179524.lean`

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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

11 declarations, 5 contain sorry, 0 declared as axioms

Central binomial sum $a(n) = \sum_{k=0}^n (-4)^k \binom{n}{k}^2 \binom{n-k}{k}^2$

The sequence is defined by $$a(n) = \sum_{k=0}^n (-4)^k \binom{n}{k}^2 \binom{n-k}{k}^2.$$

References:

A179524
Z.-W. Sun, "Open Conjectures on Congruences", arXiv preprint arXiv:0911.5665 [math.NT], 2009-2011.

Imports (1)

FormalConjecturesUtil

def OeisA179524.a The sequence $a(n) = \sum_{k=0}^n (-4)^k \binom{n}{k}^2 \binom{n-k}{k}^2$ line 36 source
Source documentation
The sequence $a(n) = \sum_{k=0}^n (-4)^k \binom{n}{k}^2 \binom{n-k}{k}^2$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat, Int
Proof uses: Finset.sum, Finset.range, HAdd.hAdd, OfNat.ofNat, HMul.hMul, HPow.hPow, Neg.neg, Nat.choose, HSub.hSub
16 instances, coercions and auxiliary constants left out.
```
def a (n : ℕ) : ℤ :=
…
```
theorem OeisA179524.a_0 line 40 source
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Int, OeisA179524.a, OfNat.ofNat, Nat
Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl
5 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA179524.a_1 line 43 source
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Int, OeisA179524.a, OfNat.ofNat, Nat
Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl
5 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA179524.a_2 line 46 source
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Int, OeisA179524.a, OfNat.ofNat, Nat, Neg.neg
Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl
7 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA179524.a_3 line 49 source
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Int, OeisA179524.a, OfNat.ofNat, Nat, Neg.neg
Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl
7 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA179524.a_4 line 52 source
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Int, OeisA179524.a, OfNat.ofNat, Nat
Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl
5 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA179524.conjecture1 If $p$ is a prime with $p \equiv 1, 9 \pmod{20}$ and $p = x^2 + 5y^2$ with $x, y$ integers, then $\sum_{k=0}^{p-1} a(k) \equiv 4x^2 - 2p \pmod{p^2}$ sorry line 60 source
Source documentation
If $p$ is a prime with $p \equiv 1, 9 \pmod{20}$ and $p = x^2 + 5y^2$ with $x, y$ integers, then $\sum_{k=0}^{p-1} a(k) \equiv 4x^2 - 2p \pmod{p^2}$.
- Zhi-Wei Sun, Jul 01 2010
depends on sorry what this means
Statement uses: Nat, Nat.Prime, Or, Int.ModEq, OfNat.ofNat, Int, Eq, HAdd.hAdd, HPow.hPow, HMul.hMul, Finset.sum, Finset.range, OeisA179524.a, HSub.hSub
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
30 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture1 (p : ℕ) (hp : p.Prime)
    (h_mod : (p : ℤ) ≡ 1 [ZMOD 20] ∨ (p : ℤ) ≡ 9 [ZMOD 20])
    (x y : ℤ) (h_sq : (p : ℤ) = x ^ 2 + 5 * y ^ 2) :
    ∑ k ∈ Finset.range p, a k ≡ 4 * x ^ 2 - 2 * (p : ℤ) [ZMOD (p : ℤ) ^ 2] := by
…
```
Mentions: OeisA179524.a
theorem OeisA179524.conjecture2 If $p$ is a prime with $p \equiv 3, 7 \pmod{20}$ and $2p = x^2 + 5y^2$ with $x, y$ integers, then $\sum_{k=0}^{p-1} a(k) \equiv 2x^2 - 2p \pmod{p^2}$ sorry line 72 source
Source documentation
If $p$ is a prime with $p \equiv 3, 7 \pmod{20}$ and $2p = x^2 + 5y^2$ with $x, y$ integers, then $\sum_{k=0}^{p-1} a(k) \equiv 2x^2 - 2p \pmod{p^2}$.
- Zhi-Wei Sun, Jul 01 2010
depends on sorry what this means
Statement uses: Nat, Nat.Prime, Or, Int.ModEq, OfNat.ofNat, Int, Eq, HMul.hMul, HAdd.hAdd, HPow.hPow, Finset.sum, Finset.range, OeisA179524.a, HSub.hSub
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
30 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture2 (p : ℕ) (hp : p.Prime)
    (h_mod : (p : ℤ) ≡ 3 [ZMOD 20] ∨ (p : ℤ) ≡ 7 [ZMOD 20])
    (x y : ℤ) (h_sq : 2 * (p : ℤ) = x ^ 2 + 5 * y ^ 2) :
    ∑ k ∈ Finset.range p, a k ≡ 2 * x ^ 2 - 2 * (p : ℤ) [ZMOD (p : ℤ) ^ 2] := by
…
```
Mentions: OeisA179524.a
theorem OeisA179524.conjecture3 If $p$ is a prime with $p \equiv 11, 13, 17, 19 \pmod{20}$, then $\sum_{k=0}^{p-1} a(k) \equiv 0 \pmod{p^2}$ sorry line 84 source
Source documentation
If $p$ is a prime with $p \equiv 11, 13, 17, 19 \pmod{20}$, then $\sum_{k=0}^{p-1} a(k) \equiv 0 \pmod{p^2}$.
- Zhi-Wei Sun, Jul 01 2010
depends on sorry what this means
Statement uses: Nat, Nat.Prime, Or, Int.ModEq, OfNat.ofNat, Int, HPow.hPow, Finset.sum, Finset.range, OeisA179524.a
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
18 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture3 (p : ℕ) (hp : p.Prime)
    (h_mod : (p : ℤ) ≡ 11 [ZMOD 20] ∨ (p : ℤ) ≡ 13 [ZMOD 20] ∨
      (p : ℤ) ≡ 17 [ZMOD 20] ∨ (p : ℤ) ≡ 19 [ZMOD 20]) :
    ∑ k ∈ Finset.range p, a k ≡ 0 [ZMOD (p : ℤ) ^ 2] := by
…
```
Mentions: OeisA179524.a
theorem OeisA179524.conjecture4 $\sum_{k=0}^{n-1}(20k+17)a(k) \equiv 0 \pmod n$ for all $n=1,2,3,\dots$ sorry line 95 source
Source documentation
$\sum_{k=0}^{n-1}(20k+17)a(k) \equiv 0 \pmod n$ for all $n=1,2,3,\dots$.
- Zhi-Wei Sun, Jul 01 2010
depends on sorry what this means
Statement uses: Nat, LE.le, OfNat.ofNat, Dvd.dvd, Int, Finset.sum, Finset.range, HMul.hMul, HAdd.hAdd, OeisA179524.a
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
22 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture4 (n : ℕ) (hn : 1 ≤ n) :
    (n : ℤ) ∣ ∑ k ∈ Finset.range n, ((20 * (k : ℤ) + 17) * a k) := by
…
```
Mentions: OeisA179524.a
theorem OeisA179524.conjecture5 $\sum_{k=0}^{p-1}(20k+17)a(k) \equiv p(10(-1/p)+7) \pmod{p^2}$ for any odd prime $p$ sorry line 104 source
Source documentation
$\sum_{k=0}^{p-1}(20k+17)a(k) \equiv p(10(-1/p)+7) \pmod{p^2}$ for any odd prime $p$.
- Zhi-Wei Sun, Jul 01 2010
depends on sorry what this means
Statement uses: Nat, Fact, Nat.Prime, Ne, OfNat.ofNat, Int.ModEq, HPow.hPow, Int, Finset.sum, Finset.range, HMul.hMul, HAdd.hAdd, OeisA179524.a, legendreSym, Neg.neg
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
28 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture5 (p : ℕ) [hp : Fact p.Prime] (_hp2 : p ≠ 2) :
    (∑ k ∈ Finset.range p, ((20 * (k : ℤ) + 17) * a k)) ≡
      (p : ℤ) * (10 * legendreSym p (-1) + 7) [ZMOD (p : ℤ) ^ 2] := by
…
```
Mentions: OeisA179524.a