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module
public import FormalConjecturesUtilDenominators of coefficients in Stirling's expansion for $\log(\Gamma(z))$
The $n$-th term is the denominator of $\frac{B_{2n}}{2n(2n-1)}$ where $B_{2n}$ is the $2n$-th Bernoulli number.
References:
@[expose] public sectionnamespace OeisA46969Denominators of coefficients in Stirling's expansion for $\log(\Gamma(z))$.
def a (n : ℕ) : ℕ :=
if n = 0 then 0
else
let m := 2 * n
let k := m * (m - 1)
(bernoulli m / (k : ℚ)).den@[category test, AMS 11]
theorem a_0 : a 0 = 0 := ⊢ a 0 = 0 All goals completed! 🐙⊢ (6⁻¹ / 2).den = 12
All goals completed! 🐙⊢ (-1 / 30 / 12).den = 360
All goals completed! 🐙hchoose2:Nat.choose 6 2 = 15hchoose3:Nat.choose 6 3 = 20hchoose4:Nat.choose 6 4 = 15h5:bernoulli' 5 = 0⊢ 1 - ∑ k ∈ Finset.range 6, ↑(Nat.choose 6 k) / (↑6 - ↑k + 1) * bernoulli' k = 1 / 42
All goals completed! 🐙⊢ (1 / 42 / 30).den = 1260
All goals completed! 🐙hchoose2:Nat.choose 8 2 = 28hchoose3:Nat.choose 8 3 = 56hchoose4:Nat.choose 8 4 = 70hchoose5:Nat.choose 8 5 = 56hchoose6:Nat.choose 8 6 = 28h3:bernoulli' 3 = 0h5:bernoulli' 5 = 0h7:bernoulli' 7 = 0⊢ 1 - ∑ k ∈ Finset.range 8, ↑(Nat.choose 8 k) / (↑8 - ↑k + 1) * bernoulli' k = -1 / 30
All goals completed! 🐙⊢ (-1 / 30 / 56).den = 1680
All goals completed! 🐙hchoose2:Nat.choose 10 2 = 45hchoose3:Nat.choose 10 3 = 120hchoose4:Nat.choose 10 4 = 210hchoose5:Nat.choose 10 5 = 252hchoose6:Nat.choose 10 6 = 210hchoose7:Nat.choose 10 7 = 120hchoose8:Nat.choose 10 8 = 45h3:bernoulli' 3 = 0h5:bernoulli' 5 = 0h7:bernoulli' 7 = 0h9:bernoulli' 9 = 0⊢ 1 - ∑ k ∈ Finset.range 10, ↑(Nat.choose 10 k) / (↑10 - ↑k + 1) * bernoulli' k = 5 / 66
All goals completed! 🐙⊢ (5 / 66 / 90).den = 1188
All goals completed! 🐙Conjecture I: if $n > 2$, then $\frac{a(\text{A005382}(n))}{12}$ is prime, where A005382 is the sequence of primes $p$ such that $2p-1$ is also prime. Since $\text{A005382}(1) = 2$, $\text{A005382}(2) = 3$ and $\text{A005382}(3) = 7$, this says that $\frac{a(p)}{12}$ is prime for every prime $p > 3$ such that $2p-1$ is also prime.
Lorenzo Sauras Altuzarra, Oct 13 2020
@[category research open, AMS 11]
theorem conjecture1 (p : ℕ) (hp : p.Prime) (hp' : (2 * p - 1).Prime) (h3 : 3 < p) :
(a p / 12).Prime := p:ℕhp:Nat.Prime php':Nat.Prime (2 * p - 1)h3:3 < p⊢ Nat.Prime (a p / 12)
All goals completed! 🐙Conjecture II: if $\frac{a(n)}{12}$ is prime, then $\frac{a(n-1)}{12} - (n-1)$, $\frac{a(n)}{12} - n$ and $\frac{a(n+2)}{12} - (n+2)$ are multiples of 6.
Lorenzo Sauras Altuzarra, Oct 13 2020
This is false for $n = 236791$.
@[category research solved, AMS 11,
formal_proof using lean4 at "https://github.com/epoch-research/LeanOpenProblems-results/blob/f02efd9a8c5fc6a735d2a90c33e24f7278ce0ffc/runs/oeis-full-50usd-ant-j0j0g4uzligm1k41/oeis_a046969_conjecture_2/Submission/Spec.lean#L845"]
theorem conjecture2 :
¬ ∀ (n : ℕ), 2 ≤ n → 12 ∣ a n → Nat.Prime (a n / 12) →
12 ∣ a (n - 1) → 12 ∣ a (n + 2) →
6 ∣ ((a (n - 1) / 12 : ℤ) - (n - 1 : ℤ)) ∧
6 ∣ ((a n / 12 : ℤ) - (n : ℤ)) ∧
6 ∣ ((a (n + 2) / 12 : ℤ) - (n + 2 : ℤ)) := ⊢ ¬∀ (n : ℕ),
2 ≤ n →
12 ∣ OeisA46969.a n →
Nat.Prime (OeisA46969.a n / 12) →
12 ∣ OeisA46969.a (n - 1) →
12 ∣ OeisA46969.a (n + 2) →
6 ∣ ↑(OeisA46969.a (n - 1)) / 12 - (↑n - 1) ∧
6 ∣ ↑(OeisA46969.a n) / 12 - ↑n ∧ 6 ∣ ↑(OeisA46969.a (n + 2)) / 12 - (↑n + 2)
All goals completed! 🐙end OeisA46969