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%I #58 Feb 23 2023 18:55:25
%S 1,1,2,1,6,9,1,24,72,12,16,1,120,600,300,200,50,25,1,720,5400,5400,
%T 2400,450,1800,450,60,90,36,1,5040,52920,88200,29400,22050,44100,7350,
%U 4410,2940,4410,882,245,147,49,1,40320,564480,1411200,376320,705600,940800
%N Triangle read by rows: T(n,k) are the coefficients of the Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its reciprocal, n >= 1, k = 1..A000041(n-1).
%C Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
%C Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series. A refinement of A141618 with row sums A000272.
%C Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
%C g(t)/t = [ 1 (0') ]
%C + [ 1 (0') (1') ] * t
%C + [ 2 (0') (1')^2 + 1 (0')^2 (2') ] * t^2/2!
%C + [ 6 (0') (1')^3 + 9 (0')^2 (1') (2') + 1 (0')^3 (3') ] * t^3/3!
%C + [24 (0') (1')^4 + 72 (0')^2 (1')^2 (2') + (0')^3 [12 (2')^2
%C + 16 (1') (3')] + (0')^4 (4')] * t^4/4!
%C + [120 (0')(1')^5 + 600 (0')^2 (1')^3(2') + (0')^3 [300 (1')(2')^2 + 200 ( 1')^2(3')] + (0')^4 [50 (2')(3') + 25 (1')(4')] + (0')^5 (5')] * t^5/5! + [720 (0')(1')^6 + (0')^2 (1')^4(2')+(0')^3 [5400 (1')^2(2')^2 + 2400 (1')^3(3')] + (0')^4 [450 (2')^3+ 1800 (1')(2')(3') + 450( 1')^2(4')]+ (0')^5 [60 (3')^2 + 90 (2')(4') + 36 (1')(5')] + (0')^6 (6')] * t^6/6! + ...
%C ..........
%C From _Tom Copeland_, Oct 28 2014: (Start)
%C Expressing g(t) as a Taylor series or formal e.g.f. in the indeterminates h_n generates a refinement of A055302, which enumerates the number of labeled root trees with n nodes and k leaves, with row sum A000169.
%C Operating with (1/n^2) d/d(1') = (1/n^2) d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
%C Multiplying the n-th partition polynomial here by (n + 1) gives the (n + 1)-th partition polynomial of A248120. (End)
%C These are also the coefficients in the expansion of a series related to the Lagrange reversion theorem presented in Wikipedia of which the Lagrange inversion formula about the origin is a special case. Cf. Copeland link. - _Tom Copeland_, Nov 01 2016
%H Andrew Howroyd, <a href="/A248927/b248927.txt">Table of n, a(n) for n = 1..2087</a> (rows 1..20)
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Tom Copeland, <a href="http://tcjpn.wordpress.com/2016/11/01/the-lagrange-reversion-theorem-and-the-lagrange-inversion-formula/">The Lagrange Reversion Theorem and the Lagrange Inversion Formula</a>
%F 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.
%F 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.
%F T(j,k) is [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
%F For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.
%F If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.
%F For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).
%F Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (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)). See A145271.
%F 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) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
%F Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
%F As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - _Tom Copeland_, Nov 25 2016
%F T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - _Andrew Howroyd_, Feb 02 2022
%e Triangle T(n,k) begins:
%e 1;
%e 1;
%e 2, 1;
%e 6, 9, 1;
%e 24, 72, 12, 16, 1;
%e 120, 600, 300, 200, 50, 25, 1;
%e 720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1;
%e ...
%e For f(t) = e^t-1, h(t) = t/f(t) = t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t) = t - t^2/2 + t^3/3 + ... = log(1+t).
%o (PARI)
%o C(v)={my(n=vecsum(v), S=Set(v)); n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
%o row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
%o { for(n=1, 7, print(row(n))) } \\ _Andrew Howroyd_, Feb 02 2022
%Y Cf. A145271
%Y Cf. A134264 and A248120, "scaled" versions of this Lagrange inversion.
%Y Cf. A036038.
%K nonn,tabf
%O 1,3
%A _Tom Copeland_, Oct 16 2014
%E Name edited and terms a(31) and beyond from _Andrew Howroyd_, Feb 02 2022
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