Discriminants of k-generalized Fibonacci polynomials

TheoremWhen the absolute discriminant of a \(k\)-generalized Fibonacci polynomial is a \(k\)-generalized Fibonacci number#●ready to prove

When the absolute discriminant of a \(k\)-generalized Fibonacci polynomial is a \(k\)-generalized Fibonacci number. Let \(k \ge 2\) be a natural number and let \(F : \mathbb{Z} \to \mathbb{Z}\) satisfy \(F(n) = 0\) for every integer \(n\) with \(2 - k \le n \le 0\), \(F(1) = 1\), and \(F(n) = F(n-1) + F(n-2) + \cdots + F(n-k)\) for every integer \(n \ge 2\), so that \(F(n)\) for \(n \ge 0\) is the \(k\)-generalized Fibonacci sequence. Then the absolute value of the discriminant of the polynomial \(X^k - X^{k-1} - \cdots - X - 1 \in \mathbb{Z}[X]\) equals \(F(n)\) for some natural number \(n\) if and only if \(k = 2\) or \(k = 3\).

For \(k = 2\) the absolute value of the discriminant is \(5 = F(5)\), and for \(k = 3\) it is \(44 = F(8)\). The paper prints the target set as \(\{F_n\}\); that it means the \(k\)-generalized sequence follows from its Lemma 1 and from Batte and Luca's restatement of this theorem (arXiv 2504.02514, Introduction). With the classical Fibonacci numbers the case \(k = 3\) would fail, since \(44\) is not a Fibonacci number.

public theorem_wanted abs_discr_kGeneralizedFibonacciPolynomial_mem_iff (k : ℕ) (hk : 2 ≤ k)
    (F : ℤ → ℤ) (hFzero : ∀ n : ℤ, 2 - (k : ℤ) ≤ n → n < 1 → F n = 0) (hFone : F 1 = 1)
    (hFrec : ∀ n : ℤ, 1 < n → F n = ∑ i ∈ Finset.range k, F (n - 1 - (i : ℤ))) :
    (∃ n : ℕ, |(X ^ k - ∑ i ∈ Finset.range k, X ^ i : ℤ[X]).discr| = F n) ↔ k = 2 ∨ k = 3

Sources

• Florian Luca, On the Discriminant of the k-Generalized Fibonacci Polynomial, II, The Fibonacci Quarterly 62 (2024), no. 3, Theorem 1.

TheoremBounds on the index in Luca's discriminant equation#●ready to prove

Let \(k > 100\) be a natural number and let \(F : \mathbb{Z} \to \mathbb{Z}\) satisfy \(F(n) = 0\) for every integer \(n\) with \(2 - k \le n \le 0\), \(F(1) = 1\), and \(F(n) = F(n-1) + F(n-2) + \cdots + F(n-k)\) for every integer \(n \ge 2\) (the \(k\)-generalized Fibonacci sequence). Write \(L = (k \log k - 2 \log(k-1)) / \log 2\), where \(\log\) is the natural logarithm. If a natural number \(n\) satisfies \(F(n)\,(k-1)^2 = 2^{k+1} k^k - (k+1)^{k+1}\), then \(k + 1.99 + L < n < k + 3.01 + L\).

The equation is the paper's equation (7), \(F(n) = (2^{k+1} k^k - (k+1)^{k+1}) / (k-1)^2\), multiplied by \((k-1)^2\).

public theorem_wanted kGeneralizedFibonacci_discr_eq_index_bounds (k : ℕ) (hk : 100 < k)
    (F : ℤ → ℤ) (hFzero : ∀ n : ℤ, 2 - (k : ℤ) ≤ n → n < 1 → F n = 0) (hFone : F 1 = 1)
    (hFrec : ∀ n : ℤ, 1 < n → F n = ∑ i ∈ Finset.range k, F (n - 1 - (i : ℤ)))
    (n : ℕ)
    (h7 : F n * ((k : ℤ) - 1) ^ 2 = 2 ^ (k + 1) * (k : ℤ) ^ k - ((k : ℤ) + 1) ^ (k + 1)) :
    (k : ℝ) + 1.99 + ((k : ℝ) * Real.log k - 2 * Real.log ((k : ℝ) - 1)) / Real.log 2 <
        (n : ℝ) ∧
      (n : ℝ) < (k : ℝ) + 3.01 +
        ((k : ℝ) * Real.log k - 2 * Real.log ((k : ℝ) - 1)) / Real.log 2

Sources

• Florian Luca, On the Discriminant of the k-Generalized Fibonacci Polynomial, II, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 1. Equation (7) is written after multiplying by (k-1)^2; F^{(k)} is given by its defining properties as in abs_discr_kGeneralizedFibonacciPolynomial_mem_iff. By Theorem 1 the equation has no solution with k > 100, so this intermediate bound is implied by that statement.

TheoremAbsolute bounds on solutions of Luca's discriminant equation#●ready to prove

Let \(k > 100\) be a natural number and let \(F : \mathbb{Z} \to \mathbb{Z}\) satisfy \(F(n) = 0\) for every integer \(n\) with \(2 - k \le n \le 0\), \(F(1) = 1\), and \(F(n) = F(n-1) + F(n-2) + \cdots + F(n-k)\) for every integer \(n \ge 2\) (the \(k\)-generalized Fibonacci sequence). If a natural number \(n\) satisfies \(F(n)\,(k-1)^2 = 2^{k+1} k^k - (k+1)^{k+1}\), then \(k < 2 \cdot 10^{16}\) and \(n < 1.5 \cdot 10^{18}\).

The equation is the paper's equation (7) multiplied by \((k-1)^2\).

public theorem_wanted kGeneralizedFibonacci_discr_eq_bounds (k : ℕ) (hk : 100 < k)
    (F : ℤ → ℤ) (hFzero : ∀ n : ℤ, 2 - (k : ℤ) ≤ n → n < 1 → F n = 0) (hFone : F 1 = 1)
    (hFrec : ∀ n : ℤ, 1 < n → F n = ∑ i ∈ Finset.range k, F (n - 1 - (i : ℤ)))
    (n : ℕ)
    (h7 : F n * ((k : ℤ) - 1) ^ 2 = 2 ^ (k + 1) * (k : ℤ) ^ k - ((k : ℤ) + 1) ^ (k + 1)) :
    k < 2 * 10 ^ 16 ∧ 2 * n < 3 * 10 ^ 18

Sources

• Florian Luca, On the Discriminant of the k-Generalized Fibonacci Polynomial, II, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2. Equation (7) is written after multiplying by (k-1)^2, as in kGeneralizedFibonacci_discr_eq_index_bounds; by Theorem 1 the hypotheses are never met.

3 ready to prove

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