Let $a(n)$ be A071585, i.e., numerator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of $4n$, with the exponents of $2$ being listed in descending order.
In the formulas section Yosu Yurramendi gives a nice recurrence: $$a(2^m+k)=a(k)+[0\leqslant k<2^{m-1}]\cdot a(k+2^{m-1})+[2^{m-1}\leqslant k<2^m]\cdot a(k-2^{m-1}), a(0)=1, a(1)=2$$
I conjecture that there also exists another recurrence $$a(n)=a(\left\lfloor\frac{n}{2}\right\rfloor)+a(k)-a(k_1)+a(k_2)-\cdots+(-1)^ma(k_m), a(0)=1$$ where $$n=(2k+1)2^{t_1}$$ $$k=(2k_1+1)2^{t_2}$$ $$k_1=(2k_2+1)2^{t_3}$$ $$\cdots$$ $$k_{m-1}=(2k_m+1)2^{t_{m+1}}$$ $$k_m=0$$ Also obviously $$n=2^{t_1}(1+2^{1+t_2}(1+\dots(1+2^{1+t_\ell}))\dots)$$
Is there a way to prove it?
