12 problems

The real-part boundedness conjecture. The set $\{\operatorname{Re}w:w\in\mathcal S\}$ is bounded above but is not bounded below. This conjecture asserts an asymmetric distribution…

Let $S_n(\lambda)$ be the Stern polynomial sequence defined by $S_0(\lambda)=0$, $S_1(\lambda)=1$, and … Let $\mathcal S$ denote the set of all complex zeros of the Stern polynomia…

For $n\in\mathbb{N}$, define … Let $B_{h_n}(t)$ denote the associated Stern polynomial. No-real-roots conjecture. For each $n\in\mathbb{N}$, the polynomial $B_{h_n}(t)$ is complete…

Let $s_{i,n}$, for $i\in\{0,1,2,3\}$, be the sequences defined earlier in the paper, and let $B_{s_{i,n}}(t)$ be their associated Stern polynomials. Conjectures for $B_{s_{i,n}}$.…

Conjectures for $B_{p_{k,n}}$. (1) If $k\equiv0\pmod 2$, there is $c_k\in\mathbb{N}_{+}$ such that $B_{p_{k,n}}(t)$ has exactly one real root for $n\in\mathbb{N}_{\geq c_k}$; under…

Let $(h_n)_{n\in\mathbb{N}}$ be the sequence defined by … Let … . The claim is prompted by two computationally identified solutions that were extended to infinite sequences; the so…

Parametric-family conjecture. For each $k\in\mathbb{N}_{+}$, this choice of $p=2^k$, $q=-3\cdot2^{k-1}$, and $u=2^k-3$ leads to a non-trivial infinite family of solutions of the co…

Let … and define … Then the following equality holds: … where $k$ is equal to $n/2$ if $n$ is even and $(n-1)/2$ if $n$ is odd. This conjecture arose from investigations of the seq…

Let $B_n(t)$ denote the Stern polynomial indexed by the positive integer $n$. For a positive integer $n$, say that it has exactly $k$ prime divisors, with the source's convention f…

Let $B_p(t)$ be the Stern polynomial indexed by a prime number $p$. The prime-index irreducibility conjecture. For every prime number $p$, the polynomial $B_p(t)$ is irreducible. T…

The characterization conjecture.

Let $B_n(t)$ denote the Stern polynomial associated with the positive integer $n$. A rational number $a$ is a zero of $B_n(t)$ if $B_n(a)=0$ for some positive integer $n$. The rati…