The total sum of the terms of a sequence defined iteratively. Consider the sequence of vectors defined as follows:
$v_{1} = (1)$, $v_{2} = (2)$, $v_{3} =(2,3)$ and for $i \geq 4$, let $v_{i-1} = (x_{1},\ldots,x_{p})$, then
$$ v_{i} = (2,3,4\ldots, x_{1}+1, 2,3,\ldots,x_{2},2,3\ldots x_{3}+1, \ldots, x_{p}+1)$$
In other words, $v_{i}$ is the concatenation of the vectors $(2,\ldots,x_{1}+1),\ldots,(2,\ldots,x_{p}+1)$. We have $v_{4} = (2,3,2,3,4)$, for example.
Let $S_{m}$ denote the sum of coordinates of $v_{m}$ for $m \geq 1$.
I need to show that  $S_{m} = \frac{1}{m+1} {2m \choose m}$. First I observed that the number of coordinates of $v_{m}$ is $S_{m-1}$. I tried to found recurrences that hold for both, $S_{m}$ and $\frac{1}{m+1} {2m \choose m}$ however I didn't find it. This problem appeared when I was trying to count some structures, then maybe my question is wrong. Any help would be appreciated.
 A: So consider the following numbers
$$A_{n,k}=\text{# of times k is in }v_n.$$ First notice that the maximum element in $v_n$ is $n$ (induction i guess) and denote $l_n$ the length of $v_n,$ then we have first that $$l_n=\sum _{k=2}^nT_{n,k}$$ and also that
$$S_n=\sum _{k=2}^nk\cdot T_{n,k}.$$
As you have notice and is easy to show by construction $l_n=S_{n-1}.$ So we will focus just in the sum. 
Notice also that $$A_{n,k}=\sum _{j=k-1}^{n-1}A_{n-1,j}$$ because in your construction you will go from $2$ to $j+1$ and so every time we see a number greater than $k-1$ we should pass by $k$ in the next generation. Consider now the change of variable $T_{n,k}=A_{n,n-k+1}$ this will allow you to have the recursion
$$T_{n,k}=\sum _{j=1}^{k-1}T_{n-1,j}$$ then a table of this numbers looks like

1 1
2 [1, 1]
3 [1, 2, 2]
4 [1, 3, 5, 5]
5 [1, 4, 9, 14, 14]
6 [1, 5, 14, 28, 42, 42]
7 [1, 6, 20, 48, 90, 132, 132]
8 [1, 7, 27, 75, 165, 297, 429, 429]

I generate this table using your construction:
L =[2]
for k in range(1,8):
        LL=[]
        for n in range(0,len(L)):
                for m in range(2,L[n]+2):
                        LL.append(m)
        LLL = [0]*(k+1)
        for i in range(0,len(LL)):
                LLL[k+2-LL[i]]+=1
        print(k+1,LLL)
        L = LL

It turns out that this numbers are in the OEIS A009766.
There you will see that they are counting some structures with particular properties, and when you add this structures you get a structure counted by Catalan numbers.
