How to prove these two sequences (from "rounding up" and "sieving") are identical?

Question

(This is related to my recent question about "rounding up to $\pi$".)

Definition ("rounding up" sequence $(a_n)_{n=1,2,3,\dots}$):

Starting with positive integer $n$, round up to the nearest multiple of $n−1$, then up to the nearest multiple of $n−2$, etc., up to the nearest multiple of $1$. Let $a_n$ denote the result.

E.g., for $a_5$ we find $5 \xrightarrow{4} 8 \xrightarrow{3} 9 \xrightarrow{2} 10 \xrightarrow{1} 10 =:a_{5}.$ The sequence is found to begin as $(1, 2, 4, 6, 10, 12, 18, 22, 30, 34, ...)$ .

Definition ("sieving" sequence $(b_n)_{n=1,2,3,\dots}$):

Starting with the sequence of positive integers, seq.$1$ $=(1,2,3,\dots),$ let seq.$(i+1)$ be the result of deleting every $(i+1)$st element from seq.$i$, for $i=1,2,3,\dots.$ Let $b_n$ denote the first element of the $n$th sequence.

seq.1:  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ...
seq.2:    2   4   6   8   10    12    14    16    18    20    22 ...
seq.3:        4   6       10    12          16    18          22 ...
seq.4:            6       10    12                18          22 ...
seq.5:                    10    12                18          22 ... 
etc.
       -------------------------------------------------------------
        1 2   4   6       10 ...                  

Thus, for example, $a_5=b_5=10,$ and as far as I'm aware it happens that $a_n=b_n$ for all $n$ that anyone has ever checked by computation; in fact, comments at OEIS claim the sequences are identical -- but I haven't found a proof of this (or even any kind of rationale for it) in any reference.

Question: How to prove that $a_n=b_n$ for all $n$? (Has this ever been proved?)

(A similar question applies to the presumed identity of a number of other OEIS sequences that seem to arise from both a "rounding up" procedure and from a "sieving" procedure.)

Answer 1

This is almost exactly proved in the Erdős & Jabotinsky (1958) paper linked in the previous question.

Specifically, that paper proves (as a special case of equation (1), used in an example at the bottom of p. 117) that $$b_n = \left\lceil \frac21 \left\lceil \frac32 \left\lceil \frac43 \cdots \left\lceil \frac{n-1}{n-2} \left\lceil \frac n{n-1}\right\rceil \right\rceil \cdots \right\rceil \right\rceil \right\rceil.$$ We can think of this expression as $f_1(f_2(f_3(f_4(\cdots (f_{n-1}(n))\cdots))))$ where $f_k(m) = k \lceil \frac mk \rceil$; all it takes is mentally splitting up the fractions in your head. For example, the application of $f_{n-1}$ is highlighted in red below: $$b_n = \left\lceil \frac21 \left\lceil \frac32 \left\lceil \frac43 \cdots \left\lceil \frac{\color{red}{n-1}}{n-2} \color{red}{\left\lceil \frac {\color{black}n}{n-1}\right\rceil} \right\rceil \cdots \right\rceil \right\rceil \right\rceil.$$ But $f_k(m)$ is exactly the operation that rounds a number up to the next multiple of $k$, so $a_n = f_1(f_2(f_3(f_4(\cdots (f_{n-1}(n))\cdots))))$ by definition, and therefore $a_n = b_n$.

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The proof of equation (1) is very much a blink-and-you'll-miss-it moment in the paper, so let me spell it out in a bit more detail. Let $A_i(j)$ be the $j^{\text{th}}$ term of sequence $i$ in the sieving process, so that $b_n$ is defined to be $A_n(1)$. Then $A_{i}$ consists of the terms numbered $2, 3, \dots, i, i+2, i+3, \dots, 2i, \dots$ from $A_{i-1}$: skipping terms $1$, $i+1$, $2i+1$, and so forth. One way to express it is that $$A_i(j) = A_{i-1}\left(\left\lceil\frac{i}{i-1} \cdot j\right\rceil\right),$$ since the sequence $\lceil\frac{i}{i-1} \cdot j\rceil$ (for $j=1,2,\dots$) is exactly the sequence of the positive integers excluding $1, i+1, 2i+1, \dots$. Nesting this recurrence exactly gives the formula for $b_n$ above.

In general, equation (1) exactly proves that the result of some generalized sieving procedure is equal to the result of a corresponding rounding-up procedure, and the argument is the same.