Touchard polynomials

The value at 1 of the nth Touchard polynomial is the nth Bell number, i.e., the number of partitions of a set of size n:

${\displaystyle T_{n}(1)=B_{n}.}$

If X is a random variable with a Poisson distribution with expected value λ, then its nth moment is E(Xⁿ) = T_n(λ), leading to the definition:

${\displaystyle T_{n}(x)=e^{-x}\sum _{k=0}^{\infty }{\frac {x^{k}k^{n}}{k!}}.}$

Using this fact one can quickly prove that this polynomial sequence is of binomial type, i.e., it satisfies the sequence of identities:

${\displaystyle T_{n}(\lambda +\mu )=\sum _{k=0}^{n}{n \choose k}T_{k}(\lambda )T_{n-k}(\mu ).}$

The Touchard polynomials constitute the only polynomial sequence of binomial type with the coefficient of x equal 1 in every polynomial.

The Touchard polynomials satisfy the Rodrigues-like formula:

${\displaystyle T_{n}\left(e^{x}\right)=e^{-e^{x}}{\frac {d^{n}}{dx^{n}}}\;e^{e^{x}}.}$

The Touchard polynomials satisfy the recurrence relation

${\displaystyle T_{n+1}(x)=x\left(1+{\frac {d}{dx}}\right)T_{n}(x)}$

and

${\displaystyle T_{n+1}(x)=x\sum _{k=0}^{n}{n \choose k}T_{k}(x).}$

In the case x = 1, this reduces to the recurrence formula for the Bell numbers.

Using the umbral notation Tⁿ(x)=T_n(x), these formulas become:

${\displaystyle T_{n}(\lambda +\mu )=\left(T(\lambda )+T(\mu )\right)^{n},}$

${\displaystyle T_{n+1}(x)=x\left(1+T(x)\right)^{n}.}$

The generating function of the Touchard polynomials is

${\displaystyle \sum _{n=0}^{\infty }{T_{n}(x) \over n!}t^{n}=e^{x\left(e^{t}-1\right)},}$

which corresponds to the generating function of Stirling numbers of the second kind.

Touchard polynomials have contour integral representation:

${\displaystyle T_{n}(x)={\frac {n!}{2\pi i}}\oint {\frac {e^{x({e^{t}}-1)}}{t^{n+1}}}\,dt.}$

All zeroes of the Touchard polynomials are real and negative. This fact was observed by L. H. Harper in 1967.[5]

The smallest zero is bounded from below (in absolute value) by[6]

${\displaystyle {\frac {1}{n}}{\binom {n}{2}}+{\frac {n-1}{n}}{\sqrt {{\binom {n}{2}}^{2}-{\frac {2n}{n-1}}\left({\binom {n}{3}}+3{\binom {n}{4}}\right)}},}$

although it is conjectured that the smallest zero grows linearly with the index n.

The Mahler measure ${\displaystyle M(T_{n})}$of the Touchard polynomials can be estimated as follows:[7]

${\displaystyle {\frac {\lbrace \textstyle {n \atop \Omega _{n}}\rbrace }{\binom {n}{\Omega _{n}}}}\leq M(T_{n})\leq {\sqrt {n+1}}\left\{{n \atop K_{n}}\right\},}$

where ${\displaystyle \Omega _{n}}$ and ${\displaystyle K_{n}}$ are the smallest of the maximum two k indices such that ${\displaystyle \lbrace \textstyle {n \atop k}\rbrace /{\binom {n}{k}}}$ and ${\displaystyle \lbrace \textstyle {n \atop k}\rbrace }$ are maximal, respectively.