1. FormalConjectures
  2. OEIS
  3. «357513»
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Numerator of a sum involving binomial coefficients

$a(n)$ is the numerator of $\sum_{k = 1}^n \frac{1}{k^3} \binom{n}{k}^2 \binom{n+k}{k}^2$ for $n \ge 1$ with $a(0) = 0$.

Reference: A357513

namespace OeisA357513 open Nat

A357513: $a(n) = \text{numerator of } \sum_{k = 1..n} \frac{1}{k^3} \binom{n}{k}^2 \binom{n+k}{k}^2 \text{ for } n \ge 1 \text{ with } a(0) = 0$.

def a (n : ℕ) : ℕ := ∑ k ∈ (Finset.Icc 1 n), ((n.choose k : ℚ) ^ 2 * ((n + k).choose k : ℚ) ^ 2) / k ^ 3 |>.num.natAbs @[category test, AMS 11] theorem a_0 : a 0 = 0 := rfl @[category test, AMS 11] theorem a_1 : a 1 = 4 := ⊢ a 1 = 4 ⊢ (∑ k ∈ Finset.Icc 1 1, ↑(choose 1 k) ^ 2 * ↑((1 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs = 4 All goals completed! 🐙 @[category test, AMS 11] theorem a_2 : a 2 = 81 := ⊢ a 2 = 81 ⊢ (∑ k ∈ Finset.Icc 1 2, ↑(choose 2 k) ^ 2 * ↑((2 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs = 81 All goals completed! 🐙 @[category test, AMS 11] theorem a_3 : a 3 = 14651 := ⊢ a 3 = 14651 ⊢ (∑ k ∈ Finset.Icc 1 3, ↑(choose 3 k) ^ 2 * ↑((3 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs = 14651 All goals completed! 🐙 @[category test, AMS 11] theorem a_4 : a 4 = 956875 := ⊢ a 4 = 956875 ⊢ (∑ k ∈ Finset.Icc 1 4, ↑(choose 4 k) ^ 2 * ↑((4 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs = 956875 All goals completed! 🐙 @[category test, AMS 11] theorem a_5 : a 5 = 1335793103 := ⊢ a 5 = 1335793103 ⊢ (∑ k ∈ Finset.Icc 1 5, ↑(choose 5 k) ^ 2 * ↑((5 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs = 1335793103 All goals completed! 🐙

We have $a(p-1) \equiv 0 \pmod{p^4}$ for all primes $p \ge 3$ except $p=7$.

proved by AlphaProof

@[category research solved, AMS 11, formal_proof using formal_conjectures at "https://github.com/google-deepmind/formal-conjectures/commit/9c7f21e7d4445637538bc1817b058b9b3f31bd2b"] theorem declaration uses 'sorry'a357513_supercongruence (p : ℕ) (hp : Nat.Prime p) (h_ge3 : p ≥ 3) (h_neq7 : p ≠ 7) : (a (p - 1) : ℤ) ≡ 0 [ZMOD (p : ℤ) ^ 4] := p:ℕhp:Nat.Prime ph_ge3:p ≥ 3h_neq7:p ≠ 7⊢ ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4] All goals completed! 🐙

Let m be a nonnegative integer and set $u(n) = $$the numerator of $$\sum{k=1}^{n} \frac{1}{k^{2m+1}} \binom{n}{k}^2 \binom{n+k}{k}^2$$ (seems like a typo in the OEIS entry: the sum starts with $k=0$ there. In order to avoid a division by zero, we replace start the sum at $k=1$.)

noncomputable def u (m : ℕ) (n : ℕ) : ℕ := ∑ k ∈ (Finset.Icc 1 n), ((n.choose k : ℚ) ^ 2 * ((n + k).choose k : ℚ) ^ 2) / k ^ (2 * m + 1) |>.num.natAbs

We conjecture that $u(p-1) == 0 (mod p^4)$ for all primes $p$, with a finite number of exceptions that depend on $m$.

@[category research open, AMS 11] theorem declaration uses 'sorry'general_supercongruence (m : ℕ) : ∃ (exceptions : Finset ℕ), ∀ p, p.Prime → p ∉ exceptions → u m (p - 1) = (0 : ZMod (p ^ 4)) := m:ℕ⊢ ∃ exceptions, ∀ (p : ℕ), Nat.Prime p → p ∉ exceptions → ↑(u m (p - 1)) = 0 All goals completed! 🐙 @[category test, AMS 11] theorem general_supercongruence_one_of_a357513_supercongruence : type_of% a357513_supercongruence → type_of% (general_supercongruence 1) := ⊢ (∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]) → ∃ exceptions, ∀ (p : ℕ), Nat.Prime p → p ∉ exceptions → ↑(u 1 (p - 1)) = 0 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]⊢ ∃ exceptions, ∀ (p : ℕ), Nat.Prime p → p ∉ exceptions → ↑(u 1 (p - 1)) = 0 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]⊢ ∀ (p : ℕ), Nat.Prime p → p ∉ {2, 7} → ↑(u 1 (p - 1)) = 0 intro p h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime p⊢ p ∉ {2, 7} → ↑(u 1 (p - 1)) = 0 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ ↑(u 1 (p - 1)) = 0 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ ↑(u 1 (p - 1)) = 0 ↔ ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ p ≥ 3h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ p ≠ 7 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ ↑(u 1 (p - 1)) = 0 ↔ ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4] h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ ↑(∑ k ∈ Finset.Icc 1 (p - 1), ↑((p - 1).choose k) ^ 2 * ↑((p - 1 + k).choose k) ^ 2 / ↑k ^ (2 * 1 + 1)).num.natAbs = 0 ↔ ↑(∑ k ∈ Finset.Icc 1 (p - 1), ↑((p - 1).choose k) ^ 2 * ↑((p - 1 + k).choose k) ^ 2 / ↑k ^ 3).num.natAbs ≡ 0 [ZMOD ↑p ^ 4] All goals completed! 🐙 h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ p ≥ 3 apply hp.two_le.lt_of_ne (h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ 2 ≠ p All goals completed! 🐙) h:∀ (p : ℕ), Nat.Prime p → p ≥ 3 → p ≠ 7 → ↑(a (p - 1)) ≡ 0 [ZMOD ↑p ^ 4]p:ℕhp:Nat.Prime ph_exception:p ∉ {2, 7}⊢ p ≠ 7 All goals completed! 🐙 end OeisA357513