# Original Question

Consider an infinite tree of constant degree $k$. For such a tree we can consider the total number of nodes at depth $n$, $g(f)$, and the total number of paths from the root, $p(f)$, to be a function of the constant function, $f=k$. We define $G(f)$ to be the resulting infinite tree. Now let us generalize this idea to functions $f(n)$, with the normal convention that the root has depth $n=0$.

Some examples:

$g(1)=1$

$p(1)=n+1$

$G(1)$ is the infinite tree (path) of constant degree 1

$g(2) = 2^n$

$p(2) = 2^{n+1}-1$

$G(2)$ is the infinite complete binary tree

$g(a) = a^n$

$p(a) = \frac{a^{n+1}-1}{(a-1)}$ for a>1

$G(a)$ is the infinite complete tree of constant degree a

$g(f=n) = n!$

$p(f=n) = !n$

$G(f=n)$ is the infinite complete tree of incremental degree

$g(f=n+1)=(n+1)!$

$g(f=2n)=2n!!$

where !! is the double-factorial

$g(f=3n)=3n!!!$

where !!! is the triple-factorial

$g(f=n^2) = n!^2$

$g(f=n^a) = n!^a$

$g(f=an^b) = a^{n}n!^b$ ; Sequences not in Sloan for a>1

$g(f=n^2+n+1)$ = ? ; Related to absolute values for Sloan A130031

$p(f=n^2+n+1)$ = ? ; Sequence: [1, 2, 7, 62, 1107, 31412, 1273917, ... ]

$g(f=2^n)=2^{((n+1)^2-(n+1))/2}$

$p(f=2^n)= ?$ ; Sequence: [1,3,11,75,1099,33867, .... ]

$g(f=a^n)=a^{((n+1)^2-(n+1))/2}$

$g (f=n!)$ = Sloan A000178

$g (f=2n!)$ = ? ; Related to Sloan A156926. Sequence: [1,2,8,96,4608,1105920,....]

$p (f=2n!)$ = ? ; Sequence: [1,3,11,107,4715,....]

$g (f=3n!)$ = ? ; Sequence: [1,3,18,324,23328,8398080,....]

$p (f=3n!)$ = ? ; Sequence: [1,4,22,346, 23674, 8421754,....]

$g(f=a^{a^n})= a^{a^{n+1}-a(3^n + 1)/2}$

My questions are: is this graph construction well known? I would be interested in any references to similar functional transformations on graphs.

Also, could anyone tell me what is the cardinality of the set of all paths in G(f=n)? Clearly it has at least Continuum cardinality. Since the factorial grows faster than $a^n$, yet slower than $2^{2^n}$, I would think it lands in the Continuum. I am not sure, though...

# Addendum

We can express p and g as functional equations:

$g(f(n))=f(n) g(f(n-1))$

$g(f(0))=1$

$p(f(n))=\sum_{i=1}^{n} g(f(i))$

If we extend to the complex domain and consider the special case f(z)=z we have the functional equation:

$g(z)= z g(z-1)$

Which has the rather well known solution $g(z) = \Gamma (z+1)$