1

Consider the mapping-graphs given by $\mathcal{M}=Cycle(\mathcal{T})^{[∗]}$ (I'm not sure if this is read as $(Cycle(T))^{[∗]}$ or as $Cycle((T)^{[∗]})$? Which one does make more sense?), whereby $\mathcal{T}$ is the family of trees $\mathcal{T}={O}∗SET(\mathcal{T})={O}∗(\mathcal{T}^{[∗]})$. I'd like to determine the formula for the exponential generating function M(z) with the symbolic method and calculate the number of mapping-graphs of size n.

I know that the corresponding equation for $T$ is $T(z)=z \exp(T(z))$, but I don't know how to combine this with the symbolic formula of $\mathcal{M}$, because it seems to me as if that expression would be quite complicated. So I'd appreciate any help to get me on the right track!

RobPratt
  • 45,619
Studentu
  • 599

1 Answers1

1

I know this is old, but...

We have $$ \mathcal M = SET\left(CYC\left(\mathcal T\right)\right) $$ so $$ M(z) = \exp\left(\log\frac{1}{1-T(z)}\right) = \frac{1}{1-T(z)} = \frac{1}{1-ze^{T(z)}} $$ The GF for $M$ involves the Lambert $W$ function, but we can find the coefficients using Lagrange inversion. Note that $T=ze^T$ implies $\frac{1}{1-ze^T} = zT'/T$. With $[z^n]$ meaning the coefficient at $z^n$, we get $$ [z^n]M(z) = [z^n]z\frac{T'}{T} = [z^{n-1}]\frac{T'}{T} = [z^{n-1}](\log T)' = n[z^n]\log T $$ Now Lagrange inversion / the Lagrange-Bürmann formula gives $$ [z^n]M(z) = [u^{n-1}]\frac{1}{u}e^{nu} = [u^n]e^{nu} = \frac{n^n}{n!} $$ Et voilà(!), the number of mappings from $n$ elements to $n$ elements is equal to $n^n$. (What a waste of energy to get that answer, huh?)

Milten
  • 7,031