Numerator of a sum involving binomial coefficients

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import FormalConjecturesUtil

/-!
# 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 $a(0) = 0$.

*References:*
- [A357513](https://oeis.org/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 := by
  unfold a
  norm_num only [Nat.choose, Finset.sum_Icc_succ_top, Finset.Icc_self, Finset.sum_singleton]

@[category test, AMS 11]
theorem a_2 : a 2 = 81 := by
  unfold a
  norm_num only [Nat.choose, Finset.sum_Icc_succ_top, Finset.Icc_self, Finset.sum_singleton]

@[category test, AMS 11]
theorem a_3 : a 3 = 14651 := by
  unfold a
  norm_num only [Nat.choose, Finset.sum_Icc_succ_top, Finset.Icc_self, Finset.sum_singleton]

@[category test, AMS 11]
theorem a_4 : a 4 = 956875 := by
  unfold a
  norm_num only [Nat.choose, Finset.sum_Icc_succ_top, Finset.Icc_self, Finset.sum_singleton]

@[category test, AMS 11]
theorem a_5 : a 5 = 1335793103 := by
  unfold a
  norm_num only [Nat.choose, Finset.sum_Icc_succ_top, Finset.Icc_self, Finset.sum_singleton]

/--
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 a357513_supercongruence (p : ℕ) (hp : p.Prime) (h_ge3 : p ≥ 3) (h_neq7 : p ≠ 7) :
    (a (p - 1) : ℤ) ≡ 0 [ZMOD (p : ℤ) ^ 4] := by
  sorry

/--
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 general_supercongruence (m : ℕ) : ∃ (exceptions : Finset ℕ), ∀ p, p.Prime →
    p ∉ exceptions → u m (p - 1) = (0 : ZMod (p ^ 4)) := by
  sorry

@[category test, AMS 11]
theorem general_supercongruence_one_of_a357513_supercongruence :
  type_of% a357513_supercongruence → type_of% (general_supercongruence 1) := by
  intro h
  use {2, 7}
  intro p hp h_exception
  convert h p hp ?_ ?_
  · unfold a u
    zify [ZMod.intCast_zmod_eq_zero_iff_dvd, Int.modEq_zero_iff_dvd, ZMod.natCast_eq_zero_iff]
  · apply hp.two_le.lt_of_ne (by bound)
  · simp_all

end OeisA357513

• Formalization source ↗

On-site verification needs an executable Lean target. None is attached yet.

A source attachment or link is not a verified proof. The statement’s correspondence with the original question needs human review.