1,6

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 5-th 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, apparently A001263 is obtained, omitting the leading 1.

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

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 4-th partition under n=5-1=4 with m=3 parts in A&S.

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) = ln(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.

Sequence in context: A023579 A023577 A134557 this_sequence A125181 A049999 A126015

Adjacent sequences: A134261 A134262 A134263 this_sequence A134265 A134266 A134267

nonn

Tom Copeland (tcjpn(AT)msn.com), Jan 14 2008