Fibonacci representations with polynomial coefficients
TheoremInteger sequences with linear Fibonacci coefficients#●ready to prove
Let \(F_n\) be the \(n\)-th Fibonacci number and put \(F_{-1} = 1\), so that \(F_{n-1} = F_{n+1} - F_n\) for every \(n \ge 0\). For rational numbers \(a, b, c, d\), the number \(w_n = (an + b)F_n + (cn + d)F_{n-1}\) is an integer for every natural number \(n\) if and only if \(d\) is an integer and there are integers \(z_1, z_2, z_3\) with \(a = (-z_1 - 3z_2 + 2z_3)/5\), \(b = (6z_1 + 3z_2 - 2z_3)/5\) and \(c = (-2z_1 + 4z_2 - z_3)/5\).
The integers \(z_i\) are then determined by the sequence: \(z_1 = w_1\), \(z_2 = w_2 - d\) and \(z_3 = w_3 - d\).
public theorem_wanted fib_linear_coeff_isInt_iff (a b c d : ℚ) :
(∀ n : ℕ, ∃ k : ℤ, (k : ℚ) =
(a * n + b) * Nat.fib n + (c * n + d) * ((Nat.fib (n + 1) : ℚ) - Nat.fib n)) ↔
(∃ k : ℤ, (k : ℚ) = d) ∧ ∃ z₁ z₂ z₃ : ℤ,
a = ((-z₁ - 3 * z₂ + 2 * z₃ : ℤ) : ℚ) / 5 ∧
b = ((6 * z₁ + 3 * z₂ - 2 * z₃ : ℤ) : ℚ) / 5 ∧
c = ((-2 * z₁ + 4 * z₂ - z₃ : ℤ) : ℚ) / 5Sources
- Kálmán Liptai, László Németh, Tamás Szakács and László Szalay, On Certain Fibonacci Representations, Fibonacci Quarterly 62 (2024), no. 4, Theorem 2.1.
TheoremInteger sequences with quadratic Fibonacci coefficients#●ready to prove
Let \(F_n\) be the \(n\)-th Fibonacci number and put \(F_{-1} = 1\), so that \(F_{n-1} = F_{n+1} - F_n\) for every \(n \ge 0\). For rational numbers \(a, b, c, d, e, f\), the number \(w_n = (an^2 + bn + c)F_n + (dn^2 + en + f)F_{n-1}\) is an integer for every natural number \(n\) if and only if \(f\) is an integer and there are integers \(z_1, \dots, z_5\) with \(a = (-z_1 + 3z_2 + z_3 - 3z_4 + z_5)/10\), \(b = (-5z_1 - 75z_2 + 15z_3 + 45z_4 - 17z_5)/50\), \(c = (30z_1 + 30z_2 - 10z_3 - 15z_4 + 6z_5)/25\), \(d = (3z_1 - 4z_2 - 3z_3 + 4z_4 - z_5)/10\) and \(e = (-45z_1 + 80z_2 + 15z_3 - 40z_4 + 11z_5)/50\).
The integers \(z_i\) are then determined by the sequence: \(z_1 = w_1\), \(z_2 = w_2 - f\), \(z_3 = w_3 - f\), \(z_4 = w_4 - 2f\) and \(z_5 = w_5 - 3f\).
public theorem_wanted fib_quadratic_coeff_isInt_iff (a b c d e f : ℚ) :
(∀ n : ℕ, ∃ k : ℤ, (k : ℚ) =
(a * n ^ 2 + b * n + c) * Nat.fib n +
(d * n ^ 2 + e * n + f) * ((Nat.fib (n + 1) : ℚ) - Nat.fib n)) ↔
(∃ k : ℤ, (k : ℚ) = f) ∧ ∃ z₁ z₂ z₃ z₄ z₅ : ℤ,
a = ((-z₁ + 3 * z₂ + z₃ - 3 * z₄ + z₅ : ℤ) : ℚ) / 10 ∧
b = ((-5 * z₁ - 75 * z₂ + 15 * z₃ + 45 * z₄ - 17 * z₅ : ℤ) : ℚ) / 50 ∧
c = ((30 * z₁ + 30 * z₂ - 10 * z₃ - 15 * z₄ + 6 * z₅ : ℤ) : ℚ) / 25 ∧
d = ((3 * z₁ - 4 * z₂ - 3 * z₃ + 4 * z₄ - z₅ : ℤ) : ℚ) / 10 ∧
e = ((-45 * z₁ + 80 * z₂ + 15 * z₃ - 40 * z₄ + 11 * z₅ : ℤ) : ℚ) / 50Sources
- Kálmán Liptai, László Németh, Tamás Szakács and László Szalay, On Certain Fibonacci Representations, Fibonacci Quarterly 62 (2024), no. 4, Theorem 2.2.
TheoremInteger sequences with quadratic and linear Fibonacci coefficients#●ready to prove
Let \(F_n\) be the \(n\)-th Fibonacci number and put \(F_{-1} = 1\), so that \(F_{n-1} = F_{n+1} - F_n\) for every \(n \ge 0\). For rational numbers \(a, b, c, d, e\), the number \(w_n = (an^2 + bn + c)F_n + (dn + e)F_{n-1}\) is an integer for every natural number \(n\) if and only if \(e\) is an integer and there are integers \(z_1, z_2, z_3, z_4\) with \(a = (2z_1 - z_2 - 2z_3 + z_4)/10\), \(b = (-56z_1 - 7z_2 + 66z_3 - 23z_4)/50\), \(c = (48z_1 + 6z_2 - 28z_3 + 9z_4)/25\) and \(d = (-6z_1 + 18z_2 - 9z_3 + 2z_4)/25\).
The integers \(z_i\) are then determined by the sequence: \(z_1 = w_1\), \(z_2 = w_2 - e\), \(z_3 = w_3 - e\) and \(z_4 = w_4 - 2e\).
public theorem_wanted fib_quadratic_linear_coeff_isInt_iff (a b c d e : ℚ) :
(∀ n : ℕ, ∃ k : ℤ, (k : ℚ) =
(a * n ^ 2 + b * n + c) * Nat.fib n +
(d * n + e) * ((Nat.fib (n + 1) : ℚ) - Nat.fib n)) ↔
(∃ k : ℤ, (k : ℚ) = e) ∧ ∃ z₁ z₂ z₃ z₄ : ℤ,
a = ((2 * z₁ - z₂ - 2 * z₃ + z₄ : ℤ) : ℚ) / 10 ∧
b = ((-56 * z₁ - 7 * z₂ + 66 * z₃ - 23 * z₄ : ℤ) : ℚ) / 50 ∧
c = ((48 * z₁ + 6 * z₂ - 28 * z₃ + 9 * z₄ : ℤ) : ℚ) / 25 ∧
d = ((-6 * z₁ + 18 * z₂ - 9 * z₃ + 2 * z₄ : ℤ) : ℚ) / 25Sources
- Kálmán Liptai, László Németh, Tamás Szakács and László Szalay, On Certain Fibonacci Representations, Fibonacci Quarterly 62 (2024), no. 4, Theorem 2.3.
TheoremInteger sequences with linear Fibonacci coefficients and an alternating term#●ready to prove
Let \(F_n\) be the \(n\)-th Fibonacci number and put \(F_{-1} = 1\), so that \(F_{n-1} = F_{n+1} - F_n\) for every \(n \ge 0\). For rational numbers \(a, b, c, d, e, f\), the number \(w_n = (an + b)F_n + (cn + d)F_{n-1} + e + f(-1)^n\) is an integer for every natural number \(n\) if and only if there are integers \(v_0, \dots, v_5\) with \(a = (3v_0 + 2v_1 - 7v_2 - v_3 + 4v_4 - v_5)/5\), \(b = (-3v_0 - 2v_1 - 3v_2 + 6v_3 + 6v_4 - 4v_5)/5\), \(c = (-4v_0 - v_1 + 11v_2 - 2v_3 - 7v_4 + 3v_5)/5\), \(d = -2v_1 + v_2 + 2v_3 - v_4\), \(e = (v_0 + 3v_1 + v_2 - 3v_3 - v_4 + v_5)/2\) and \(f = (v_0 + v_1 - 3v_2 - v_3 + 3v_4 - v_5)/2\).
These integers are then the initial values of the sequence: \(v_i = w_i\) for \(0 \le i \le 5\).
public theorem_wanted fib_linear_coeff_add_alternating_isInt_iff (a b c d e f : ℚ) :
(∀ n : ℕ, ∃ k : ℤ, (k : ℚ) =
(a * n + b) * Nat.fib n + (c * n + d) * ((Nat.fib (n + 1) : ℚ) - Nat.fib n) +
e + f * (-1) ^ n) ↔
∃ w₀ w₁ w₂ w₃ w₄ w₅ : ℤ,
a = ((3 * w₀ + 2 * w₁ - 7 * w₂ - w₃ + 4 * w₄ - w₅ : ℤ) : ℚ) / 5 ∧
b = ((-3 * w₀ - 2 * w₁ - 3 * w₂ + 6 * w₃ + 6 * w₄ - 4 * w₅ : ℤ) : ℚ) / 5 ∧
c = ((-4 * w₀ - w₁ + 11 * w₂ - 2 * w₃ - 7 * w₄ + 3 * w₅ : ℤ) : ℚ) / 5 ∧
d = ((-2 * w₁ + w₂ + 2 * w₃ - w₄ : ℤ) : ℚ) ∧
e = ((w₀ + 3 * w₁ + w₂ - 3 * w₃ - w₄ + w₅ : ℤ) : ℚ) / 2 ∧
f = ((w₀ + w₁ - 3 * w₂ - w₃ + 3 * w₄ - w₅ : ℤ) : ℚ) / 2Sources
- Kálmán Liptai, László Németh, Tamás Szakács and László Szalay, On Certain Fibonacci Representations, Fibonacci Quarterly 62 (2024), no. 4, Theorem 3.1.
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