

A134685


Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion .


15



1, 1, 3, 1, 15, 10, 1, 105, 105, 15, 10, 1, 945, 1260, 280, 210, 21, 35, 1
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OFFSET

1,3


COMMENTS

Let f(t) = u(t)  u(0) = Ev[exp(u.* t)  u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[log(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 reference in A127671 and u_n = (n1)! * 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 ((m1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n1 listed in Abramowitz and Stegun on page 831 and the number of partitions in P(n,t) is given by A000041(n1).
Combinatorial interpretations are given in the link.


REFERENCES

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


LINKS

Table of n, a(n) for n=1..19.
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)
T. Copeland, Lagrange a la Lah
E. Getzler, The semiclassical approximation for modular operads (see pg. 2)


FORMULA

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)^(n1+e(1)) * [2*(n1)e(1)]! / [2!^e(2)*e(2)!*3!^e(3)*e(3)! ... n!^e(n)*e(n)! ].
From Tom Copeland, Sep 05 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u.*exp(u.*t)]
= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),
an e.g.f. for the partition polynomials of A133314
(signed A049019) with an index shift.
Then for the partition polynomials of A134685,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (Cf. A000311 and A134991)(End)
From Tom Copeland, Oct 30 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*h(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*h(d/dt) = t* 1/[u_1+(u_2)*d/dt+(u_3)*(d/dt)^2/2!+...], and
L =f(d/dt)=(u_1)*d/dt+(u_2)*(d/dt)^2/2!+(u_3)*(d/dt)^3/3!+....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n1)/n!]^n evaluated at x=0.  Tom Copeland, Jul 07 2015


EXAMPLE

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)2t1 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 = (n1)! for n>1, then f(t) =
log(1t)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 (n2)! for n>1, then
f(t) = (1+t) log(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(1t/2) and g(t) = [e.g.f. of (0,A001147)] = 1  (12t)^(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(1t) and g(t) = t * [o.g.f. of A000108] = [1  (14t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) =
(0,P(1,1),P(2,1),P(3,1),...) .


CROSSREFS

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).
Sequence in context: A039815 A147453 A147020 * A130757 A181996 A014621
Adjacent sequences: A134682 A134683 A134684 * A134686 A134687 A134688


KEYWORD

sign,tabf,more


AUTHOR

Tom Copeland, Jan 26 2008, Sep 13 2008


STATUS

approved



