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)! ].
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.
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(n-1,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)^(n-1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n-1)/n!]^n evaluated at x=0. -
Tom Copeland, Jul 07 2015
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix
A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from
A145271 and
A133314. -
Tom Copeland, Jul 15 2016
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with
A139605. -
Tom Copeland, Aug 27 2016
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. link and
A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2,..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|_{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|_{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the Bell polynomials of
A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." -
Tom Copeland, Feb 06 2018
