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

%I A133932 #87 Mar 21 2024 21:19:09
%S A133932 1,-1,3,-2,-15,20,-6,105,-210,40,90,-24,-945,2520,-1120,-1260,420,504,
%T A133932 -120,10395,-34650,25200,18900,-2240,-15120,-9072,1260,2688,3360,-720,
%U A133932 -135135,540540,-554400,-311850,123200,415800,166320,-50400,-56700,-120960,-75600,18144,20160,25920,-5040
%N A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.
%C A133932 Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
%C A133932 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'),
%C A133932 P(1,t) = (1')^(-1) * [ 1 ] * t
%C A133932 P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
%C A133932 P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
%C A133932 P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
%C A133932 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!
%C A133932 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!
%C A133932 See A134685 for more information.
%C A133932 From _Tom Copeland_, Sep 28 2016: (Start)
%C A133932 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!
%C A133932 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)
%H A133932 T. Copeland, <a href="https://tcjpn.wordpress.com/2011/04/11/lagrange-a-la-lah/">Lagrange a la Lah</a>, 2011.
%H A133932 T. Copeland, <a href="https://tcjpn.wordpress.com/2014/09/17/compositional-inverse-pairs-the-inviscid-burgers-hopf-equation-and-the-stasheff-associahedra/">Compositional inverse pairs, the Burgers-Hopf equation, and the Stasheff associahedra</a>, 2014.
%H A133932 T. Copeland, <a href="http://tcjpn.wordpress.com/2015/12/21/generators-inversion-and-matrix-binomial-and-integral-transforms/">Generators, Inversion, and Matrix, Binomial, and Integral Transforms</a>, 2015.
%H A133932 T. Copeland, <a href="https://tcjpn.wordpress.com/2018/01/23/formal-group-laws-and-binomial-sheffer-sequences/">Formal group laws and binomial Sheffer sequences</a>, 2018.
%H A133932 Jin Wang, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Wang/wang53.html">Nonlinear Inverse Relations for Bell Polynomials via the Lagrange Inversion Formula</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.8.
%F A133932 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))! ].
%F A133932 From _Tom Copeland_, Sep 06 2011: (Start)
%F A133932 Let h(t) = 1/(df(t)/dt)
%F A133932   = 1/Ev[u./(1-u.t)]
%F A133932   = 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
%F A133932   where Ev denotes umbral evaluation.
%F A133932 Then for the partition polynomials of A133932,
%F A133932   n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
%F A133932   and the compositional inverse of f(t) is
%F A133932   g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
%F A133932   Also, dg(t)/dt = h(g(t)). (End)
%F A133932 From _Tom Copeland_, Oct 20 2011: (Start)
%F A133932 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
%F A133932 R = t*h(d/dt) = t* 1/[(u_1) + (u_2)*d/dt + (u_3)*(d/dt)^2 + ...], and
%F A133932 L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2/2 + (u_3)*(d/dt)^3/3 + ....
%F A133932 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)
%F A133932 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
%F A133932 From _Tom Copeland_, Sep 19 2016: (Start)
%F A133932 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.
%F A133932 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).
%F A133932 Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
%F A133932 From _Tom Copeland_, Sep 20 2016: (Start)
%F A133932 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.
%F A133932 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.
%F A133932 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).
%F A133932 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].
%F A133932 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}.
%F A133932 (End)
%F A133932 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
%e A133932 From _Tom Copeland_, Sep 18 2014: (Start)
%e A133932 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.
%e A133932 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.
%e A133932 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)
%t A133932 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]]]}]];
%t A133932 rows[7] // Flatten (* _Andrey Zabolotskiy_, Mar 08 2024 *)
%Y A133932 Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).
%Y A133932 Cf. A008292, A086810, A094587, A133314, A134685, A139605, A263633.
%Y A133932 Cf. A036039, A133437.
%K A133932 sign,tabf
%O A133932 1,3
%A A133932 _Tom Copeland_, Jan 27 2008
%E A133932 Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by _Andrey Zabolotskiy_, Mar 08 2024