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(Notation: residue $n_0\mod 2^{\lceil i \log_23\rceil}$ = residue $b\mod2^k$ from your wiki page)

About the "discarded" 5 reaching maximum 8 (or 16), already reached by "surviving" 3:

One of the discarded sequence is the inverse V-Shape sequence which rise for $i$ steps of $f(x)=\frac{3x+1}{2}$ and then fall bellow the initial value by successive division by $2$ (See here). Of all discarded sequences $2^{\lceil i \log_23\rceil}n+n_0$ for a specific $n$ , this is the type of sequence that potentially reaches the highest value: $$(2^{\lceil i \log_23\rceil}n+n_0+1)\frac{3^i}{2^{i}}-1$$

Note: $n_0\leq 2^{\lceil i \log_23\rceil}-3$ and the exact value can be found in the link above

e.g. with $4n+1=5$ where $n_0=1$, $i=1$,$n=1$ which reaches $8$ before dropping to $4<5$

One of the surviving sequence is the straight line which rise for the whole $k={\lceil i \log_23\rceil}$ steps of $f(x)=\frac{3x+1}{2}$ . Of all surviving sequences for a specific $n$ , this is the sequence (starting from $2\cdot2^{\lceil i \log_23\rceil}n-1$ ) that reaches the highest value (limited to $k={\lceil i \log_23\rceil}$ steps): $$3^{\lceil i \log_23\rceil}(n+1)-1$$

Note: here we always have $n_0= 2^{\lceil i \log_23\rceil}-1$

e.g. with $4n+3=7$ where $i=1$,$n=1$ which reaches $17$ (in 2 steps), or with $n=0$: $3$ reaches $8$

Now it is easy to show that the highest value that can be reached by a discarded sequence at $n$ is smaller (or equal) than the highest value already reached by a surviving sequence at $n-1$

e.g with discarded $4(1)+1=5$ reaches $8$ which was already reached by surviving $4(1-1)+3=3$

Surviving highest value at $n-1$ is greater then discarded value at $n$?

$$3^{\lceil i \log_23\rceil}n-1 \geq (2^{\lceil i \log_23\rceil}n+n_0+1)\frac{3^i}{2^{i}}-1$$ and with $n_0< 2^{\lceil i \log_23\rceil}-1$, we just need to show that $$3^{\lceil i \log_23\rceil}n-1 \geq (2^{\lceil i \log_23\rceil}(n+1))\frac{3^i}{2^{i}}-1$$ $$\Big(\frac{3}{2}\Big)^{\lceil i \log_23\rceil}n \geq \Big(\frac{3}{2}\Big)^i(n+1)$$ $$\Big(\frac{3}{2}\Big)^{\lceil i \log_2\frac{3}{2}\rceil} \geq 1+\frac{1}{n}$$ which is already true for $n-1=0$ when $i\geq 3$ (manually checked for $i=1$ and $i=2$ by using the exact value of $n_0$ in those cases)

e.g. with $n-1=0$: discarded $32n+23$ reaches $188$ but surviving $32(n-1)+31$ already reached $242$

Note: you can multiply both side by 2 to get the "real" maximum (16 instead of 8).

The key idea is that even if the discarded inverse V-Shape at $n$ was at the highest possible residue $n_0= 2^{\lceil i \log_23\rceil}-3$, it would reach a smaller value than the straight line at $n-1$ (always with residue $n_0= 2^{\lceil i \log_23\rceil}-1$).

This means that record paths are always found in residue $b\mod2^k$ (in other word, at $2^k\cdot n+b$ with $n=0$)

EDIT:

even more, when sieving $2^{k+1}$: values below $2^k$ that are dropping cannot produce new path records (obviously), but value above $2^k$ that are not surviving after $2^{k+1}$ sieve are now known, and there maximum is still the RHS above: indeed the condition $n_0+2^{\lceil i \log_23\rceil}< 2^{\lceil i \log_23\rceil+1}-1$ or $n_0< 2^{\lceil i \log_23\rceil}-1$ do not change, and the value of $i$ (climbing steps) neither since the last step was a drop bellow initial value.

So even if the max value on the LHS do not climb anymore at step $k+1$, it would still be higher (the whole equation would stay the same).

This means that new record paths are only found in surviving residue $b\mod2^k$

No need to check discarded residue at all, even within the sieve range.