A sequence reminiscent of Fibonacci's recursion The sequence in question is A296768 in the Online Encyclopedia. It starts with
1, 3, 5, 9, 11, 17, 24, 32, 36, 46, ...
It is obtained by starting with the positive integers in order, (b(i)= i for all positive integers i),and permuting them again and again. On the k-th pass, we exchange b(k+1) with b(b(k+1) + b(k)). The sequence is the limit of these sequences as k goes to infinity.
Is it true that the number 2 gets carried off to infinity, and does not appear in the final sequence? This seems to be the case, based on the first 67 values. More generally, is it true that this sequence is increasing? If it is increasing, what is its growth rate?
 A: Consider the following set of conditions.  Before one swaps $b(k+1)$,

*

*$b(i)$ is an increasing with $i$  sequence for $i$ less than $k+1$. (And for large enough $k$, $b(k)$ is greater than $k$.)

*For $i$ larger than $k$, $b(i)$ is at most $i$.

*$b(i)$ is less than $i$ exactly when $i$ is greater than $k$ and $i$ is part of the increasing sequence ($i$ is one of $b(1)$ or $b(2)$ or ... or $b(k)$).

Thus $b(k+1)$ is a value at most $k+1$ which gets swapped with the value $b(k)+b(k+1)=b(b(k)+b(k+1))$.
We see the conditions 1,2, and 3 hold for $k+1$ in place of $k$ after the swap.
As a result, $b(k)$ stabilizes into an increasing sequence.  It seems to be the case that  the $k$th triangular number $T(k)=b(k)+b(1)+b(2)+\cdots+b(j)$ with $b(j)$ largest value less than or equal to $k$, or something close to that.  If this is true, then $b(k)$ grows like $k(k+1)/2$ minus a term like $O(k^{3/2})$.
Edit 2020.08.05
Here is an alternate definition of the sequence. Define $pr(k)$ to be $b(i)$ when $k$ equals $b(i+1)$, otherwise $pr(k)$ is 0. Then
$$b(k+1)=b(k)+k+1-pr(k+1).$$
Of course $i$ is positive and $b(1)=1$, or define $b(0)$ to be 0 to take one step back.
End Edit 2020.08.05.
Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.
