Bumping Series and its Formula I was thinking of random series, popping up in mind, when I thought of one possible series in my head. It is as follows:

The basic idea is, take a line of natural numbers $\mathbb{N}$ which goes till infinity, and add them. One visible thing here is that the most maximum highest number $\mathbb{N}_{max}$ would be $\mathbb{N}_{i}$. In basic words, if we go till number 5, $\mathbb{N}_5$ then the height it reaches by summation is 5.
Further, carrying on, we can get:

The basic implication here is that we bump the numbers by specific $\mathbb{N}$. At start, we take the beginning number, in our case it is 1, we move once up and then once down. Then we do it twice, thrice and so on. So 1 3 2 according to my diagram is one bump. At the ending $\mathbb{N}$ which is 2 here, we will jump it by 2 and make it low by 2. So it gets 2 5 12 7 4. Here, assume $\mathbb{N}_i$ as the number of incrementation, before it was 1, now it is 2. We get different sets, with different terms, but total number of terms we get through this would be $2 \mathbb{N}_i + 1$. Now, it will start from 4, go on making 3 jumps before landing by three terms. By this, we get series highlighted by circles in that triangular array as:
1, 3, 2, 5, 12, 7, 4, 9, 20, 44, 24, 13, 7, 15, 32, 68, 144, 76, 40, 21, 11, 23, 48, 100, 208, 432, 224, 116, 60, 31, 16...

The series seem to be divergent, my specific question here is how to represent this series in Mathematical terms. Can there be a formula to guess the next number or any probability test?

The $m$-th term of the $n$-th line of that "triangle" is $u_{n,m} = \sum_{k=0}^n (m+k){n\choose k} = 2^{n-1}(2m+n)$ (you can check this by induction).
Then all that is left is writing $m$ and $n$ in terms of the $k$-th number of the series so that $v_k = u_{n(k), m(k)}$.
