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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'),

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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 normalized by the factorials, these are signed, refined face polynomials of the associahedra.

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Terms ordered according to the reversed AmramowitzAbramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 07 2024

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

P(5,t) = (1')^(-9) * [ 1680 (2')^4 - 2520 (1') (2')^2 (3') + 360 (1')^2 (3')^2 + 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 (23')(54') + 5040 (1')^3 (32')(45') - 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') + 60480 (3')^3 + 181440 (2')^2(5')] + (1')^4 [20160 (4')^2 + 40320 (23')(65') + 40320 (3')(5') + 20160 (4')^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') + 1814400 (3')^2(4')] + (1')^5 [362880 (24')(75') + 362880 (3')(6') + 362880 (42')(57')] - 40320 (1')^6(8')] * t^8 / 8!

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 heaxagon 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)

N. Arkani-Hamed, Y. Bai, S. He, and G. Yan, <a href="https://arxiv.org/abs/1711.09102">Scattering forms and the positive geometry of kinematics, color, and the worldsheet </a>, arXiv:1711.09102 [hep-th], 2017.

A. Frabetti and D. Manchon, <a href="https://arxiv.org/abs/1402.5551">Five interpretations of Fa`a di Bruno’s formula </a>, arXiv:1402.5551 [math.CO], 2014, p. 11.

X. Gao, S. He, and Y. Zhan, <a href="https://arxiv.org/abs/1708.08701">Labelled tree graphs, Feynman diagrams and disk integrals </a>, arXiv:1708.08701 [hep-th], 2017.

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 *)

Terms ordered according to the reversed Amramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 07 2024

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