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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. 9
 1, -1, 3, -2, -15, 20, -6, 105, -210, 90, 40, -24, -945, 2520, -1120, -1260, 504, 420, -120, 10395, -34650, 25200, 18900, -15120, -2240, -9072, 3360, 2688, 1260, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -56700, -120960, -75600, -50400, 25920, 20160, 18144, -5040 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 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') + 90 (1')^2 (2') (4') + 40 (1')^2 (3')^2 - 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') + 504 (1')^3 (2')(5') + 420 (1')^3 (3')(4') - 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 [15120 (2')(3')(4') + 2240 (3')^3 + 9072 (2')^2(5')] + (1')^4 [3360 (2')(6') + 2688 (3')(5') + 1260 (4')^2] - 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 [56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6') + 50400 (3')^2(4')] + (1')^5 [25920 (2')(7') + 20160 (3')(6') + 18144 (4')(5')] - 5040 (1')^6(8')] * t^8 / 8! (End) LINKS T. Copeland, Lagrange a la Lah, 2011. T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015. T. Copeland, Formal group laws and binomial Sheffer sequences, 2018. Jin Wang, Nonlinear Inverse Relations for Bell Polynomials via the Lagrange Inversion Formula, J. Int. Seq., Vol. 22 (2019), Article 19.3.8. 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,...,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 EXAMPLE 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)). Cf. A008292, A086810, A094587, A133314, A134685, A139605, A263633. Cf. A036039. Sequence in context: A302845 A291251 A223523 * A111999 A286947 A190961 Adjacent sequences:  A133929 A133930 A133931 * A133933 A133934 A133935 KEYWORD sign AUTHOR Tom Copeland, Jan 27 2008 STATUS approved

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Last modified November 17 05:27 EST 2019. Contains 329217 sequences. (Running on oeis4.)