

A134264


Coefficients, T(j,k), of a partition transform for Lagrange inversion.


5



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
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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)=j1) [ 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 j1) [ T(j,k) * H(j1,k ; h_0,h_1,...) ] }, where H(j1,k ; h_0,h_1,...) is the kth partition for h_1 through h_(j1) corresponding to n=j1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(jm)=(h_0)^(n+1m) 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.


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].


FORMULA

For j>1, there are P(j,m;a...) = j! / [ (jm)! (a_1)! (a_2)! ... (a_(j1))! ] permutations of h_0 through h_(j1) in which h_0 is repeated (jm) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j1) * a_(j1) = j1, then each distinct combination of these arrangements is correlated with a partition of j1.
T(j,k) is [ P(j,m;a...) / j ] for the kth partition of j1 as described in the comments.
For example from g(t) above, T(5,4) = [ 5! / ((53)! * 2!) ] / 5 = 6 for the 4th partition under n=51=4 with m=3 parts in A&S.
Contribution from Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)1+:1/(1h.*x):]^2 / {(h_0):[h.x/(1h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)(h_2)x^22(h_3)x^33(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_(n1)]=(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)
Contribution 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(n1,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)^22(h_3)(d/dt)^33(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 / (t1), then h(t) = (1t), giving h_0 = 1, h_1 = 1 and h_n = 0 for n>1 . Then g(t) = t  t^2  t^3  ... = t / (t1).
2) With f(t) = t*(1t), then h(t) = 1 / (1t), 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(j1).
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) = ln(1a*t) / (a) = sum(n=1,...) a^(n1) * t^n / n. Therefore with h_n = Bernoulli(n) * (a)^n / n!, { sum(over all permutations with s(1)+s(2)+... +s(j)=j1) [ h_s(1) * h_s(2) * ... h_s(j) ] } = { j * sum(k=1,..., # of partitions for j1) [ T(j,k) * H(j1,k ; h_0,h_1,...) ] } = a^(j1). Note, in turn, sum(a=1 to m) a^(j1) = [ Bernoulli(j,m+1)  Bernoulli(j) ] / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x)= t/[x1+1/(1t)], then h(t,x)=[x1+1/(1t)], giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = {1(1x)*tsqrt[12*(1+x)*t+[(x1)*t]^2]}/2 , a shifted o.g.f. in t for the Narayana polynomials in x of A001263.


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


AUTHOR

Tom Copeland, Jan 14 2008


STATUS

approved



