FormalConjectures/OEIS/3161.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.
- L4DEL
- Lean4 Domain Explanation Language
Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.
Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
8 declarations, 1 contain sorry, 0 declared as axioms
A binomial coefficient sum
A binomial coefficient sum: $$a(n) = \sum_{k=0}^{\lfloor n/2 \rfloor} \left( \binom{n}{k} - \binom{n}{k-1} \right)^3$$ where $\binom{n}{-1} = 0$.
References:
Imports (1)
- def
OeisA3161.aA binomial coefficient sum: $a(n) = \sum_{k=0}^{\lfloor n/2 \rfloor} (\binom{n}{k} - \binom{n}{k-1})^3$ line 36 sourceSource documentation
A binomial coefficient sum: $a(n) = \sum_{k=0}^{\lfloor n/2 \rfloor} (\binom{n}{k} - \binom{n}{k-1})^3$.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
NatProof uses:
Finset.sum,Finset.range,HAdd.hAdd,HDiv.hDiv,OfNat.ofNat,Int,HSub.hSub,Nat.choose,ite,Eq,Int.toNat,HPow.hPow17 instances, coercions and auxiliary constants left out.
def a (n : ℕ) : ℕ := …
- def
OeisA3161.bAuxiliary sequence $b(n) = a(2n-1)$ line 42 sourceSource documentation
Auxiliary sequence $b(n) = a(2n-1)$.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
NatProof uses: OeisA3161.a,
HSub.hSub,HMul.hMul,OfNat.ofNat5 instances, coercions and auxiliary constants left out.
def b (n : ℕ) : ℕ := …
- theorem
OeisA3161.a_0Value of the sequenceaat 0 line 47 sourceSource documentation
Value of the sequence
aat 0.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Eq,Nat, OeisA3161.a,OfNat.ofNatProof uses:
Eq.refl2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem a_0 : a 0 = 1 := by rfl …
Mentions: OeisA3161.a
- theorem
OeisA3161.a_1Value of the sequenceaat 1 line 51 sourceSource documentation
Value of the sequence
aat 1.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Eq,Nat, OeisA3161.a,OfNat.ofNatProof uses:
Eq.refl2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem a_1 : a 1 = 1 := by rfl …
Mentions: OeisA3161.a
- theorem
OeisA3161.a_2Value of the sequenceaat 2 line 55 sourceSource documentation
Value of the sequence
aat 2.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Eq,Nat, OeisA3161.a,OfNat.ofNatProof uses:
Eq.refl2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem a_2 : a 2 = 2 := by rfl …
Mentions: OeisA3161.a
- theorem
OeisA3161.a_3Value of the sequenceaat 3 line 59 sourceSource documentation
Value of the sequence
aat 3.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Eq,Nat, OeisA3161.a,OfNat.ofNatProof uses:
Eq.refl2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem a_3 : a 3 = 9 := by rfl …
Mentions: OeisA3161.a
- theorem
OeisA3161.a_4Value of the sequenceaat 4 line 63 sourceSource documentation
Value of the sequence
aat 4.standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Eq,Nat, OeisA3161.a,OfNat.ofNatProof uses:
Eq.refl2 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem a_4 : a 4 = 36 := by rfl …
Mentions: OeisA3161.a
- theorem
OeisA3161.conjectureLet $b(n) = a(2n-1)$ sorry line 70 sourceSource documentation
Let $b(n) = a(2n-1)$. Then the supercongruence $b(n p^k) \equiv b(n p^{k-1}) \pmod{p^{3k}}$ holds for positive integers $n$ and $k$ and all primes $p \ge 5$.
- Zhi-Wei Sun, Nov 16 2019
depends on sorry what this means
Statement uses:
Nat,LT.lt,OfNat.ofNat,Nat.Prime,LE.le,Int.ModEq,HPow.hPow,Int,HMul.hMul, OeisA3161.b,HSub.hSubProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous28 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem conjecture (n k p : ℕ) (hn : 0 < n) (hk : 0 < k) (hp : p.Prime) (hp_ge : 5 ≤ p) : (b (n * p ^ k) : ℤ) ≡ (b (n * p ^ (k - 1)) : ℤ) [ZMOD (p : ℤ) ^ (3 * k)] := by …Mentions: OeisA3161.b