Let $a(n)$ be A103318, number of solutions $i$ in range $[0,n-1]$ to $i \equiv 0 \pmod {2^{n-i}}$: the sequence begins with $$1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2$$

Also let's consider $$\ell(n)=\left\lfloor\log_{2}(n)\right\rfloor$$ and $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ Here $T(n,k)$ is the $(k+1)$-th bit from the right side in binary expansion of $n$.

Then we have an integer sequence given by $$b(n)=\sum\limits_{k=0}^{\ell(n)}\sum\limits_{j=0}^{\ell(n)-k}2^{j}T(n-k,j+k)$$ The sequence begins with $$1, 2, 4, 5, 7, 9, 11, 12, 14, 16, 19, 20, 22, 24, 26, 27, 29, 31, 34, 36$$

I conjecture that $a(n)=b(n)-b(n-1)$ with $a(1)=1$.

Is there a way to prove it?

Let $a(n)$ be A103318, number of solutions $i$ in range $[0,n-1]$ to $i \equiv 0 \pmod {2^{n-i}}$: the sequence begins with $$1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2$$

Also let's consider $$\ell(n)=\left\lfloor\log_{2}(n)\right\rfloor$$ and $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ Here $T(n,k)$ is the $(k+1)$-th bit from the right side in binary expansion of $n$.

Then we have an integer sequence given by $$b(n)=\sum\limits_{k=0}^{\ell(n)}\sum\limits_{j=0}^{\ell(n)-k}2^{j}T(n-k,j+k)=\sum\limits_{k=0}^{\ell(n)}\left\lfloor\frac{n-k}{2^k}\right\rfloor$$ The sequence begins with $$1, 2, 4, 5, 7, 9, 11, 12, 14, 16, 19, 20, 22, 24, 26, 27, 29, 31, 34, 36$$

I conjecture that $a(n)=b(n)-b(n-1)$ with $a(1)=1$.

Is there a way to prove it?