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The solutions look like a mess, so it's not too surprising that you always end up with one. If we follow Iosif Pinelis in dividing all by the last constraint by $\lambda$ and substituting $r = \frac{1 …

It turns out to be more straightforward than I expected. Let $C(z) = \sum_{i \ge 0} c_i z^i$ be the g.f. for $c_i$, excluding $c_{-1}$ since that doesn't show up in your recurrence. Starting with $$\f …

I'm going to use $\operatorname{msb}$ (for most significant bit) as an alias of $f$. Since $q_i$ is a permutation, the property that $q_i(n)<2^k$ iff $n < 2^k$ is equivalent to $\operatorname{msb}(q_i …

Cleaning up the notation a bit, $$b_{m,k}(n) = m\, b_{m,k}(n-2^{\ell(n)}) + k \sum_{j=0}^{\ell(n)-1} [n \,\&\, 2^j = 0] \,b_{m,k}(n - 2^{\ell(n)} + 2^j)$$ where $\&$ is bitwise AND. $$s_{m,k}(n) = \su …

Generalise $a$: $a(n, k)$ is the number of steps to perform this process with $n+k$ boxes and balls starting with $n \ge 1$ balls in the first box and one ball each in the next $k$ boxes. Then the ori …

The second half is already given in the question, so really what you're asking is whether $$b(2n-1)=b(2n-3)+b(n-1)-2(b(2n-3)\bmod b(n-1))$$ But as noted in OEIS (quoted with relabelling), Moshe Newma …

Expressing the linear recurrence in matrix form as $$\begin{bmatrix} a_{n+k+1} & a_{n+k} \\ a_{n+k} & a_{n+k-1} \end{bmatrix} = \begin{bmatrix} s & t \\ 1 & 0 \end{bmatrix}^k \begin{bmatrix} a_{n+1} & …

Experimenting with a CAS suggests an induction. In order to handle the induction, we need to consider the forms of the numbers involved. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ alternate …