We have general recurrence for A243499 (which is product of parts of integer partitions as enumerated in the table A125106) $$a(n)=(1+b(n))a(t(n)), a(0)=1$$ where $b(n)$ is A023416 (which is number of 0's in binary expansion of n) $$b(2n+1)=b(n), b(2n)=b(n)+1, b(0)=1, b(1)=0$$ and $t(n)$ is A053645 (which is distance to largest power of 2 less than or equal to n) $$t(n)=n2^{\left\lfloor\log_2{n}\right\rfloor}$$ There exist binary recurrence $$a(2n+1)=a(n), a(2n)=a(n2^{f(n)})+a(2n2^{f(n)}), a(0)=1$$ where $f(n)$ is A007814 (which is exponent of highest power of 2 dividing n) $$f(2n+1)=0, f(2n)=f(n)+1$$ Is there proof of binary recurrence from general recurrence?

$\begingroup$ Please explain what these sequences are, beyond a reference to their OEIS numbers. $\endgroup$– Sam HopkinsJun 9 '21 at 13:06

$\begingroup$ @SamHopkins, done. $\endgroup$– NotamathematicianJun 9 '21 at 13:33
Let, $n=2^{r_m}+2^{r_{m1}}+....+2^{r_1}$. Then using the given equation $a(n)=(1+b(n))a(t(n))$ we get the following representation for $a(n)$.
$a(n)=(r_m+2m)(r_{m1}+3m)....(r_1+m+1m)$.....$(1)$
[It can be seen that $t(n)=2^{r_{m1}}+....+2^{r_1}$]
So, $$a(2n)=(r_m+3m)(r_{m1}+4m)...(r_1+m+2m)$$
Now, $n2^{f(n)}=2^{r_m}+2^{r_{m1}}+.....+2^{r_2}$ and $2n2^{f(n)}=2^{r_m+1}+2^{r_{m1}+1}+....2^{r_2+1}+2^{r_1}$, as $f(n)=r_1$.
Using $(1)$, we get $$a(n2^{f(n)})=(r_m+3m)(r_{m1}+4m)...(r_2+m+1m)$$ ...(2) as here are $(m1)$ $1$s
And, $$a(2n2^{f(n)})=(r_m+3m)(r_{m1}+4m)...(r_2+m+1m)(r_1+m+1m)$$...(3)
Adding (2) and (3) we get that $$a(n2^{f(n)})+a(2n2^{f(n)})=(r_m+3m)(r_{m1}+4m)...(r_1+m+2m)=a(2n)$$
While the relation $a(2n+1)=a(n)$ can be proved using induction as follows. Let, for all $n<m$ this happens. Then, $a(2m+1)=(1+b(2m+1))a(t(2m+1))$.
Now, $t(2n+1)=2t(n)+1 \rightarrow a(t(2n+1))=a(2t(n)+1)=a(t(n))$ as $t(n)<n$ and it is given that $b(2m+1)=b(m)$. Hence, proved by induction.