Is there even a formula? 🤔

Let's just look at the left half of the line. So there's a $2$ on the left and a $1$ on the right. Every value written down in between is going to be an integer linear combination of these boundary values. Let's call them $\alpha$ and $\beta$ instead of $2$ and $1$ for now.

The first value we write down is $\alpha + \beta.$ The next two are $2\alpha + \beta$ and $\alpha + 2\beta.$ The next four are $3\alpha + \beta, 3\alpha + 2\beta, 2\alpha + 3\beta, \alpha + 3\beta.$ And so on. If we represent $m\alpha + n\beta$ as $\begin{bmatrix} m\\n \end{bmatrix},$ then successive steps yield $\begin{bmatrix} 1\\1 \end{bmatrix}, \\ \begin{bmatrix} 2\\1 \end{bmatrix}, \begin{bmatrix} 1\\2 \end{bmatrix}, \\ \begin{bmatrix} 3\\1 \end{bmatrix}, \begin{bmatrix} 3\\2 \end{bmatrix}, \begin{bmatrix} 2\\3 \end{bmatrix}, \begin{bmatrix} 1\\3 \end{bmatrix}.$ This is exactly how the terms of the Farey sequence are generated. From two adjacent fractions $\frac{m_1}{n_1}, \frac{m_2}{n_2},$ we generate $\frac{m_1+m_2}{n_1+n_2}.$ It is well-known that iterating this process will generate every fraction in lowest terms exactly once.

Hence the number of $100$ s will be the number of ways to write $m\alpha + n\beta = 100$ where $m,n$ are relatively prime. Let's substitute back now: $2m+n=100.$ Now the general answer will follow from a lemma:

Lemma: If $N \ge 3,$ the number of solutions in nonnegative coprime integers $m,n$ to the equation $2m+n = N$ is $\phi(N)/2.$

Proof: There must be a fancy way to do this, but I don't see it. Take cases. If $N$ is odd, then every positive integer $m < N/2$ that is coprime to $N$ will produce one solution. There are $\phi(N)/2$ of these. If $N$ is even, then $n=2k$ and we get $m+k=N/2$ with $m$ odd. Then if $N/2$ is odd, then $k$ is even and we are in the previous case, so there are $\phi(N/2)/2$ solutions, which equals $\phi(N)/2.$ If $N/2$ is even, then every positive integer $m \le N$ that is coprime to $N/2$ will produce a solution, so the answer is $\phi(N/2) = \phi(N)/2.$

At any rate, the lemma shows that the number of $N$ s on the left half of the line is $\phi(N)/2,$ so the total number of $N$ s is twice that, or $\phi(N).$ For $N=100$ this gives $\fbox{40}.$

At each stage, Billy and Tom are creating successive rows in what is called the Steiner Diatomic Array. This paper records a result that the number $n$ occurs $\phi(n)$ times in the $n$ th row of the SDA, where $\phi$ is the Euler totient function. Since every element of the SDA (apart from the $1$ s in the first row) are obtained by adding two positive integers together, any new number (one that did not appear in the row above) in the $(n+1)$ st or later row will be larger than $n$ . Thus the last time that $n$ will be added to the SDA will be in the $n$ th row.

Thus we want to know the number of times $100$ occurs in the $100$ th row, which is $\phi(100) = \boxed{40}$ .

here is a Mathematica algorithm that finds the nth step

for example if p=6 we get the 6th step (just change the value of p to find the step you want)

p=6;s={2,1,2};z=0;While[z<p-1,t=Tr/@Partition[s,2,1];s=Riffle[s,t];z++];Print@s

{2,11,9,16,7,19,12,17,5,18,13,21,8,19,11,14,3,13,10,17,7,18,11,15,4,13,9,14,5,11,6,7,1,7,6,11,5,14,9,13,4,15,11,18,7,17,10,13,3,14,11,19,8,21,13,18,5,17,12,19,7,16,9,11,2}

to solve this puzzle we use the above algorithm, raising the value of "limit" until we get the right result

limit = 40; s = {2, 1, 2}; While[Count[s, 100] < limit, t = Tr /@ Partition[s, 2, 1]; s = Riffle[s, t]] Count[s, 100]

here is another solution

FixedPoint[Count[Riffle[#,Tr/@Partition[#,2,1]]&,100],{2,1,2}]