Return to Answer

Consider the following set of conditions. Before one swaps $b(k+1)$,

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 $k$.)
2. For $i$ larger than $k$, $b(i)$ is at most $i$.
3. $b(i)$ 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 $O(k^{3/2})$.

Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.

Consider the following set of conditions. Before one swaps $b(k+1)$,

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

Consider the following set of conditions. Before one swaps $b(k+1)$,

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 $k$.)
2. For $i$ larger than $k$, $b(i)$ is at most $i$.
3. $b(i)$ 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 $O(k^{3/2})$.

Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.

Consider the following set of conditions. Before one swaps $b(k+1)$,

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 $k$.)
2. For $i$ larger than $k$, $b(i)$ is at most $i$.
3. $b(i)$ 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 $O(k^{3/2})$.

Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.

Consider the following set of conditions. Before one swaps b(k+1),

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.)
2. For i larger than k, b(i) is at most i.
3. 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 kth triangular number T(k)=b(k)+b(1)+b(2)+...+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}).

Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.

Consider the following set of conditions. Before one swaps $b(k+1)$,

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 $k$.)
2. For $i$ larger than $k$, $b(i)$ is at most $i$.
3. $b(i)$ 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 $O(k^{3/2})$.

Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.