Let $f(n)$ be [A153733](https://oeis.org/A153733), remove all trailing ones in binary representation of $n$. Here
\begin{align}
f(2n)& = 2n\\
f(2n+1)& = f(n)\\
\end{align}


Then we have an integer sequence given by
\begin{align}
a(0)& = 1\\
a(2n+1)& = a(n)\\
a(4n+2)& = 2a(n)\\
a(4n)& = 2a(2n)-a(n)
\end{align}
Here $a(n)$ is [A124758](https://oeis.org/A124758), product of the parts of the compositions in standard order.


Let
$$b(n) = \sum\limits\_{k=0}^{n}(\binom{n}{k}\operatorname{mod} 2)a(k)$$
Then I conjecture that
\begin{align}
b(0)& = 1\\
b(2n+1)& = b(n) + b(2n)\\
b(2n)& = b(n) + b(f(n-1))\\
\end{align}
Is there a way to prove it?



Modulo $2$ binomial transform of A124758
The first relation $b(0)=0$ is trivial.


By the condition, $a(n)$ is the product of number of consecutive zeroes plus $1$, if we write $n$ in binary. For instance, if $n=1220$, we have $n=100 1100 0100$ in binary, thus $a(n)=3\times 4\times 3=36$. Notice that if $n=2^u\times v$ where $v$ is an odd number, we have $a(n)=(u+1)a(v)$


We need a lemma: Lucas' Theorem: $\binom{a}{b}$ is odd if and only if $a\&b=b$ where $\&$ is bitwise AND operation. Let $S\_n$ be the set of $i$ if $i\&n=i$.


Now the second relation is clear. Notice that if $k\in S\_{2n+1}$ if there exist $k'\in S\_n$ such that $k=2k'$ or $k=2k'+1$. The even ones consists of exactly the elements in $S\_{2n}$. So we can divide $S\_{2n+1}$ into odd part and even part, We have


$$b(2n+1)=\sum\_{k\in S\_{2n+1}}a(k)=\sum\_{k\in S\_{2n+1},k\equiv 0\mod 2}a(k)+\sum\_{k\in S\_{2n+1},k\equiv 1\mod 2}a(k)$$
We know that
$$\sum\_{k\in S\_{2n+1},k\equiv 0\mod 2}a(k)=\sum\_{k\in S\_{2n}}a(k)=b(2n)$$


And since the odd numbers, when deleting the ending $1$ the $a(\cdot)$ value won't change. Thus,
$$\sum\_{k\in S\_{2n+1},k\equiv 1\mod 2}a(k)=\sum\_{\in S\_{2n}}a(k)=b(n)$$
Therefore we have $b(2n+1)=b(2n)+b(n)$


To prove the third one, we state some of the properties of
$S\_{n}$: Let $n=a\_ma\_{m-1}\dots a\_{k+1}10\dots 0$ ($k$ zeros in the end). Then, $f(n-1)=a\_ma\_{m-1}\dots a\_{k+1}0$. Notice that there are exactly one $1$ less from $f(n-1)$ to $n$, so we do a ``bijection''. Notice that $S(2n)$ can be divided into $b\_mb\_{m-1}\dots b\_{k+1}10\dots 0$ ($k$ zeros in the end) and $b\_mb\_{m-1}\dots b\_{k+1}00\dots 0$ ($k=1$ zeros in the end), differed by the $k+1$th from the last digit. Lett a certain pair of them be $u=b\_mb\_{m-1}\dots b\_{k+1}10\dots 0$, $v=b\_mb\_{m-1}\dots b\_{k+1}00\dots 0$ and $w=b\_mb\_{m-1}\dots b\_{k+1}0$. What we only need to prove is $(a(u)+a(v))=a((u/2)+a(v/2))+a(w)$. Summing over these pairs we have those $a(u)+a(v)$ sums to $b(2n)$, $a(u/2)+a(v/2)$ sums to $b(n)$ and $a(w)$ sums to $b(f(n-1))$.


Now we finish the proof. Let $b\_mb\_{m-1}\dots b\_{k+1}$ end in $q$ zeroes, that is, $b\_{k+q+1}=1$, $b\_{k+q}=\dots=b\_{k+1}=0$. So


$$a(u)-a(u/2)=a(b\_mb\_{m-1}\dots b\_{k+1}10\dots 0)\text{($k$ zeroes)}-a(b\_mb\_{m-1}\dots b\_{k+1}10\dots 0)\text{($k-1$ zeroes)}=(k+1)a(b\_mb\_{m-1}\dots b\_{k+1}1)-ka(b\_mb\_{m-1}\dots b\_{k+1}1)=a(b\_mb\_{m-1}\dots b\_{k+1}1)=a(b\_mb\_{m-1}\dots b\_{k+1})=qa(b\_mb\_{m-1}\dots b\_{k+q+1})$$


$$a(v)-a(v/2)=a(b\_mb\_{m-1}\dots b\_{k+1}00\dots 0)\text{($k+1$ zeroes, $k+q+1$ zeroes in the end totally)}-a(b\_mb\_{m-1}\dots b\_{k+1}00\dots 0)\text{($k$ zeroes, $k+q$ zeroes in the end totally)}=(k+q+1)a(b\_mb\_{m-1}\dots b\_{k+q+1})-(k+q)a(b\_mb\_{m-1}\dots b\_{k+q+1})=a(b\_mb\_{m-1}\dots b\_{k+q+1})$$


$$a(w)=a(b\_mb\_{m-1}\dots b\_{k+1}0)\text{($q+1$ zeroes in the end totally)}=(q+1)a(b\_mb\_{m-1}\dots b\_{k+q+1})$$


So we have $a(u)-a(u/2)+a(v)-a(v/2)=a(w)$, which yields $(a(u)+a(v))=a((u/2)+a(v/2))+a(w)$.


