

A133437


Irregular triangle of coefficients C(j,k) of a partition transform for direct Lagrange inversion of an o.g.f., complementary to A134685 for an e.g.f.


16



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

1,2


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') + 720 (1')^2 (2') (4') + 360 (1')^2 (3')^2  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 (2')(5') + 5040 (1')^3 (3')(4')  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 [362880 (2')(3')(4') + 60480 (3')^3 + 181440 (2')^2(5')] + (1')^4 [40320 (2')(6') + 40320 (3')(5') + 20160 (4')^2]  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 (2')(4')^2 + 3628800 (2')(3')(5') + 1814400 (2')^2(6') + 1814400 (3')^2(4')] + (1')^5 [362880 (2')(7') + 362880 (3')(6') + 362880 (4')(5')]  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 3D associahedron with 14 vertices (0D faces), 21 edges (1D faces), 6 pentagons (2D faces), 3 rectangles (2D faces), 1 3D polytope (3D 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', 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


REFERENCES

G. Ferraro, The Rise and Development of the Theory of Series up to the Early 1820s, Springer Science and Business Media, 2007


LINKS

Table of n, a(n) for n=1..45.
F. Brown and J. Bergstrom, Inversion of series and the cohomology of the moduli spaces M_(o,n), arXiv:0910.0120 [math.AG], 2009.
T. Copeland, Lagrange a la Lah , 2011.
T. Copeland Compositional inverse pairs, the BurgersHopf equation, and the Stasheff associahedra,, 2014.
T. Copeland, Compositional inversion and generating functions in algebraic geometry, MathOverflow question, 2014.
T. Copeland, Guises of the Stasheff polytopes, associahedra for the Coxeter A_n root system?, MathOverflow question, 2014.
T. Copeland, Why is there a connection between enumerative geometry and nonlinear waves?, MathOverflow answer, 2014.
T. Copeland, Important formulas in combinatorics, MathOverflow answer, 2015.
T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015.
S. Devadoss, B. Fehrman, T. Heath, and A. Vashist, Moduli spaces of punctured PoincarĂ© discs, arXiv:1109.2830 [math.AT], 2011.
S. Devadoss, T. Heath, and W. Vipismakul, Deformation of bordered surfaces and convex polytopes, Notices of the AMS, April 2011, Volume 58, Issue 04.
C. Dupont and B. Vallette Browns' moduli spaces of curves and the gravity operad, arXiv:1590.08840 [math.AG], p. 15, 2015.
S. Forcey, The Hedra Zoo [From Tom Copeland, Jan 09 2017]
J. Golden, M. Paulos, M. Spradlin, and A. Volovich, Cluster polylogarithms for scattering amplitudes, arxiv: 1401.6446 (v2) [hep.th], 2014.
D. Jackson, A. Kempf, A. Morales, A robust generalization of the Legendre transform for QFT, arXiv:1612.0046 [hepth], 2017.
D. Kreimer and K. Yeats, Diffeomorphisms of quantum fields, arXiv:1610.01837 [mathph], 2016. (pg. 7)
M. Liu, Moduli of Jholomorphic curves with lagrangian boundary conditions and open GromovWitten invariants for an S^1 pair, arXiv:math/0210257 [math.SG], 20022004. (see Fig. 2 pg. 13)
J.L. Loday, The Multiple Facets of the Associahedron [From Tom Copeland, Sep 29 2008]
MathOverflow, Loday's characterization and enumeration of faces of associahedra (Stasheff polytopes), an MO question posed by T. Copeland and answered by R. Davis, 2017.
J. McCammond, Noncrossing Partitions in Surprising Locations, American Mathematical Monthly 113 (2006) 598610. [From Tom Copeland, Sep 29 2008]
Alison Schuetz and Gwyneth Whieldon, Polygonal Dissections and Reversions of Series, arxiv: 1401.7194 [math.CO], 2014.
J. Zhou, Quantum deformation theory of the Airy curve and the mirror symmetry of a point, arXiv preprint arXiv:1405.5296 [math.AG], 2014.


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)^(n1+e(1)) * [2*(n1)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./(1u.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(n1,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)^(n1) 1/[u_1 + u_2 * x + u_3 * x^2 + ... + u_n * x^(n1)]^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 squarebracketed 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, ..., (K1) * bK, ...) = [(n+1)/2] * Sum_{k = 2..n1} PS(nk+1,u_1=1,u_2,...,u_(nk+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,...,(K1)*bK,..) = d/dt t^(n2)*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 mth 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+1k}, or equivalently multiply the diagonals of A132159 by u_m. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(1) * S]^(n1) 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)


CROSSREFS

Cf. A145271, (A086810,A181289) = (reduced array, associated g(x)).
Cf. A001764, A007318, A033282, A086810, A111785, A126216, A129185, A132159, A133314, A134685, A139605.
Sequence in context: A035877 A086494 A107414 * A245692 A182126 A266511
Adjacent sequences: A133434 A133435 A133436 * A133438 A133439 A133440


KEYWORD

sign,tabf


AUTHOR

Tom Copeland, Jan 27 2008


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


STATUS

approved



