Catalan identities for generalized Fibonacci polynomials

TheoremExplicit formula for Fibonacci-type generalized Fibonacci polynomials#●ready to prove

Let \(d\) and \(g\) be polynomials with integer coefficients, and let \(F_n\) be the generalized Fibonacci polynomials defined by \(F_0 = 0\), \(F_1 = 1\) and \(F_{n+2} = d F_{n+1} + g F_n\). Then for every \(n \ge 1\),

\[F_n = \sum_{i=0}^{\lfloor (n-1)/2 \rfloor} \binom{n-i-1}{i} d^{n-2i-1} g^i.\]

The source quotes this formula from Flórez and Saunders.

public theorem_wanted gfp_fib_eq_sum_choose (d g : ℤ[X]) {n : ℕ} (hn : 1 ≤ n) :
    generalizedFibonacciPolynomial 0 1 d g n =
      ∑ i ∈ range ((n - 1) / 2 + 1), ((n - i - 1).choose i : ℤ[X]) * d ^ (n - 2 * i - 1) * g ^ i
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 3.1 (1).
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremExplicit formula for Lucas-type generalized Fibonacci polynomials#●ready to prove

Let \(\alpha\) and \(p_0\) be integers with \(\alpha p_0 = 2\), let \(p_1\) and \(g\) be polynomials with integer coefficients, and put \(d = \alpha p_1\). Let \(L_n\) be the generalized Fibonacci polynomials defined by \(L_0 = p_0\), \(L_1 = p_1\) and \(L_{n+2} = d L_{n+1} + g L_n\). Then for every \(n \ge 1\), writing \(c_i = \lfloor n \binom{n-i}{i} / (n-i) \rfloor\),

\[\alpha L_n = \sum_{i=0}^{\lfloor n/2 \rfloor} c_i \, d^{n-2i} g^i.\]

The quotient \(n \binom{n-i}{i} / (n-i)\) is always an integer, so the floor does not change it; the source writes the coefficient as \(\frac{n}{n-i} \binom{n-i}{i}\) and quotes the formula from Flórez and Saunders.

public theorem_wanted gfp_lucas_eq_sum_choose {α p₀ : ℤ} (hα : α * p₀ = 2) {p₁ d : ℤ[X]}
    (hd : d = C α * p₁) (g : ℤ[X]) {n : ℕ} (hn : 1 ≤ n) :
    C α * generalizedFibonacciPolynomial (C p₀) p₁ d g n =
      ∑ i ∈ range (n / 2 + 1),
        ((n * (n - i).choose i / (n - i) : ℕ) : ℤ[X]) * d ^ (n - 2 * i) * g ^ i
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 3.1 (2).
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremWebb–Parberry binomial identity#●ready to prove

For natural numbers \(m < n\),

\[\binom{n-m-1}{m} = 2^{2m-n+1} \sum_{j=m}^{\lfloor (n-1)/2 \rfloor} \binom{j}{m} \binom{n}{2j+1},\]

an identity of rational numbers in which the exponent \(2m-n+1\) may be negative.

public theorem_wanted choose_sub_sub_one_eq_two_zpow_mul_sum {m n : ℕ} (hmn : m < n) :
    ((n - m - 1).choose m : ℚ) =
      (2 : ℚ) ^ (2 * (m : ℤ) - n + 1) *
        ∑ j ∈ Icc m ((n - 1) / 2), (j.choose m : ℚ) * (n.choose (2 * j + 1) : ℚ)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 3.2 (1).

TheoremBinomial coefficients of odd lower index as an alternating sum#●ready to prove

For natural numbers \(n\) and \(k\) with \(k < \lfloor (n-1)/2 \rfloor\),

\[\binom{n}{2k+1} = \sum_{i=k}^{\lfloor (n-1)/2 \rfloor} (-1)^{i-k} \, 2^{n-2i-1} \binom{n-i-1}{i} \binom{i}{k}.\]
public theorem_wanted choose_two_mul_add_one_eq_alternating_sum {n k : ℕ}
    (hk : k < (n - 1) / 2) :
    (n.choose (2 * k + 1) : ℤ) =
      ∑ i ∈ Icc k ((n - 1) / 2),
        (-1) ^ (i - k) * 2 ^ (n - 2 * i - 1) * ((n - i - 1).choose i : ℤ) * (i.choose k : ℤ)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 3.2 (2).

TheoremBinomial coefficients of even lower index as an alternating sum#●ready to prove

For natural numbers \(n\) and \(k\) with \(k < \lfloor n/2 \rfloor\),

\[\binom{n}{2k} = n \sum_{i=k}^{\lfloor n/2 \rfloor} \frac{(-1)^{i-k}}{n-i} \, 2^{n-2i-1} \binom{n-i}{i} \binom{i}{k},\]

an identity of rational numbers in which the exponent \(n-2i-1\) may be \(-1\).

public theorem_wanted choose_two_mul_eq_alternating_sum {n k : ℕ} (hk : k < n / 2) :
    (n.choose (2 * k) : ℚ) =
      n * ∑ i ∈ Icc k (n / 2),
        (-1) ^ (i - k) / ((n : ℚ) - i) * (2 : ℚ) ^ ((n : ℤ) - 2 * i - 1) *
          ((n - i).choose i : ℚ) * (i.choose k : ℚ)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 3.2 (3).

TheoremCatalan identity for Fibonacci-type generalized Fibonacci polynomials#●ready to prove

Let \(d\) and \(g\) be polynomials with integer coefficients, and let \(F_n\) be the generalized Fibonacci polynomials defined by \(F_0 = 0\), \(F_1 = 1\) and \(F_{n+2} = d F_{n+1} + g F_n\). Then for every natural number \(n\) (for \(n = 0\) both sides are \(0\)),

\[2^{n-1} F_n = \sum_{i=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2i+1} d^{n-2i-1} (4g + d^2)^i.\]
public theorem_wanted gfp_fib_catalan (d g : ℤ[X]) (n : ℕ) :
    2 ^ (n - 1) * generalizedFibonacciPolynomial 0 1 d g n =
      ∑ i ∈ range ((n - 1) / 2 + 1),
        (n.choose (2 * i + 1) : ℤ[X]) * d ^ (n - 2 * i - 1) * (4 * g + d ^ 2) ^ i
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Theorem 3.3 (1).
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremCatalan identity for Lucas-type generalized Fibonacci polynomials#●ready to prove

Let \(\alpha\) and \(p_0\) be integers with \(\alpha p_0 = 2\), let \(p_1\) and \(g\) be polynomials with integer coefficients, and put \(d = \alpha p_1\). Let \(L_n\) be the generalized Fibonacci polynomials defined by \(L_0 = p_0\), \(L_1 = p_1\) and \(L_{n+2} = d L_{n+1} + g L_n\). Then for every natural number \(n\),

\[2^n \alpha L_n = 2 \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i} d^{n-2i} (4g + d^2)^i.\]

Dividing both sides by \(2\) gives the source's form, with \(\alpha 2^{n-1} L_n\) on the left; the factor \(2\) is cleared so that \(n = 0\) is included.

public theorem_wanted gfp_lucas_catalan {α p₀ : ℤ} (hα : α * p₀ = 2) {p₁ d : ℤ[X]}
    (hd : d = C α * p₁) (g : ℤ[X]) (n : ℕ) :
    2 ^ n * C α * generalizedFibonacciPolynomial (C p₀) p₁ d g n =
      2 * ∑ i ∈ range (n / 2 + 1), (n.choose (2 * i) : ℤ[X]) * d ^ (n - 2 * i) * (4 * g + d ^ 2) ^ i
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Theorem 3.3 (2).
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremAlternating sum of shifted binomial coefficients#●ready to prove

Let \(\ell\), \(n\), \(j\) and \(s\) be natural numbers with \(1 \le \ell \le n\), \(j \ge 1\) and \(\ell j + s \le n\). Then

\[\sum_{i=0}^{\ell} (-1)^i \binom{\ell}{i} \binom{n-i}{\ell j+s} = \binom{n-\ell}{\ell(j-1)+s}.\]

The paper states this for \(\lfloor (1-s)/\ell \rfloor \le j \le \lfloor (n-s)/\ell \rfloor\); only \(j \ge 1\) is stated here, since for \(j \le 0\) the right side can have a negative lower index. For \(j \ge 1\) the bound \(j \le \lfloor (n-s)/\ell \rfloor\) is the condition \(\ell j + s \le n\).

public theorem_wanted alternating_sum_choose_mul_choose_sub {ℓ n j s : ℕ} (hℓ : 1 ≤ ℓ)
    (hℓn : ℓ ≤ n) (hj : 1 ≤ j) (hjn : ℓ * j + s ≤ n) :
    ∑ i ∈ range (ℓ + 1), (ℓ.choose i : ℤ) * (-1) ^ i * ((n - i).choose (ℓ * j + s) : ℤ) =
      ((n - ℓ).choose (ℓ * (j - 1) + s) : ℤ)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 4.1.

TheoremRecurrence for the floor-exponent binomial sums#●ready to prove

Let \(R\) be a commutative ring, \(t \in R\) and \(\ell \ge 1\) an integer, and for \(m \ge 0\) let \(T_\ell(m; t) = \sum_{j=0}^{m} \binom{m}{j} t^{\lfloor j/\ell \rfloor}\). Then for every integer \(n \ge 0\),

\[\sum_{i=0}^{\ell} (-1)^i \binom{\ell}{i} T_\ell(n+\ell-i; t) - t\, T_\ell(n; t) = 0.\]
public theorem_wanted floorPowBinomialSum_recurrence {R : Type*} [CommRing R] (t : R) {ℓ : ℕ}
    (hℓ : 1 ≤ ℓ) (n : ℕ) :
    ∑ i ∈ range (ℓ + 1), (ℓ.choose i : R) * (-1) ^ i *
        ∑ j ∈ range (n + ℓ - i + 1), ((n + ℓ - i).choose j : R) * t ^ (j / ℓ) -
      t * ∑ j ∈ range (n + 1), (n.choose j : R) * t ^ (j / ℓ) = 0
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Theorem 4.2.

TheoremAlternating sum of binomial row sums#●ready to prove

For all natural numbers \(\ell\) and \(n\),

\[\sum_{i=0}^{n} (-1)^i \binom{\ell}{i} \sum_{j=0}^{n-i} \binom{n-i}{j} = \sum_{i=0}^{n} (-1)^i \binom{i+\ell-n-1}{i},\]

where on the right \(\binom{a}{i} = a(a-1)\cdots(a-i+1)/i!\) is the binomial coefficient with integer upper index \(a\). For \(n < \ell\) the upper index \(i + \ell - n - 1\) is nonnegative and this is the ordinary binomial coefficient, the case used in Theorem 4.4.

public theorem_wanted alternating_choose_mul_sum_choose (ℓ n : ℕ) :
    ∑ i ∈ range (n + 1), (ℓ.choose i : ℤ) * (-1) ^ i *
        ∑ j ∈ range (n - i + 1), ((n - i).choose j : ℤ) =
      ∑ i ∈ range (n + 1), (-1) ^ i * Ring.choose ((i : ℤ) + ℓ - n - 1) i
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Lemma 4.3.

TheoremGenerating function of the floor-exponent binomial sums#●ready to prove

Let \(R\) be a commutative ring, \(t \in R\) and \(\ell \ge 1\) an integer, and for \(n \ge 0\) let \(T_\ell(n; t) = \sum_{i=0}^{n} \binom{n}{i} t^{\lfloor i/\ell \rfloor}\). Then in the ring \(R[[x]]\) of formal power series,

\[\Big(\sum_{n \ge 0} T_\ell(n; t)\, x^n\Big) \big((1-x)^\ell - t x^\ell\big) = \sum_{k=0}^{\ell-1} \Big(\sum_{i=0}^{k} (-1)^i \binom{i+\ell-k-1}{i}\Big) x^k.\]

So the generating function of the \(T_\ell(n; t)\) is this polynomial divided by the series \((1-x)^\ell - t x^\ell\), which is invertible since its constant term is \(1\).

public theorem_wanted floorPowBinomialSum_generatingFunction {R : Type*} [CommRing R] (t : R)
    {ℓ : ℕ} (hℓ : 1 ≤ ℓ) :
    PowerSeries.mk (fun n => ∑ i ∈ range (n + 1), (n.choose i : R) * t ^ (i / ℓ)) *
        ((1 - PowerSeries.X) ^ ℓ - PowerSeries.C t * PowerSeries.X ^ ℓ) =
      ∑ k ∈ range ℓ, PowerSeries.C (∑ i ∈ range (k + 1),
        (-1 : R) ^ i * ((i + ℓ - k - 1).choose i : R)) * PowerSeries.X ^ k
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Theorem 4.4.

TheoremGenerating function of the binomial section sums#●ready to prove

Let \(R\) be a commutative ring, \(t \in R\), and let \(\ell\) and \(s\) be natural numbers with \(s < \ell\). Put \(T_{\ell,s}(n; t) = \sum_{i=0}^{n} \binom{n}{\ell i + s} t^i\). Then, as formal power series in \(x\) over \(R\),

\[\Big( \sum_{n \ge 0} T_{\ell,s}(n; t) x^n \Big) \big( (1-x)^\ell - t x^\ell \big) = \sum_{k=0}^{\ell-1} \Big( \sum_{i=0}^{k} (-1)^i \binom{\ell}{i} \sum_{r=0}^{k} \binom{k-i}{r\ell+s} t^r \Big) x^k.\]

That is, the generating function of \(T_{\ell,s}(n; t)\) is this polynomial divided by \((1-x)^\ell - t x^\ell\).

The printed numerator in the source has \(t^t\), a typo for \(t^r\) (\(r\) is the summation index), and the range \(s < \ell\) is the one of (4.4). For \(s \ge \ell\) the identity fails: for \(\ell = 1\), \(s = 1\), \(t = 2\) the left side is \(x/(1-x)\) and the right side is \(0\).

public theorem_wanted sectionBinomialSum_generatingFunction {R : Type*} [CommRing R] (t : R)
    {ℓ s : ℕ} (hs : s < ℓ) :
    PowerSeries.mk (fun n => ∑ i ∈ range (n + 1), (n.choose (ℓ * i + s) : R) * t ^ i) *
        ((1 - PowerSeries.X) ^ ℓ - PowerSeries.C t * PowerSeries.X ^ ℓ) =
      ∑ k ∈ range ℓ, PowerSeries.C (∑ i ∈ range (k + 1), (-1 : R) ^ i * (ℓ.choose i : R) *
        ∑ r ∈ range (k + 1), ((k - i).choose (r * ℓ + s) : R) * t ^ r) * PowerSeries.X ^ k
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Theorem 4.5.

TheoremIntegral representation of Fibonacci-type generalized Fibonacci polynomials#●ready to prove

Let \(d\) and \(g\) be polynomials with integer coefficients, and let \(F_n\) be the generalized Fibonacci polynomials defined by \(F_0 = 0\), \(F_1 = 1\) and \(F_{n+2} = d F_{n+1} + g F_n\). If \(x\) is a real number with \(d(x)^2 + 4 g(x) \ge 0\), then for every \(n \ge 2\),

\[F_n(x) = \frac{n}{2^n} \int_{-1}^{1} \left( d(x) + y \sqrt{d(x)^2 + 4 g(x)} \right)^{n-1} dy.\]
public theorem_wanted gfp_fib_eval_eq_integral (d g : ℤ[X]) (x : ℝ)
    (hx : 0 ≤ aeval x d ^ 2 + 4 * aeval x g) {n : ℕ} (hn : 2 ≤ n) :
    aeval x (generalizedFibonacciPolynomial 0 1 d g n) =
      n / 2 ^ n * ∫ y in (-1 : ℝ)..1,
        (aeval x d + y * Real.sqrt (aeval x d ^ 2 + 4 * aeval x g)) ^ (n - 1)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Proposition 5.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremStewart's integral formula for the Fibonacci numbers#●ready to prove

For every integer \(n \ge 1\), the \(n\)-th Fibonacci number \(F_n\) (with \(F_0 = 0\) and \(F_1 = 1\)) satisfies

\[F_n = \frac{n}{2^n} \int_{-1}^{1} \left(1 + y\sqrt{5}\right)^{n-1} dy.\]
public theorem_wanted fib_eq_integral_one_add_mul_sqrt_five {n : ℕ} (hn : 1 ≤ n) :
    (Nat.fib n : ℝ) = n / 2 ^ n * ∫ y in (-1 : ℝ)..1, (1 + y * Real.sqrt 5) ^ (n - 1)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Corollary 5.1.

TheoremBinomial sum of generalized Fibonacci polynomials as an integral#●ready to prove

Let \(d\) and \(g\) be polynomials with integer coefficients, let \(F_m\) be the generalized Fibonacci polynomials with \(F_0 = 0\), \(F_1 = 1\) and \(F_{m+2} = d F_{m+1} + g F_m\), and let \(x\) be a real number with \(d(x)^2 + 4 g(x) \ge 0\) and \(g(x) \ne 0\). Let \(\Phi \colon \mathbb{Z} \to \mathbb{R}\) satisfy \(\Phi(m) = F_m(x)\) for \(m \ge 0\) and \(\Phi(m+2) = d(x) \Phi(m+1) + g(x) \Phi(m)\) for every integer \(m\). Then for every real \(k \ge 2\) and every integer \(n \ge 1\),

\[\sum_{i=0}^{n} (-1)^{n+1} \binom{n}{i} (k+1)^i (k-1)^{n-i} g(x)^i \, \Phi(n-2i) = n \int_{-k}^{k} \Big(y \sqrt{d(x)^2 + 4 g(x)} + d(x)\Big)^{n-1} dy.\]

The extension \(\Phi\) of \(m \mapsto F_m(x)\) to negative indices exists only when \(g(x) \ne 0\): if \(g(x) = 0\), the recurrence at \(m = -1\) forces \(F_1(x) = d(x) F_0(x) = 0\).

public theorem_wanted gfp_fib_binomial_sum_eq_integral (d g : ℤ[X]) (x : ℝ)
    (hx : 0 ≤ aeval x d ^ 2 + 4 * aeval x g) (hg : aeval x g ≠ 0) (F : ℤ → ℝ)
    (hF : ∀ m : ℕ, F m = aeval x (generalizedFibonacciPolynomial 0 1 d g m))
    (hrec : ∀ m : ℤ, F (m + 2) = aeval x d * F (m + 1) + aeval x g * F m)
    {k : ℝ} (hk : 2 ≤ k) {n : ℕ} (hn : 1 ≤ n) :
    ∑ i ∈ range (n + 1), (-1) ^ (n + 1) * (n.choose i : ℝ) * (k + 1) ^ i * (k - 1) ^ (n - i) *
        aeval x g ^ i * F ((n : ℤ) - 2 * i) =
      n * ∫ y in -k..k, (y * Real.sqrt (aeval x d ^ 2 + 4 * aeval x g) + aeval x d) ^ (n - 1)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Proposition 5.2.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremBinomial sum of Fibonacci numbers as an integral#●ready to prove

Let \(F \colon \mathbb{Z} \to \mathbb{R}\) be a function whose value \(F_m\) at every \(m \ge 0\) is the \(m\)-th Fibonacci number (with \(F_0 = 0\) and \(F_1 = 1\)) and which satisfies \(F_{m+2} = F_{m+1} + F_m\) for every integer \(m\). Then for every real \(k \ge 2\) and every integer \(n \ge 1\),

\[\sum_{i=0}^{n} (-1)^{n+1} \binom{n}{i} (k+1)^i (k-1)^{n-i} F_{n-2i} = n \int_{-k}^{k} (\sqrt{5}\, y + 1)^{n-1} \, dy.\]
public theorem_wanted fib_binomial_sum_eq_integral (F : ℤ → ℝ) (hF : ∀ m : ℕ, F m = Nat.fib m)
    (hrec : ∀ m : ℤ, F (m + 2) = F (m + 1) + F m) {k : ℝ} (hk : 2 ≤ k) {n : ℕ} (hn : 1 ≤ n) :
    ∑ i ∈ range (n + 1), (-1) ^ (n + 1) * (n.choose i : ℝ) * (k + 1) ^ i * (k - 1) ^ (n - i) *
        F ((n : ℤ) - 2 * i) =
      n * ∫ y in -k..k, (Real.sqrt 5 * y + 1) ^ (n - 1)
Sources
  • Maribel Diaz Noguera, Rigoberto Flórez, José L. Ramírez, and Martha Romero Rojas, Catalan Identities for Generalized Fibonacci Polynomials, The Fibonacci Quarterly 62 (2024), Article florez11012023, Corollary 5.2.
16 ready to prove

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