1,3

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = ln{Ev[(exp(z.* t))/z_0]} = Ev[-ln(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in ref. in A127671 and u_n = (n-1)! * a_n .

If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = sum(j=1,...) P(j,t) where, with u_n denoted by (n') for brevity,

P(1,t) = (1')^(-1) * [ 1 ] * t

P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!

P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!

P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!

P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 15 (1')^2 (2') (4') + 10 (1')^2 (3')^2 - (1')^3 (5') ] * t^5 / 5!

P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 21 (1')^3 (2')(5') + 35 (1')^3 (3')(4') - (1')^4 (6') ] * t^6 / 6!

...

Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 and the number of partitions in P(n,t) is given by A000041(n-1).

Combinatorial interpretations are given in the link.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.

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].

T. Copeland, Short Note on Lagrange Inversion, (posted Sept 2008)

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)! ].

Examples and checks:

1) Let u_1 = -1 and u_n = 1 for n>1, then f(t) = exp(u.*

t) - u(0) = exp(t)-2t-1 and g(t) = [e.g.f. of signed A000311];

therefore the row sums of unsigned [C(j,k)] are A000311 =

(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...) .

2) Let u_1 = -1 and u_n = (n-1)! for n>1, then f(t) =

-ln(1-t)-2t and g(t) = [e.g.f. of signed (0,A032188)] with (0,A032188)

= (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...) .

3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then

f(t) = (1+t) ln(1+t) - 2t and g(t) = [e.g.f. of signed (0,A074059)]

with (0,A074059) = (0,1,1,2,7,34,213,...) =

(0,-P(1,1),2!*P(2,1),-3!*P(3,1),...) .

4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2, then f(t)

= t(1-t/2) and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)

with (0,A001147) = (0,1,1,3,15,105,945...) =

(0,P(1,1),2!*P(2,1),3!*P(3,1),...) .

5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2, then f(t)

= t(1-t) and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2

with (0,A000108) = (0,1,1,2,5,14,42,...) =

(0,P(1,1),P(2,1),P(3,1),...) .

Sequence in context: A039815 A147453 A147020 this_sequence A130757 A014621 A144006

Adjacent sequences: A134682 A134683 A134684 this_sequence A134686 A134687 A134688

sign,tabf,more

Tom Copeland (tcjpn(AT)msn.com), Jan 26 2008, Sep 13 2008