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Recurrences like these often times have conserved quantities. For your particular case the quantity $$A_n=\frac{s_{n+1}+qs_n+qs_{n-k+1}+s_{n-k}}{s_ns_{n-1}\cdots s_{n-k+1}}$$ remains constant for any …

Let's define two auxiliary sequences $c_n=a_{n+2}-a_{n+1}-2$ and $d_n=b_{n+2}-b_{n+1}$ for $n\geq 1$. One can prove with an induction argument that the sequence $c_n$ takes values in $\{0,1,2,3\}$ and …

Before this becomes another forgotten open problem on MO, let me record here a comment. An equivalent way to state the sequence is as follows: Let $x(1)=2s-1$ and look at the recurrence equation: $$x( …

The elements of your sequence are $$a_n=\left(\frac{\alpha^n-\beta^n}{2\sqrt{3}}\right)\left(\frac{\alpha^{n+1}+\beta^{n+1}}{2}\right)$$ where $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$. Notice that b …

The following was conjectured by D. Reutter in problem 664A, Elemente der mathematik 27 and proved by H. Harborth in Solution to problem 664A, Elemente der mathematik 29, 14-15 The maximum number …

In response to Mark's comment, it is possible to determine that the probability of $X_n=0$ decays exponentially directly, and this is in fact easier than the theorems about the growth of these random …

Let's call A001936(n) by $a(n)$. Here is a sketch of why $$a(n)\equiv \sigma(4n+1)\pmod{4}$$ Firs note that the generating function of $a(n)$ is $$A(x)=\sum_{n\geq 0}a(k)x^n=\prod_{k\geq 1}\left(\fra …