The Discriminant of the Characteristic Polynomial of the $k$th Fibonacci sequence is not a member of the $k$th Lucas sequence

TL;DR

This work proves there are no integers $n\ge0$ and $k\ge2$ with $L_n^{(k)}=|\mathrm{Disc}(g_k)|$, where $g_k(X)=X^k-X^{k-1}-\cdots-1$ and $|\mathrm{Disc}(g_k)|=\dfrac{2^{k+1}k^k-(k+1)^{k+1}}{(k-1)^2}$. The authors combine detailed $2$-adic analysis of $L_n^{(k)}$ and of $\mathrm{Disc}(g_k)$ with linear forms in logarithms (Matveev) and a refinement by Bugeaud–Laurent to bound the parameters, deriving $n$ in a tight interval in terms of $k$, and ultimately establishing global bounds $k<7\cdot10^{16}$ and $n<4\cdot10^{18}$. A case split on $r=n\bmod(k+1)$ reduces the problem to finitely many subcases: $r=0$, $r=1$, $r=2$, and $3\le r\le k$. Each branch leads to contradictions either by parity/valuation arguments or by explicit computational and modular checks, so no solutions exist. The result extends the analogy with the Fibonacci–Lucas setting and demonstrates the discriminant of the characteristic polynomial of the $k$-generalized Fibonacci sequence rarely appears as a Lucas term, highlighting the power of combining $2$-adic valuations with linear forms in logarithms in Diophantine problems.

Abstract

Let $k\ge 2$ and $\{L_n^{(k)}\}_{n\geq 2-k}$ be the sequence of $k$-generalized Lucas numbers whose first $k$ terms are $0,\ldots,0,2,1$ and each term afterwards is the sum of the preceding $k$ terms. In this paper, we show that this sequence does not contain the discriminant of its characteristic polynomial.

Theorems & Definitions (12)

• Theorem 1.1
• Lemma 2.1
• Lemma 3.1
• proof
• Lemma 4.1
• Theorem 5.1: Matveev, MAT
• Lemma 5.1: Bugeaud and Laurent, BL
• Lemma 6.1
• Lemma 6.2
• proof