Let $m\geq 2$ be a fixed integer.

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.

Then we have an integer sequence given by \begin{align} a(0)=a(1)&=1\\ a(2n)& = a(n)+a(n-2^{f(n)})+a(2n-2^{f(n)})\\ a(2n+1) &= ma(n) \end{align}

Let $$s(n,m)=\sum\limits_{k=0}^{2^n-1}a(k)$$

Then I conjecture that for any $m\geq 2$ $$s(n,m)=(n+m)s(n-1,m)+((m+1)^2-4)\frac{(n+m-1)(g(n+m-2)-g(m+1))}{(m+3)(m+1)!}, s(0,m)=1$$ where $$g(n)=\sum\limits_{k=0}^{n-1}k!$$ is A003422, left factorials.

What is nice here, it is the fact that if we take large adjacent terms of the sequence (for any $m\geq 2$) in any number system $p$ where $p\geq 2$, then we have some identical digits at the end (or from the right side).

You can easily verified it using simple recurrence above and pari function Vecrev(digits(s(n,m), p)).

For example, if we take $s(15,2)$ and $s(16,2)$ in base $2$ we have $$110100001110010101011010101101001{\color{red}{10101110100111}}$$ $$1110101101110101100110011101100010010{\color{red}{10101110100111}}$$ Same values in base $6$: $$1044125322511{\color{red}{512155}}$$ $$32220524330415{\color{red}{512155}}$$ Same values in base $12$: $$10(10)69123{\color{red}{82865(11)}}$$ $$174359527{\color{red}{82865(11)}}$$

Is there any explanation of this interesting fact?