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import FormalConjecturesUtilDenominators of $\sum_{k=1}^n \frac{1}{k 2^k}$
$a(n)$ is the denominator of the sum $$\sum_{i=1}^n \frac{1}{i} \binom{2n-i-1}{i-1}$$
References:
arxiv/2605.22763 Advancing Mathematics Research with AI-Driven Formal Proof Search by George Tsoukalas et al.
namespace OeisA175386$a(n)$ is the denominator of the sum $$\sum_{i=1}^n \frac{1}{i} \binom{2n-i-1}{i-1}$$
def a (n : ℕ) : ℕ :=
(Finset.sum (Finset.Icc 1 n) fun i : ℕ =>
-- The upper index is $2n - i - 1$, which is equivalent to
-- $2n - (i+1)$ in $\mathbb{N}$ for $i \le n$.
-- The lower index $i-1$ is standard subtraction in $\mathbb{N}$.
let num : ℕ := Nat.choose (2 * n - (i + 1)) (i - 1)
(num : ℚ) / (i : ℚ)
).den@[category test, AMS 11]
lemma a_1 : a 1 = 1 := ⊢ a 1 = 1 All goals completed! 🐙@[category test, AMS 11]
lemma a_2 : a 2 = 2 := ⊢ a 2 = 2 All goals completed! 🐙@[category test, AMS 11]
lemma a_3 : a 3 = 6 := ⊢ a 3 = 6 All goals completed! 🐙@[category test, AMS 11]
lemma a_4 : a 4 = 4 := ⊢ a 4 = 4 All goals completed! 🐙@[category test, AMS 11]
lemma a_5 : a 5 = 5 := ⊢ a 5 = 5 All goals completed! 🐙We conjecture that $\sum_{i=1}^{n} \frac{1}{i} \binom{2n-i-1}{i-1}$ is not an integer for $n > 1$.
A formal proof has been found with the methods described in arxiv/2605.22763.
@[category research solved, AMS 11, formal_proof using formal_conjectures at
"https://github.com/mo271/formal-conjectures/blob/a32396489dcb8f86c3549b93aa358ac6a10a3a1f/FormalConjectures/OEIS/175386.wip.lean#L304"]
theorem a_ne_one (n : ℕ) (hn : 1 < n) : a n ≠ 1 := n:ℕhn:1 < n⊢ a n ≠ 1
All goals completed! 🐙end OeisA175386