Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n-1)$ fixed and reversing every n consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ (A057030).

The sequence begins $$1, 3, 4, 6, 11, 13, 14, 22, 27, 29, 40, 42, 47, 55, 66$$

Let $$b(n,m)=(m-n)\left\lfloor\frac{b(n-1,m)}{m-n}+1\right\rfloor - b(n-1,m)\operatorname{mod}(m-n), b(0,m)=m$$

I conjecture that $$a(n)=b(n-1,n)$$

In other words, we need $n-1$ iterations to get $a(n)$ starting from $n$, for example:

$$5-7-8-10-11$$ $$10-17-23-26-28-27-25-26-28-29$$ $$15-27-38-46-53-57-60-60-59-55-60-64-65-65-66$$

Is there a way to prove it?

Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n-1)$ fixed and reversing every n consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ (A057030).

The sequence begins $$1, 3, 4, 6, 11, 13, 14, 22, 27, 29, 40, 42, 47, 55, 66$$

Let $$b(n,m)=b(n-1,m) + (m-n) - 2(b(n-1,m)\operatorname{mod}(m-n)), b(0,m)=m$$

I conjecture that $$a(n)=b(n-1,n)$$

In other words, we need $n-1$ iterations to get $a(n)$ starting from $n$, for example:

$$5-7-8-10-11$$ $$10-17-23-26-28-27-25-26-28-29$$ $$15-27-38-46-53-57-60-60-59-55-60-64-65-65-66$$

Is there a way to prove it?

Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n-1)$ fixed and reversing every n consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ (A057030).

The sequence begins $$1, 3, 4, 6, 11, 13, 14, 22, 27, 29, 40, 42, 47, 55, 66$$

Let $$b(n,m)=(m-n)\left\lfloor\frac{b(n-1,m)}{m-n}+1\right\rfloor - b(n-1,m)\operatorname{mod}(m-n), b(0,m)=m$$

I conjecture that $$a(n)=b(n-1,n)$$

Is there a way to prove it?

Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n-1)$ fixed and reversing every n consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ (A057030).

The sequence begins $$1, 3, 4, 6, 11, 13, 14, 22, 27, 29, 40, 42, 47, 55, 66$$

Let $$b(n,m)=(m-n)\left\lfloor\frac{b(n-1,m)}{m-n}+1\right\rfloor - b(n-1,m)\operatorname{mod}(m-n), b(0,m)=m$$

I conjecture that $$a(n)=b(n-1,n)$$

In other words, we need $n-1$ iterations to get $a(n)$ starting from $n$, for example:

$$5-7-8-10-11$$ $$10-17-23-26-28-27-25-26-28-29$$ $$15-27-38-46-53-57-60-60-59-55-60-64-65-65-66$$

Is there a way to prove it?