Weighted power sums and perfect powers in Lucas sequences

TheoremPowers of a root of the Lucas characteristic polynomial#●ready to prove

Let \(x\) and \(y \neq 0\) be integers and let \((U_n)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{n+2} = x U_{n+1} - y U_n\). If \(\alpha\) is a complex number with \(\alpha^2 - x\alpha + y = 0\), then \(\alpha^n = \alpha U_n - y U_{n-1}\) for every \(n \geq 1\).

public theorem_wanted lucasU_root_pow (P : LucasSequenceParams) {α : ℂ}
    (hα : α ^ 2 - P.p * α + P.q = 0) {n : ℕ} (hn : 1 ≤ n) :
    α ^ n = α * lucasU P n - P.q * lucasU P (n - 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.1 (i), (ii).
Uses 2

TheoremThe companion Lucas sequence in terms of the fundamental one#●ready to prove

Let \(x\) and \(y \neq 0\) be integers and let \((U_n)\) and \((V_n)\) be the Lucas sequences with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\), \(V_0 = 2\), \(V_1 = x\) and \(W_{n+2} = x W_{n+1} - y W_n\) for \(W = U, V\). Then \(V_n = U_{n+1} - y U_{n-1}\) for every \(n \geq 1\).

public theorem_wanted lucasV_eq_lucasU_succ_sub (P : LucasSequenceParams) {n : ℕ} (hn : 1 ≤ n) :
    lucasV P n = lucasU P (n + 1) - P.q * lucasU P (n - 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.1 (iii).
Uses 3

TheoremAddition formula for the Lucas sequence \(U\)#●ready to prove

Let \(x\) and \(y \neq 0\) be integers and let \((U_n)\) and \((V_n)\) be the Lucas sequences with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\), \(V_0 = 2\), \(V_1 = x\) and \(W_{n+2} = x W_{n+1} - y W_n\) for \(W = U, V\). Then for all natural numbers \(n \leq m\), \(U_{m+n} = U_m V_n - y^n U_{m-n}\).

public theorem_wanted lucasU_add_eq_mul_lucasV_sub (P : LucasSequenceParams) {m n : ℕ}
    (hnm : n ≤ m) :
    lucasU P (m + n) = lucasU P m * lucasV P n - P.q ^ n * lucasU P (m - n)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.1 (iv).
Uses 3

TheoremSquare of the companion Lucas sequence#●ready to prove

Let \(x\) and \(y \neq 0\) be integers and let \((U_n)\) and \((V_n)\) be the Lucas sequences with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\), \(V_0 = 2\), \(V_1 = x\) and \(W_{n+2} = x W_{n+1} - y W_n\) for \(W = U, V\). Then for every natural number \(n\), \(V_n^2 = (x^2 - 4y) U_n^2 + 4 y^n\).

public theorem_wanted lucasV_sq_eq (P : LucasSequenceParams) (n : ℕ) :
    lucasV P n ^ 2 = (P.p ^ 2 - 4 * P.q) * lucasU P n ^ 2 + 4 * P.q ^ n
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.1 (vi).
Uses 3

TheoremBinomial-sum formula for the Lucas sequence \(U\)#●ready to prove

Let \(x\) and \(y \neq 0\) be integers and let \((U_n)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{n+2} = x U_{n+1} - y U_n\). Then for every \(r \geq 1\),

\[ U_r = \sum_{i=0}^{\lfloor (r-1)/2 \rfloor} (-y)^i \binom{r-i-1}{i} x^{r-2i-1}. \]
public theorem_wanted lucasU_eq_sum_choose (P : LucasSequenceParams) {r : ℕ} (hr : 1 ≤ r) :
    lucasU P r = ∑ i ∈ Finset.range ((r - 1) / 2 + 1),
      (-P.q) ^ i * ((r - i - 1).choose i : ℤ) * P.p ^ (r - 2 * i - 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.1 (vii).
Uses 2

TheoremBounds on Lucas sequences by powers of the dominant root#●ready to prove

Let \(x\) and \(y \neq 0\) be integers with \(x \geq 1\), \(|y + 1| \leq x\) and \(D = x^2 - 4y > 0\), let \((U_n)\) and \((V_n)\) be the Lucas sequences with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\), \(V_0 = 2\), \(V_1 = x\) and \(W_{n+2} = x W_{n+1} - y W_n\) for \(W = U, V\), and put \(\alpha = (x + \sqrt{D})/2\). Then for every \(n \geq 2\) we have \(x \alpha^{n-2} \leq U_n\); if \(y \leq 0\), then \(U_n \leq \alpha^{n-1}\) and \(x \alpha^{n-1} \leq V_n \leq \alpha^{n+1}/x\); and if \(y \geq 0\), then \(U_n \leq \alpha^n/\sqrt{D}\) and \(\alpha^n \leq V_n \leq \alpha^{n+1}/\sqrt{D}\).

public theorem_wanted lucasU_lucasV_alpha_bounds (P : LucasSequenceParams) (hx : 1 ≤ P.p)
    (hy : |P.q + 1| ≤ P.p) (hD : 0 < P.p ^ 2 - 4 * P.q) {n : ℕ} (hn : 2 ≤ n) :
    let α : ℝ := (P.p + Real.sqrt (P.p ^ 2 - 4 * P.q)) / 2
    (P.p : ℝ) * α ^ (n - 2) ≤ lucasU P n ∧
      (P.q ≤ 0 → (lucasU P n : ℝ) ≤ α ^ (n - 1)) ∧
      (0 ≤ P.q → (lucasU P n : ℝ) ≤ α ^ n / Real.sqrt (P.p ^ 2 - 4 * P.q)) ∧
      (P.q ≤ 0 → (P.p : ℝ) * α ^ (n - 1) ≤ lucasV P n ∧ (lucasV P n : ℝ) ≤ α ^ (n + 1) / P.p) ∧
      (0 ≤ P.q → α ^ n ≤ (lucasV P n : ℝ) ∧
        (lucasV P n : ℝ) ≤ α ^ (n + 1) / Real.sqrt (P.p ^ 2 - 4 * P.q))
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.2.
Uses 3

TheoremLower bounds for ratios of consecutive Lucas terms#●ready to prove

Let \(x\) and \(y \neq 0\) be integers with \(x \geq 1\) and \(|y + 1| \leq x\), and let \((U_n)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{n+2} = x U_{n+1} - y U_n\). If \(y < 0\), then \(U_n / U_{n-1} \geq x - xy/(x^2 - y)\) for every \(n > 2\). If \(y > 0\), then \(U_n / U_{n-1} \geq (x + \sqrt{x^2 - 4y})/2\) for every \(n \geq 2\).

public theorem_wanted lucasU_ratio_lower_bound (P : LucasSequenceParams) (hx : 1 ≤ P.p)
    (hy : |P.q + 1| ≤ P.p) :
    (P.q < 0 → ∀ n : ℕ, 2 < n →
      (P.p : ℝ) - P.p * P.q / (P.p ^ 2 - P.q) ≤ (lucasU P n : ℝ) / lucasU P (n - 1)) ∧
    (0 < P.q → ∀ n : ℕ, 2 ≤ n →
      (P.p + Real.sqrt (P.p ^ 2 - 4 * P.q)) / 2 ≤ (lucasU P n : ℝ) / lucasU P (n - 1))
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.3.
Uses 2

TheoremLogarithmic bounds for weighted Lucas power sums when \(y < 0\)#●ready to prove

Let \(x \ge 1\) and \(y < 0\) be integers with \(|y + 1| \le x\), let \((U_m)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - y U_m\), and let \(\alpha = (x + \sqrt{x^2 - 4y})/2\). Let \(k\) be a natural number and \(n\), \(p\), \(q\) positive integers with \(\sum_{j=1}^k j U_j^p = U_n^q\). If \(k \ge 2\), then \(\log_\alpha k + p \log_\alpha x + (k - 2)p < (n - 1)q\). If \(k \ge 3\), then \(q \log_\alpha x + (n - 2)q < (k - 1)p + \log_\alpha k + \log_\alpha\bigl(1 + (x^2 - 2y)/(x^{p-1}(x^3 - x^2 - 2xy + y))\bigr)\).

public theorem_wanted weighted_power_sum_log_bounds_of_neg (P : LucasSequenceParams)
    (hx : 1 ≤ P.p) (hyneg : P.q < 0) (hy : |P.q + 1| ≤ P.p) {n k p q : ℕ} (hn : 0 < n)
    (hp : 0 < p) (hq : 0 < q)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    let α : ℝ := (P.p + Real.sqrt (P.p ^ 2 - 4 * P.q)) / 2
    (2 ≤ k → Real.logb α k + p * Real.logb α P.p + ((k : ℝ) - 2) * p < ((n : ℝ) - 1) * q) ∧
    (3 ≤ k → q * Real.logb α P.p + ((n : ℝ) - 2) * q <
      ((k : ℝ) - 1) * p + Real.logb α k + Real.logb α (1 + ((P.p : ℝ) ^ 2 - 2 * P.q) /
        ((P.p : ℝ) ^ (p - 1) * ((P.p : ℝ) ^ 3 - P.p ^ 2 - 2 * P.p * P.q + P.q))))
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.4.
Uses 2

TheoremLogarithmic bounds for weighted Lucas power sums when \(y > 0\)#●ready to prove

Let \(x\) and \(y > 0\) be integers with \(|y + 1| \le x\) and \(D = x^2 - 4y > 0\), let \((U_m)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - y U_m\), and let \(\alpha = (x + \sqrt{D})/2\). Let \(k\) be a natural number and \(n\), \(p\), \(q\) positive integers with \(\sum_{j=1}^k j U_j^p = U_n^q\). If \(k \ge 2\), then \(\log_\alpha k + p \log_\alpha x + (k - 2)p < q(n - \log_\alpha \sqrt{D})\). If \(k \ge 3\), then \(q \log_\alpha x + (n - 2)q < \log_\alpha k + p(k + 1 - \log_\alpha \sqrt{D}) - \log_\alpha(\alpha^p - 1)\).

The hypothesis \(D > 0\) is added to the statement of the paper: for \(x = 2\), \(y = 1\) one has \(\alpha = 1\), and \(\log_\alpha\) is undefined.

public theorem_wanted weighted_power_sum_log_bounds_of_pos (P : LucasSequenceParams)
    (hypos : 0 < P.q) (hy : |P.q + 1| ≤ P.p) (hD : 0 < P.p ^ 2 - 4 * P.q) {n k p q : ℕ}
    (hn : 0 < n) (hp : 0 < p) (hq : 0 < q)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    let D : ℝ := P.p ^ 2 - 4 * P.q
    let α : ℝ := (P.p + Real.sqrt D) / 2
    (2 ≤ k → Real.logb α k + p * Real.logb α P.p + ((k : ℝ) - 2) * p <
      q * (n - Real.logb α (Real.sqrt D))) ∧
    (3 ≤ k → q * Real.logb α P.p + ((n : ℝ) - 2) * q <
      Real.logb α k + p * ((k : ℝ) + 1 - Real.logb α (Real.sqrt D)) - Real.logb α (α ^ p - 1))
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.5.
Uses 2

TheoremValuation of \(U_n\) at a prime dividing the discriminant#●ready to prove

Let \(x \ge 1\) and \(y \ne 0\) be coprime integers with \(|y + 1| \le x\), let \((U_m)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - y U_m\), and let \(\ell\) be a prime dividing \(D = x^2 - 4y\). Write \(\nu_\ell\) for the \(\ell\)-adic valuation of an integer, with \(\nu_\ell(0) = 0\). Then for every \(n \ge 1\), \(\nu_\ell(U_n) = \nu_\ell(n) + \nu_\ell(U_\ell) - 1\) if \(\ell \mid n\), and \(\nu_\ell(U_n) = 0\) if \(\ell \nmid n\). Moreover, \(\nu_\ell(U_\ell) = 1\) if \(\ell \ge 5\).

Coprimality of \(x\) and \(y\) (Sanna's standing hypothesis) is added to the statement of the paper: without it, \(x = 6\), \(y = 3\) and \(\ell = 3 \mid D = 24\) give \(\nu_3(U_2) = \nu_3(6) = 1 \ne 0\).

public theorem_wanted padicValInt_lucasU_of_dvd_discr (P : LucasSequenceParams) (hx : 1 ≤ P.p)
    (hy : |P.q + 1| ≤ P.p) (hcop : IsCoprime P.p P.q) {ℓ : ℕ} (hℓ : ℓ.Prime)
    (hℓD : (ℓ : ℤ) ∣ P.p ^ 2 - 4 * P.q) {n : ℕ} (hn : 1 ≤ n) :
    (ℓ ∣ n → padicValInt ℓ (lucasU P n) = padicValNat ℓ n + padicValInt ℓ (lucasU P ℓ) - 1) ∧
    (¬ ℓ ∣ n → padicValInt ℓ (lucasU P n) = 0) ∧
    (5 ≤ ℓ → padicValInt ℓ (lucasU P ℓ) = 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.6.
Uses 2

TheoremA quadratic equation in the companion Lucas sequence#●ready to prove

Let \(x\) be a positive integer, let \((V_n)\) be the companion Lucas sequence with parameters \((x, -1)\), given by \(V_0 = 2\), \(V_1 = x\) and \(V_{n+2} = x V_{n+1} + V_n\), and let \(D = x^2 + 4\). Let \(p, q, k\) be natural numbers with \(p\) odd and \(q, k\) positive. If

\[ V_p k^2 + (V_p - 2) k - 1 = \pm D^{p-q} V_p^2 \]

in \(\mathbb{Q}\), where the exponent \(p - q\) may be negative, then \((p, q, k, x)\) is one of \((1, 1, 1, 1)\), \((1, 1, 2, 1)\) and \((1, 1, 2, 5)\).

public theorem_wanted lucasV_quadratic_eq_pm_discr_zpow (P : LucasSequenceParams)
    (hyP : P.q = -1) {p q k : ℕ} (hx : 0 < P.p) (hp : Odd p) (hq : 0 < q) (hk : 0 < k)
    (h : let V : ℚ := lucasV P p
      let D : ℚ := (P.p : ℚ) ^ 2 + 4
      V * k ^ 2 + (V - 2) * k - 1 = D ^ ((p : ℤ) - q) * V ^ 2 ∨
        V * k ^ 2 + (V - 2) * k - 1 = -(D ^ ((p : ℤ) - q) * V ^ 2)) :
    (p = 1 ∧ q = 1 ∧ k = 1 ∧ P.p = 1) ∨ (p = 1 ∧ q = 1 ∧ k = 2 ∧ P.p = 1) ∨
      (p = 1 ∧ q = 1 ∧ k = 2 ∧ P.p = 5)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 2.7.
Uses 2

TheoremSolutions of \(1 + 2x^p = U_n^q\) for Lucas sequences with \(y = \pm 1\)#●ready to prove

Let \(x\) be a positive integer, let \(y \in \{1, -1\}\), and let \((U_m)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - y U_m\). For positive integers \(n\), \(p\), \(q\), the equation \(1 + 2x^p = U_n^q\) holds if and only if \((n, p, q, x, y)\) is one of \((4, p, 1, 1, -1)\), \((1 + 2^{p+1}, p, 1, 2, 1)\), \((3, 2, 2, 2, 1)\), \((3, 1, 1, 2, -1)\) and \((5, 3, 1, 3, 1)\), where \(p\) is arbitrary in the first two.

Since \(U_1 = 1\) and \(U_2 = x\), this is the case \(k = 2\) of the equation \(\sum_{j=1}^k j U_j^p = U_n^q\).

public theorem_wanted one_add_two_mul_pow_eq_lucasU_pow_iff (P : LucasSequenceParams)
    (hy : P.q = 1 ∨ P.q = -1) (hx : 0 < P.p) {n p q : ℕ} (hn : 0 < n) (hp : 0 < p)
    (hq : 0 < q) :
    1 + 2 * P.p ^ p = lucasU P n ^ q ↔
      (n = 4 ∧ q = 1 ∧ P.p = 1 ∧ P.q = -1) ∨ (n = 1 + 2 ^ (p + 1) ∧ q = 1 ∧ P.p = 2 ∧ P.q = 1) ∨
        (n = 3 ∧ p = 2 ∧ q = 2 ∧ P.p = 2 ∧ P.q = 1) ∨
        (n = 3 ∧ p = 1 ∧ q = 1 ∧ P.p = 2 ∧ P.q = -1) ∨
        (n = 5 ∧ p = 3 ∧ q = 1 ∧ P.p = 3 ∧ P.q = 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Theorem 1.1 and Proposition 3.1.
Uses 2

TheoremNo solutions of \(\sum_{j=1}^k j U_j^p = U_n^q\) when \(y = -1\) and \(x \ge 2\) or \(y = 1\) and \(x \ge 3\), with \(k \ge 3\) and exponents at most \(11\)#●ready to prove

No solutions of \(\sum_{j=1}^k j U_j^p = U_n^q\) when \(y = -1\) and \(x \ge 2\) or \(y = 1\) and \(x \ge 3\), with \(k \ge 3\) and exponents at most \(11\). Let \(y \in \{1, -1\}\), let \(x\) be an integer with \(x \ge 2\) if \(y = -1\) and \(x \ge 3\) if \(y = 1\), and let \((U_m)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - y U_m\). If \(n\), \(p\), \(q\) are positive integers with \(p \le 11\) and \(q \le 11\), and \(k \ge 3\) is an integer, then \(\sum_{j=1}^k j U_j^p \ne U_n^q\).

public theorem_wanted weighted_power_sum_ne_lucasU_pow (P : LucasSequenceParams)
    (hy : (P.q = -1 ∧ 2 ≤ P.p) ∨ (P.q = 1 ∧ 3 ≤ P.p)) {n k p q : ℕ} (hn : 0 < n)
    (hp : 0 < p) (hq : 0 < q) (hp11 : p ≤ 11) (hq11 : q ≤ 11) (hk : 3 ≤ k) :
    ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p ≠ lucasU P n ^ q
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Theorem 1.1, second part.
Uses 2

TheoremRemainder form of the weighted power-sum equation for Lucas sequences#●ready to prove

Let \(x\) be a positive integer, let \(y \in \{1, -1\}\), and let \((U_n)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{n+2} = x U_{n+1} - y U_n\). Suppose \(D = x^2 - 4y > 0\) and put \(\alpha = (x + \sqrt{D})/2\). Let \(n, p, q\) be positive integers and \(k \geq 3\) an integer with

\[ \sum_{j=1}^{k} j \, U_j^p = U_n^q. \]

Then there are real numbers \(R_1, R_2\) with \(|R_1| < k^2 \alpha^{k(p-2)}\), \(|R_2| < \alpha^{n(q-2)}\) and

\[ \frac{D^{(q-p)/2} (k\alpha^p - (k+1))}{(\alpha^p - 1)^2} \alpha^{p(k+1)} - \alpha^{qn} = D^{q/2} R_2 - D^{q/2} R_1 - \frac{D^{(q-p)/2} \alpha^p}{(\alpha^p - 1)^2}. \]

The remainders \(R_1, R_2\) are only asserted to exist; the source defines them explicitly in its proof.

public theorem_wanted exists_remainders_weighted_power_sum (P : LucasSequenceParams)
    (hy : P.q = 1 ∨ P.q = -1) (hx : 0 < P.p) (hD : 0 < P.p ^ 2 - 4 * P.q) {n k p q : ℕ}
    (hn : 0 < n) (hp : 0 < p) (hq : 0 < q) (hk : 3 ≤ k)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    let D : ℝ := P.p ^ 2 - 4 * P.q
    let α : ℝ := (P.p + Real.sqrt D) / 2
    ∃ R₁ R₂ : ℝ, |R₁| < (k : ℝ) ^ 2 * α ^ ((k : ℝ) * ((p : ℝ) - 2)) ∧
      |R₂| < α ^ ((n : ℝ) * ((q : ℝ) - 2)) ∧
      D ^ (((q : ℝ) - p) / 2) * (k * α ^ p - (k + 1)) / (α ^ p - 1) ^ 2 * α ^ (p * (k + 1)) -
          α ^ (q * n) =
        D ^ ((q : ℝ) / 2) * R₂ - D ^ ((q : ℝ) / 2) * R₁ -
          D ^ (((q : ℝ) - p) / 2) * α ^ p / (α ^ p - 1) ^ 2
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Proposition 4.1.
Uses 2

TheoremBound on the parameter in the weighted power-sum equation for Lucas sequences#●ready to prove

Let \(x\) be a positive integer, let \(y \in \{1, -1\}\), and let \((U_n)\) be the Lucas sequence with parameters \((x, y)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{n+2} = x U_{n+1} - y U_n\). If \(n, p, q\) are positive integers and \(k \geq 3\) is an integer with

\[ \sum_{j=1}^{k} j \, U_j^p = U_n^q, \]

then \(x \leq (k + 1)^2/4 + 1\).

The hypothesis \(k \geq 3\) cannot be dropped entirely: for \(k = 1\) and \(y = -1\), \((n, p, q, x) = (1, 1, 1, 5)\) is a solution and \(5 > (1 + 1)^2/4 + 1 = 2\).

public theorem_wanted le_of_weighted_power_sum_eq (P : LucasSequenceParams)
    (hy : P.q = 1 ∨ P.q = -1) (hx : 0 < P.p) {n k p q : ℕ} (hn : 0 < n) (hp : 0 < p)
    (hq : 0 < q) (hk : 3 ≤ k)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    (P.p : ℚ) ≤ ((k : ℚ) + 1) ^ 2 / 4 + 1
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 4.4.
Uses 2

TheoremRemainder bound for weighted Lucas power sums when \(y = -1\)#●ready to prove

Let \(x\) be a positive integer, let \((U_m)\) be the Lucas sequence with parameters \((x, -1)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} + U_m\), and let \(D = x^2 + 4\) and \(\alpha = (x + \sqrt{D})/2\). If positive integers \(n\), \(p\), \(q\) and an integer \(k \ge 3\) satisfy \(\sum_{j=1}^k j U_j^p = U_n^q\), then the remainder \(R_2 = U_n^q - \alpha^{nq}/D^{q/2}\) satisfies \(D^{q/2} |R_2| < k (D + 1/x)^{q+1} \alpha^{kp - 2k/q + 1}\).

public theorem_wanted remainder_bound_of_neg (P : LucasSequenceParams) (hyP : P.q = -1)
    (hx : 0 < P.p) {n k p q : ℕ} (hn : 0 < n) (hp : 0 < p) (hq : 0 < q) (hk : 3 ≤ k)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    let D : ℝ := P.p ^ 2 - 4 * P.q
    let α : ℝ := (P.p + Real.sqrt D) / 2
    D ^ ((q : ℝ) / 2) * |(lucasU P n : ℝ) ^ q - α ^ (n * q) / D ^ ((q : ℝ) / 2)| <
      k * (D + 1 / P.p) ^ (q + 1) * α ^ ((k : ℝ) * p - 2 * k / q + 1)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 4.2.
Uses 2

TheoremRemainder bound for weighted Lucas power sums when \(y = 1\)#●ready to prove

Let \(x\) be a positive integer with \(D = x^2 - 4 > 0\), let \((U_m)\) be the Lucas sequence with parameters \((x, 1)\), given by \(U_0 = 0\), \(U_1 = 1\) and \(U_{m+2} = x U_{m+1} - U_m\), let \(V_2 = x^2 - 2\) be the term of index \(2\) of the companion Lucas sequence, and let \(\alpha = (x + \sqrt{D})/2\). If positive integers \(n\), \(p\), \(q\) and an integer \(k \ge 3\) satisfy \(\sum_{j=1}^k j U_j^p = U_n^q\), then the remainder \(R_2 = U_n^q - \alpha^{nq}/D^{q/2}\) satisfies \(D^{q/2} |R_2| < k (V_2 + 1/x)^{q+1} \alpha^{kp - 2k/q + 1/2}\).

public theorem_wanted remainder_bound_of_pos (P : LucasSequenceParams) (hyP : P.q = 1)
    (hD : 0 < P.p ^ 2 - 4 * P.q) (hx : 0 < P.p) {n k p q : ℕ} (hn : 0 < n) (hp : 0 < p)
    (hq : 0 < q) (hk : 3 ≤ k)
    (hsol : ∑ j ∈ Finset.Icc 1 k, (j : ℤ) * lucasU P j ^ p = lucasU P n ^ q) :
    let D : ℝ := P.p ^ 2 - 4 * P.q
    let α : ℝ := (P.p + Real.sqrt D) / 2
    D ^ ((q : ℝ) / 2) * |(lucasU P n : ℝ) ^ q - α ^ (n * q) / D ^ ((q : ℝ) / 2)| <
      k * ((lucasV P 2 : ℝ) + 1 / P.p) ^ (q + 1) * α ^ ((k : ℝ) * p - 2 * k / q + 1 / 2)
Sources
  • Benjamin Earp-Lynch, Simon Earp-Lynch, Omar Kihel and Puntani Pongsumpun, Extension of the Equation ∑ j F_j^p = F_n^q to a Family of Lucas Sequences, The Fibonacci Quarterly 62 (2024), no. 3, Lemma 4.3.
Uses 3
17 ready to prove

The dependency explorer needs JavaScript and its graph data.

Download the complete graph data.