1,3

Row n contains partition(n) = A000041(n) terms, which are arranged in the standard Abramowitz and Stegun order.

Let f(t) = u(t) - u(0) = U. exp(u.* t) - u(0) = ln[U.(exp(z.* t))/z_0] = U. -ln(1- a.* t), where the operator U. denotes umbral evaluation of the umbral variables u., z. or a. . 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(n,t) where, with u_n denoted by (n'),

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!

...

The C(j,k) are obtained by evaluating h(t) = t / (u(t) - u(0)) in terms of the u_n using the list partition transform of A133314 and then operating on h(t) with the partition transform for indirect Lagrange inversion in A134264.

Conjectures for general formulae for n>1:

1: 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 length of each row of [C(j,k)] is given by A000041.

2: Each partition of P(n,t) in the brackets corresponds to a unique partition of 2(n-1) into n-1 parts.

3: Equivalently, the partitions of P(n,t) with component (1')^m satisfy m+e(2)+e(3)+...+e(n) = n-1 and m+2e(2)+3e(3)+...+n*e(n) = 2(n-1) with e(k) denoting the exponent of (k').

4: The sign of C(n,k) for each partition for P(n,t) associated to (1')^m is that of (-1)^(n+m-1).

5: The magnitude of C(n,k) for each partition [ (1')^m (2')^e(2)...(n')^e(n) ] corresponds to the # of ways [ 2e(2)+3e(3)+...+n*e(n) ] labeled items can be separated without regard to order into e(2) groups of size 2, e(3) groups of size 3,..., and e(n) groups of size n. E.g., C(5,4), from C(5,4) (1')^2 (3')^2 = 10 (1')^2 (3')^2, corresponds to the # of ways 6 items may be grouped into two groups of equal size, whereas C(5,3) from C(5,3) (1')^2 (2') (4') = 15 (1')^2 (2') (4') corresponds to the # of ways 6 items may be distributed into two groups of size 2 and 4.

6: Summing over the coefficients of the partitions of P(n,t) for fixed (1')^m gives the associated Stirling number of the second kind, S[2(n-1)-m,n-1-m] (see Comtet ref. in A008299). E.g., for P(5,t) for m=2, C(5,3)+C(5,4) = 25 = S(6,2), and P(4,t) = (1')^(-7) * [ -S(6,3) (2')^3 + S(5,2) (1')(2')(3') - S(4,1) (1')^2 (4') ] * t^4 / 4! . So, the unsigned C(n,k) are a refinement of A008299.

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

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].

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: A134144 A035342 A039815 this_sequence A130757 A014621 A113378

Adjacent sequences: A134682 A134683 A134684 this_sequence A134686 A134687 A134688

sign,uned,tabf

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