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A134264 Coefficients, T(j,k), of a partition transform for Lagrange inversion. 6
1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,6

COMMENTS

Matrix begins

1;

1;

1, 1;

1, 3, 1;

1, 4, 2, 6, 1;

1, 5, 5, 10, 10, 10, 1;

1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1;

...

Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal 0, form h(t) = t / f(t) and denote h_n as the coefficient of t^n in h(t).

Lagrange inversion gives the compositional inverse about t=0 as g(t) = sum(j=1,2,...) { t^j * (1/j) * sum(over all permutations with s(1)+s(2)+... +s(j)=j-1) [ h_s(1) * h_s(2) * ... h_s(j) ] } = t * T(1,1) * h_0 + sum(j=2,3,...) { t^j * sum(k=1,..., # of partitions for j-1) [ T(j,k) * H(j-1,k ; h_0,h_1,...) ] }, where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.

Denoting h_n by (n') for brevity, to 5th order in t,

g(t) = t * (0') + t^2 * [ (0') (1') ] + t^3 * [ (0')^2 (2') + (0') (1')^2 ] + t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ] + t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ] + ...

A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.

If the coefficients of partitions with the same number of h_0 are summed within rows, A001263 is obtained, omitting the leading 1.

From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014

LINKS

Table of n, a(n) for n=1..30.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

F. Ardila, F. Rincon, L. Williams, Positroids and non-crossing partitions, arXiv preprint arXiv:1308.2698v2, 2013 (pg. 25).

FORMULA

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.

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.

T(j,k) is [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.

For example from g(t) above, T(5,4) = [ 5! / ((5-3)! * 2!) ] / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.

From Tom Copeland, Sep 30 2011: (Start)

Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}

  = [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}

  = [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.

Then for the partition polynomials of A134264,

  Poly[n;h_0,...,h_(n-1)]=(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)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1. (End)

From Tom Copeland, Oct 20 2011: (Start)

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) = t* [(h_0)+(h_1)d/dt+(h_2)(d/dt)^2+...]^2 / [(h_0)-(h_2)(d/dt)^2-2(h_3)(d/dt)^3-3(h_4)(d/dt)^4+...],  and

L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2+...]

Then  P(n,t) = (t^n/n!) dPS(n,z)/dz  eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)(End)

EXAMPLE

1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).

2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).

3) With f(t) = [ exp(-a*t)-1 ] / (-a), then h(t) = sum(n=0,1,...) Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = sum(n=1,...) a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, { sum(over all permutations with s(1)+s(2)+... +s(j)=j-1) [ h_s(1) * h_s(2) * ... h_s(j) ] } = { j * sum(k=1,..., # of partitions for j-1) [ T(j,k) * H(j-1,k ; h_0,h_1,...) ] } = a^(j-1). Note, in turn, sum(a=1 to m) a^(j-1) = [ Bernoulli(j,m+1) - Bernoulli(j) ] / j for the Bernoulli polynomials and numbers, for j>1.

4) With f(t,x)= t/[x-1+1/(1-t)], then h(t,x)=[x-1+1/(1-t)], giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = {1-(1-x)*t-sqrt[1-2*(1+x)*t+[(x-1)*t]^2]}/2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.

5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)

CROSSREFS

Cf. A145271, (A001263,A119900) = (reduced array, associated g(x)).

Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).

Sequence in context: A188139 A134557 A219842 * A125181 A157076 A049999

Adjacent sequences:  A134261 A134262 A134263 * A134265 A134266 A134267

KEYWORD

nonn,tabf,more,changed

AUTHOR

Tom Copeland, Jan 14 2008

STATUS

approved

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Last modified October 31 12:07 EDT 2014. Contains 248864 sequences.