6') ] * t^6 / 6!
See A134685 for more information.
From Tom Copeland, Sep 28 2016: (Start)
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 34650 (1')(2')^4(3') + (1')^2 [25200 (2')^2(3')^2 + 18900 (2')^3(4')] - (1')^3 [2240 (3')^3 + 15120 (2')(3')(4') + 9072 (2')^2(5')] + (1')^4 [1260 (4')^2 + 2688 (3')(5') + 3360 (2')(6')] - 720 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 540540 (1')(2')^5(3') - (1')^2 [554400 (2')^3(3')^2 + 311850 (2')^4(4')] + (1')^3 [123200 (2')(3')^3 + 415800 (2')^2(3')(4') + 166320 (2')^3(5')] - (1')^4 [50400 (3')^2(4') + 56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6')] + (1')^5 [18144 (4')(5') + 20160 (3')(6') + 25920 (2')(7')] - 5040 (1')^6(8')] * t^8 / 8! (End)

Examples

			From _Tom Copeland_, Sep 18 2014: (Start)
Let f(x) = log((1-ax)/(1-bx))/(b-a) = -log(1-u.*x) = x + (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 + (a^3+a^2b+ab^2+a^3)x^4/4 + ... . with (u.)^n = u_n = h_(n-1)(a,b) the complete homogeneous polynomials in two indeterminates.
Then the inverse g(x) = (e^(ax)-e^(bx))/(a*e^(ax)-b*e^(bx)) = x - (a+b)x^2/2! + (a^2+4ab+b^2)x^3/3! - (a^3+11a^2b+11ab^2+b^3)x^4/4! + ... , where the bivariate polynomials are the Eulerian polynomials of A008292.
The inversion formula gives, e.g., P(3,x) = 3(u_2)^2 - 2u_3 = 3(h_1)^2 - 2h_2 = 3(a+b)^2 - 2(a^2 + ab + b^2) = a^2 + 4ab + b^2. (End)
		

Crossrefs

Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k/(k+1), {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

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 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)]
= 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133932,
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) + (u_2)*d/dt + (u_3)*(d/dt)^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
From Tom Copeland, Sep 19 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-1)! 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 A094587 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.
With u_1 = 1, the first column of UP^(-1) with u_1 = 1 (with initial indices [0,0]) is composed of the row polynomials n! * OP_n(-u_2,...,-u_(n+1)), where OP_n(x[1],...,x[n]) are the row polynomials of A263633 for n > 0 and OP_0 = 1, which are related to those of A133314 as reciprocal o.g.f.s are related to reciprocal e.g.f.s; e.g., UP^(-1)[0,0] = 1, Up^(-1)[1,0] = -u_2, UP^(-1)[2,0] = 2! * (-u_3 + u_2^2) = 2! * OP_2(-u_2,-u_3).
Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
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 = -24 u5, b2 = 90 u2 u4 + 40 u3^2, b3 = -210 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers's 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 * 210 u2^2 u3 + 1 * 90 u2 u4 + 1 * 40 u3^2 - 0 * -24 u5 = 315 u2^4 - 420 u2^2 u3 + 90 u2 u4 + 40 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 + 20 u2 u3 - 6 u4) + 6!/(3!*3!) * (3 u2^2 - 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 refined Stirling polynomials of the first kind A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 08 2024

A276816 Irregular triangle read by rows: T(n,m) = coefficients in power/Fourier series expansion of an arbitrary anharmonic oscillator's exact period.

Original entry on oeis.org

-24, 480, -120, 6720, 3360, -241920, 1774080, -560, 40320, 40320, -1774080, 20160, -3548160, 61501440, -591360, 92252160, -1845043200, 8364195840, -2520, 221760, 221760, -11531520, 221760, -23063040, 461260800, 110880, -23063040, -11531520, 1383782400, -15682867200, -11531520, 691891200, 1383782400, -62731468800, 476759162880
Offset: 1

Views

Author

Bradley Klee, Sep 18 2016

Keywords

Comments

The phase space trajectory A276738 has phase space angular velocity A276814 and differential time dependence A276815. We calculate the period K = Int dt over the range [2*Pi, 0], trivial to compute from A276815 using A273496. Then K/(2*Pi) = 1 + sum b^(2n)*T(n,m)*f'(n,m); where the sum runs over n = 1, 2, 3 ... and m = 1, 2, 3, ... A000041(2n), and f'(n,m) = f(2n,m) of A276738 with Q=1/2. Choosing one point from the infinite dimensional coefficient space--v_i=0 for odd i, v_i=(-1)^(i/2-1)/2/(i!) otherwise--setting b^2 = 4*k, and summing over the entire table obtains the EllipticK expansion 2*A038534/A038533. For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Examples

			n/m   1     2     3         4         5
------------------------------------------
1  | -24   480
2  | -120  6720  3360   -241920   1774080
------------------------------------------
For pendulum values, f'(1,*)={(-1/384), 0}, f'(2,*) = {1/46080, 0, 1/294912, 0, 0}. Then K/(2Pi) = 1+(-1/384)*(-24)*4*k+((1/46080)*(-120)+(1/294912)*3360)*16*k^2=1+(1/4)*k + (9/64)*k^2, the first few terms of EllipticK.
		

Crossrefs

Programs

  • Mathematica
    RExp[n_]:=Expand[b Plus[R[0], Total[b^# R[#] & /@ Range[n]]]]
    RCalc[n_]:=With[{basis =Subtract[Tally[Join[Range[n + 2], #]][[All, 2]],Table[1, {n + 2}]] & /@ IntegerPartitions[n + 2][[3 ;; -1]]},
    Total@ReplaceAll[Times[-2, Multinomial @@ #, v[Total[#]],Times @@ Power[RSet[# - 1] & /@ Range[n + 2], #]] & /@ basis, {Q^2 -> 1, v[2] -> 1/4}]]
    dt[n_] := With[{exp = Normal[Series[-1/(1 + x)/.x -> Total[(2 # v[#] RExp[n - 1]^(# - 2) &/@Range[3, n + 2])], {b, 0, n}]]},
    Expand@ReplaceAll[Coefficient[exp, b, #] & /@ Range[n], R -> RSet]]
    RingGens[n_] :=Times @@ (v /@ #) & /@ (IntegerPartitions[n]/. x_Integer :> x + 2)
    tri[m_] := MapThread[Function[{a, b},Times[-# /. v[n_] :> Q^n /. Q^n_ :>  Binomial[n, n/2],(1/2) Coefficient[a, #]] & /@ b], {dt[2 m][[2 #]] & /@ Range[m], RingGens[2 #] & /@ Range[m]}]
    RSet[0] = 1; Set[RSet[#], Expand@RCalc[#]] & /@ Range[2*7];
    tri7 = tri[7]; tri7 // TableForm
    PeriodExpansion[tri_, n_] := ReplaceAll[ 1 + Dot[MapThread[ Dot, {tri,
      2 RingGens[2 #] & /@ Range[n]}], (2 h)^(Range[n])], {v[m_] :> (v[m]*(1/2)^m)}]
    {#,SameQ[Normal@Series[(2/Pi)*EllipticK[k],{k,0,7}],#]}&@ReplaceAll[
    PeriodExpansion[tri7,7],{v[n_/;OddQ[n]]:>0,v[n_]:> (-1)^(n/2-1)/2/(n!),h->2 k}]

A201509 Irregular triangle read by rows: T(n,k) = 2*T(n-1,k) + T(n-2,k-1) with T(0,0) = 0, T(n,0) = T(1,1) = 1 and T(n,k) = 0 if k < 0 or if n < k.

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 1, 8, 12, 4, 16, 28, 13, 1, 32, 64, 38, 6, 64, 144, 104, 25, 1, 128, 320, 272, 88, 8, 256, 704, 688, 280, 41, 1, 512, 1536, 1696, 832, 170, 10, 1024, 3328, 4096, 2352, 620, 61, 1, 2048, 7168
Offset: 0

Views

Author

Paul Curtz, Dec 02 2011

Keywords

Comments

This is the pseudo-triangle whose successive lines are of the type T(n,0), T(n,1)+T(n-1,0), T(n,2)+T(n-1,1), ... T(n,k)+T(n-1,k-1), without 0's, with T=A201701. [e-mail, Philippe Deléham, Dec 04 2011]

Examples

			Triangle starts:
    1   1
    2   2
    4   5   1
    8  12   4
   16  28  13  1
   32  64  38  6
   64 144 104 25 1
  128 320 272 88 8
  ...
Triangle begins (full version):
    0
    1,   1
    2,   2,   0
    4,   5,   1,  0
    8,  12,   4,  0, 0
   16,  28,  13,  1, 0, 0
   32,  64,  38,  6, 0, 0, 0
   64, 144, 104, 25, 1, 0, 0, 0
  128, 320, 272, 88, 8, 0, 0, 0, 0
		

Crossrefs

Cf. A052542 (row sums).

Formula

T(n,k) = 2*T(n-1,k) + T(n-2,k-1) with T(0,0) = 0, T(n,0) = T(1,1) = 1 and T(n,k) = 0 if k < 0 or if n < k. - Philippe Deléham, Dec 05 2011
The n-th row polynomial appears to equal Sum_{k = 1..floor((n+1)/2)} binomial(n,2*k-1)*(1+t)^k. Cf. A034867. - Peter Bala, Sep 10 2012
Aside from the first two rows below, the signed coefficients appear in the expansion (b*x - 1)^2 / (a*b*x^2 - 2a*x + 1) = 1 + (2 a - 2 b)x + (4 a^2 - 5 a b + b^2)x^2 + (8 a^3 - 12 a^2b + 4 ab^2)x^3 + ..., the reciprocal of the derivative of x*(1-a*x) / (1-b*x). This is related to A263633 via the expansion (a*b*x^2 - 2a*x + 1) / (b*x - 1)^2 = 1 + (b - a) (2x + 3b x^2 + 4b^2 x^3 + ...). See also A201780. - Tom Copeland, Oct 30 2023

Extensions

Edited and new name using Philippe Deléham's formula, Joerg Arndt, Dec 13 2023

A356144 Coefficients of the set of partition polynomials [RT] = [P][E]; i.e., coefficients of polynomials resulting from using the set of refined Eulerian polynomials, [E], of A145271 as the indeterminates of the set of permutahedra polynomials, [P], of A133314. Irregular triangle read by rows with lengths given by A000041.

Original entry on oeis.org

1, -1, 1, -1, -1, 2, -1, 1, -3, 2, 1, -1, -1, 4, -4, -2, 5, -1, -1, 1, -5, 8, 2, -4, -2, -4, 5, 4, -4, -1, -1, 6, -12, -3, 8, 18, -6, -14, 13, 2, -16, 14, 0, -8, -1, 1, -7, 18, 3, -20, 0, -15, 8, 18, 57, 6, -54, -15, -12, 84, -30, -48, 14, 14, -8, -13, -1, -1, 8, -24, -4, 32, 51, -27, -16, -6, 171, -42, -177, 50, 90, -18, 456, -276, -246, -15, 30, 154, -42, 124, -166, -113, 42, 6, -21, -19, -1
Offset: 0

Views

Author

Tom Copeland, Jul 27 2022

Keywords

Comments

I stipulate that the row lengths are A000041, but this imposes the insertion of a zero as a coefficient of a monomial for the polynomial RT_7 and for RT_8. The number of nonzero coefficients in each higher order polynomial remains to be determined. The monomials of the partition polynomials are arranged in the order (bottom to top) in Abramowitz and Stegun (starting on p. 831, link in A000041).
The analytic interpretation of these coefficients is related to the e.g.f.s of reciprocals of the derivatives (slopes of tangents) of a pair of compositionally inverse e.g.f.s as explicitly shown in the formulas.
With the notation introduced in the formula section, this set of partition polynomials, [RT], is the e.g.f. counterpart to the special Schur expansion coefficients [b], or [K], of A355201 for o.g.f.s. and is conjugate dual to the Lagrange inversion polynomials [L] of A134685.
For example, as shown in the formulas, [RT] = [P][E] = [P][L][P] where [P] is the set of polynomials of A133314, the refined Euler characteristic polynomials of the permutahedra; [E], the set A145271, the refined Eulerian polynomials; and [L], the set A134685, the classic Lagrange inversion polynomials--all related to transformations of e.g.f.s, or Taylor series, for which [RT], [L], and [E] can each be used to give the compositional inverse and [P], the multiplicative inverse, or reciprocal.
On the other hand, as shown in formulas for A355201, [K] = [R][N] = [R][A][R] where [R] is the set A263633 (mod signs), refined Pascal polynomials; [N], the set A134264, the refined Narayana, or noncrossing partition, polynomials; and [A], the set A133437, the refined Euler characteristic polynomials of the associahedra--all related to transformations of o.g.f.s, or power series, for which [K], [A], and [N] can each be used to give the compositional inverse and [R], the multiplicative inverse, or reciprocal. This is related to three pairs of compositionally inverse series--two pairs of Laurent series and one pair of power series.

Examples

			Arranged by rows, the coefficients are
0)  1;
1) -1;
2)  1, -1;
3) -1, 2, -1;
4)  1, -3, 2, 1, -1;
5) -1, 4, -4, -2, 5, -1, -1;
6)  1, -5, 8, 2, -4, -2, -4, 5, 4, -4, -1;
7) -1, 6, -12, -3, 8, 18, -6, -14, 13, 2, -16, 14, 0, -8, -1;
8)  1, -7, 18, 3, -20, 0, -15, 8, 18, 57, 6, -54, -15, -12, 84, -30, -48, 14, 14, -8, -13, -1;
. . .
The first few partition polynomials are
RT_0 =  1,
RT_1 = -a1,
RT_2 = a1^2  - a2,
RT_3 = -a1^3 + 2 a1 a2 - a3,
Rt_4 = a1^4 - 3 a1^2 a2 + 2 a2^2 + a1 a3 - a4,
RT_5 = -a1^5 + 4 a1^3 a2 - 4 a1 a2^2 - 2 a1^2 a3 + 5 a2 a3 - a1 a4 - a5,
RT_6 = a1^6 - 5 a1^4 a2 + 8 a1^2 a2^2 + 2 a1^3 a3 - 4 a2^3 - 2 a1 a2 a3 - 4 a1^2 a4 + 5 a3^2 + 4 a2 a4 - 4 a1 a5 - a6,
RT_7 = -a1^7 + 6 a1^5 a2 - 12*a1^3 a2^2 - 3 a1^4 a3 + 8 a1 a2^3 + 18 a1^2 a2 a3 - 6 a1^3 a4 - 14 a2^2 a3 + 13 a1 a3^2 + 2 a1 a2 a4 - 16 a1^2 a5 + 14 a3 a4 + 0 a2 a5 - 8 a1 a6 - a7,
RT_8 =  a1^8 - 7 a1^6 a2 + 18 a1^4 a2^2 + 3 a1^5 a3 - 20 a1^2 a2^3 + 0 a1^3 a2 a3 - 15 a1^4 a4 + 8 a2^4 + 18 a1 a2^2 a3 + 57 a1^2 a3^2 + 6 a1^2 a2 a4 - 54 a1^3 a5 - 15 a2 a3^2 - 12 a2^2 a4 + 84 a1 a3 a4 - 30 a1 a2 a5 - 48 a1^2 a6 + 14 a4^2 + 14 a3 a5 - 8 a2 a6 - 13 a1 a7 - a8.
		

Crossrefs

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = 1 / D[InverseSeries[Integrate[1/(1 + Sum[c[k] x^k/k!, {k, nn}] + O[x]^(nn+1)), x]], x]}, Table[Coefficient[n! s, x^n Product[c[t], {t, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Feb 17 2024 *)
  • SageMath
    B. = PolynomialRing(ZZ)
    A. = PowerSeriesRing(B)
    f = 1/(1 + a1*x + a2*x^2/factorial(2) + a3*x^3/factorial(3) + a4*x^4/factorial(4) + a5*x^5/factorial(5) + a6*x^6/factorial(6) + a7*x^7/factorial(7) + a8*x^8/factorial(8) + a9*x^9/factorial(9) + a10*x^10/factorial(10) )
    g = integrate(f)
    h = g.reverse()
    w = derivative(h,x)
    I = 1 / w
    # Added by # Peter Luschny, Feb 17 2024:
    # The list of coefficients in sparse format (i.e. without the zeros):
    for n, c in enumerate(I.list()[:10]):
        print(f"RT[{n}]", (factorial(n)*c).coefficients())

Formula

Denote this set of partition polynomials by [RT], the permutahedra polynomials of A133314 by [P], the refined Eulerian polynomials of A145271 by [E], and the Lagrange inversion polynomials of A134685 for e.g.f.s by [L]. Let the typically noncommutative product of two sets, e.g., [P][E], represent the substitution of the polynomials of [E] for the indeterminates of [P], i.e., a composition at the level of the indeterminates (see A356145 for examples). Let [I] be the substitutional identity transformation, and mark the substitutional inverse with the superscript -1. Then the following relations hold.
[RT] = [P][E] = [P][L][P] = [P]^{-1}[L][P] = [P][L][P]^{-1} since [P] is an involution, i.e., [P]^2 = [I], or [P] = [P]^{-1}, so [RT] and [L] are conjugate duals.
[RT]^{-1} = ([P][E])^{-1} = [E]^{-1}[P] = ([P][L][P])^{-1} = [P][L][P] = [RT], with [E]^{-1} = A356145, since [L] and [P] are involutions, so is [RT], i.e., [RT]^2 = [I].
RT_n(a_1,a_2,...,a_n) = D_{x=0}^n 1 / [ D_x f^{(-1)}(x)] for which D_x is the derivative w.r.t. x and the indeterminates are defined by 1 / [D_x f(x)] = 1 + a_1 x + a_2 x^2/2! + a_3 x^3/3! + ... with f(x) and f^{(-1)}(x) a compositional inverse pair of formal Taylor series, or e.g.f.s. This is the analytic equivalent of the algebraic relation [RT] = [P][E]. In words, the partition polynomials of row n (initial row is 0) is the n-th coefficient of the formal Taylor series of the reciprocal of the derivative of the compositional inverse of a function in terms of the Taylor series coefficients of the reciprocal of the derivative of that function. Note the correspondence with the analytic interpretation of [E]^{-1} of A356145, consistent with the algebraic identities above.
RT_n(a_1,a_2,...,a_n) = D_{x=0}^n f'(f^{(-1)}(x)) also, by the inverse function theorem, where the prime denotes differentiation with respect to the argument of the function.
With all a_k = (-1)^k, RT_0 = RT_1 = 1, otherwise RT_n = 0. This is determined with f(x) = e^{x}-1 and f^{(-1)}(x) = log(1+x).
With all a_k = 1, RT_0 = 1, RT_1 = -1, otherwise RT_n=0. This is determined with f(x) =1-e^{-x} and f^{(-1)}(x) = -log(1-x).
With all a_k = -1, RT_0 = 1 and RT_n = 2^(n-1) otherwise. This is determined with f(x) = (x - log(2-e^x))/2 and f^{(-1)}(x) = x - log(cosh(x)). (Careful, these are not the row sums of the absolute values of the numerical coefficients, which for the first ten polynomials are 1, 1, 2, 4, 8, 18, 40, 122, 446, and 2428.)
With a_k = k! 2^k, RT_0 = 1 and RT_n = -2*(2(n-1))! / (n-1)! = -2*n!*A000108(n-1) otherwise. This is determined with f(x) = x - x^2 and f^{(-1)}(x) = (1 - sqrt(1-4x))/2. Similar relations hold for the Fuss-Catalan sequences with f(x) = x - x^{m+1} for m > 1.

Extensions

Order of terms in rows 4-6 corrected by Andrey Zabolotskiy, Feb 17 2024

A355201 Normalized Schur self-convolution expansion coefficients K_{n+1}^n / n giving the coefficients of the Laurent series (compositionally) inverse to f(z) = c_0 z + c_1 + c_2 / z + c_3 / z^2 + ... . Irregular triangle for partition polynomials, with row lengths A000041(n) - 1 except for the first two, which are both of length 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 3, 3, 1, 1, 6, 4, 2, 12, 6, 2, 4, 4, 1, 1, 10, 5, 10, 30, 10, 10, 10, 20, 10, 5, 5, 5, 1, 1, 15, 6, 30, 60, 15, 5, 60, 30, 60, 20, 15, 15, 30, 30, 15, 3, 6, 6, 6, 1, 1, 21, 7, 70, 105, 21, 35, 210, 70, 140, 35, 35, 105, 105, 105, 105, 35, 7, 42, 21, 21, 42, 42, 21, 7, 7, 7, 7, 1
Offset: 0

Views

Author

Tom Copeland, Jun 23 2022

Keywords

Comments

For the formal Laurent series f(z) = a_0 z + a_1 + a_2 / z + a_3 / z^2 + ..., the formal compositional inverse is g(z) = b_0 z + b_1 + b_2 / z + b_3 / z^2 + ..., whose coefficients are partition polynomials whose numerical factors are those of this irregular triangle T(n,k). For the Schur coefficients defined in the formula section, -b_n = K_{n}^{n-1} / (n-1) for n > 1.
Analytic proofs of the relationship between the partition polynomials of the compositional inverse pair of Laurent series and Schur's self-convolution expansion coefficients are given in Schur (beware of a sign error) and in Airault and Ren.
Explicit examples (with a_0=1) of K_{n}^{n-1} up through n=5 are in Airault and Bouali on p. 182.
Various formulas for the b_n in terms of the associahedra (A133437), noncrossing (A134264), reciprocal (A263633), and Faber partition (A263916) polynomials are given in Copeland as well as a derivation of the explicit multi-factorial expression in the formula section and a combinatorial model.

Examples

			Triangle begins:
1) 1
2) 1
3) 1
4) 1,  1
5) 1,  1,  2,  1
6) 1,  3,  3,  3,  3,  1
7) 1,  6,  4,  2, 12,  6,  2,  4,  4,  1
8) 1, 10,  5, 10, 30, 10, 10, 10, 20, 10,  5,  5,  5,  1
  ...
The first few partition polynomials, with the monomials in the order of the partitions on p. 831 of Abramowitz and Stegun, are
b0 =    1 / a0
b1 = - a1 / a0
b2 = - a2
b3 = -(a1 a2 + a0 a3)
b4 = -(a1^2 a2 + a0 a2^2 + 2 a0 a1 a3 + a0^2 a4)
b5 = -(a1^3 a2+ 3 a0 a1 a2^2 + 3 a0 a1^2 a_3 + 3 a0^2 a2 a3 + 3 a0^2 a1 a4
      + a0^3 a_5)
b6 = -(a1^4 a2 + 6 a0 a1^2 a2^2 + 4 a0 a1^3 a3 + 2 a0^2 a2^3 + 12 a0^2 a1 a2 a3
      + 6 a0^2 a1^2 a4  + 2 a0^3 a3^2 + 4a0^3 a2 a4 + 4 a0^3 a1 a5 + a0^4 a6)
b7 = -(a1^5 a2 + 10 a_0 a1^3 a2^2 + 5 a0 a1^4 a3 + 10 a0^2 a1 a2^3
      + 30 a0^2 a1^2 a2 a3 + 10 a0^2 a1^3 a4 + 10 a0^3 a2^2 a3 + 10 a0^3 a1 a3^2
      + 20 a0^3 a1 a2 a4 + 10 a0^3 a1^2 a5 + 5 a0^4 a3 a4 + 5 a0^4 a2 a5
      + 5 a0^4 a1 a6 + a0^5 a7)
_____________________
		

Crossrefs

Programs

  • Mathematica
    row[0] = row[1] = {1};
    row[n_] := With[{s = Expand[Coefficient[Sum[c[k] x^k, {k, 0, n}]^(n-1), x, n] / (n-1)]}, Table[Coefficient[s, Product[c[t], {t, p}]], {p, Reverse[Sort[Sort /@ IntegerPartitions[n, {n-1}, Range[0, n]]]]}]];
    Table[row[n], {n, 0, 8}] // Flatten (* Andrey Zabolotskiy, Feb 05 2023 *)

Formula

The index notations b(n), b_n, and bn are used interchangeably in this entry for indeterminates.
For n > 1, b_n(a_0,a_1,...,a_n) is a sum of monomials of the form a0^{e0} a1^{e1} a2^{e2} ... an^{en} with e1 * 1 + e2 * 2 + ... + en * n = n. When a_0 is not set to unity, e0 + e1 + ... + en = n - 1. (a1^n is not present.)
From a combinatorial argument in Copeland, the unsigned numerical coefficient of the monomial is given by (n-2)! / [(n - 1 - (e1 + e2 + ... + en))! e1! e2! ... en!].
The partition polynomials are generated by a subset of the Schur self-convolution expansion coefficients as -b_n = K_{n}^{n-1} / (n-1) =(D_{x=0}^n / n!) (a_0 + a_1 x + a_2 x^2 + ... + a_n x^n)^{n-1} / (n-1).
Row sums are the Catalan numbers A000108, ignoring the overall sign, for b_1 onwards.
Reduces to the Narayana triangle A001263 with a_0 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
Reduces to A091869 (reversed A091187) with a_1 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
b_n(c_1,...,c_n) = - Sum_{k=0}^{n-1} b_k(c_1,...,c_k) N_{n-k}(c_1,...,c_{n-k}) with c_0 = 1 and N_k(c_1,...,c_k) the noncrossing partition polynomials of A134264.
[b] = [R][N], representing the substitution of the noncrossing partition polynomials of A134264 for the indeterminates of the signed reciprocal polynomials of A263633 defined by R_n = 1 / (1 + c_1 x + c_2 x^2 + ...).
Conversely, [R][b] = [N] since the substitution transformation denoted by [R] is an involution, i.e., [R]^2 = [I], the identity substitution.
[b] = [R][A][R], a substitutional conjugation of the set of associahedra partition polynomials of A133147, or A111785, with re-indexing and (1') = 1, e.g., A_0 = 1, A_1 = -c_1, and A_2 = 2 c_1^2 - c_2.
Conversely, [A] = [R][b][R].

Extensions

Rows 8-9 added by Andrey Zabolotskiy, Feb 05 2023
Showing 1-10 of 10 results.