51903.lean`

Explanation legend Lean4 Domain Explanation Language (L4DEL)

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L4DEL: Lean4 Domain Explanation Language

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

9 declarations, 3 contain sorry, 0 declared as axioms

Maximum exponent in the prime factorization of $n$

References:

A051903

Imports (1)

FormalConjecturesUtil

def OeisA51903.a Maximum exponent in the prime factorization of $n$ line 31 source
Source documentation
Maximum exponent in the prime factorization of $n$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat
Proof uses: List.foldr, Max.max, OfNat.ofNat, List.map, List.count, Nat.primeFactorsList
4 instances, coercions and auxiliary constants left out.
```
def a (n : ℕ) : ℕ :=
…
```
theorem OeisA51903.a_1 line 35 source
extra axioms: _private.0.OeisA51903.a_1._native.decide.ax_1 what this means
Statement uses: Eq, Nat, OeisA51903.a, OfNat.ofNat
Proof uses: of_decide_eq_true
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA51903.a_2 line 39 source
extra axioms: _private.0.OeisA51903.a_2._native.decide.ax_1 what this means
Statement uses: Eq, Nat, OeisA51903.a, OfNat.ofNat
Proof uses: of_decide_eq_true
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA51903.a_3 line 43 source
extra axioms: _private.0.OeisA51903.a_3._native.decide.ax_1 what this means
Statement uses: Eq, Nat, OeisA51903.a, OfNat.ofNat
Proof uses: of_decide_eq_true
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA51903.a_4 line 47 source
extra axioms: _private.0.OeisA51903.a_4._native.decide.ax_1 what this means
Statement uses: Eq, Nat, OeisA51903.a, OfNat.ofNat
Proof uses: of_decide_eq_true
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA51903.a_5 line 51 source
extra axioms: _private.0.OeisA51903.a_5._native.decide.ax_1 what this means
Statement uses: Eq, Nat, OeisA51903.a, OfNat.ofNat
Proof uses: of_decide_eq_true
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem OeisA51903.conjecture1 Are there composite numbers $n > 4$ such that $n \equiv a(n) \pmod{\phi(n)}$? sorry line 69 source
Source documentation
Are there composite numbers $n > 4$ such that $n \equiv a(n) \pmod{\phi(n)}$?
- Thomas Ordowski, Dec 02 2019
This question is equivalent to Lehmer's totient problem LehmerTotient.lehmer_totient; a positive answer here falsifies the universal statement asked about in Erdos828.erdos_828.variants.lehmer_conjecture. Any composite $n$ with $\phi(n) \mid n - 1$ is squarefree, so $a(n) = 1$ and the condition here holds. Conversely, let $n > 4$ be composite with $\phi(n) \mid n - a(n)$ and $e = a(n) \ge 2$, and pick $p$ with $p^e \mid n$. Then $p^{e-1} \mid \phi(n) \mid n - e$ and $p^{e-1} \mid n$, so $p^{e-1} \mid e$, which forces $p = 2$ and $e = 2$. Now $n = 4m$ with $m$ odd and squarefree, and $2\phi(m) \mid 4m - 2$ makes $\phi(m)$ odd, so $m = 1$ and $n = 4$. Hence $e = 1$ and the condition is $\phi(n) \mid n - 1$.
depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, LT.lt, OfNat.ofNat, Not, Nat.Prime, Dvd.dvd, Nat.totient, HSub.hSub, OeisA51903.a
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
10 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture1 :
    answer(sorry) ↔ ∃ n : ℕ, 4 < n ∧ ¬ n.Prime ∧ n.totient ∣ (n - a n) := by
…
```
Mentions: OeisA51903.a
theorem OeisA51903.conjecture2 Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\lambda(n)}$? sorry line 79 source
Source documentation
Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\lambda(n)}$? (Equivalently, odd numbers $n$ such that $a(n) > 1$ and $b^n \equiv b^{a(n)} \pmod n$ for all $b$.)
- Thomas Ordowski, Dec 02 2019
depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, Odd, LT.lt, OfNat.ofNat, OeisA51903.a, Nat.ModEq, HPow.hPow
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
14 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture2 :
    answer(sorry) ↔ ∃ n : ℕ, Odd n ∧ 1 < a n ∧ ∀ b : ℕ, b ^ n ≡ b ^ (a n) [MOD n] := by
…
```
Mentions: OeisA51903.a
theorem OeisA51903.conjecture3 Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\operatorname{ord}_n(2)}$? sorry line 89 source
Source documentation
Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\operatorname{ord}_n(2)}$? (Equivalently, odd numbers $n$ such that $a(n) > 1$ and $2^n \equiv 2^{a(n)} \pmod n$.)
- Thomas Ordowski, Dec 02 2019
depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, Odd, LT.lt, OfNat.ofNat, OeisA51903.a, Nat.ModEq, HPow.hPow
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
14 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture3 :
    answer(sorry) ↔ ∃ n : ℕ, Odd n ∧ 1 < a n ∧ 2 ^ n ≡ 2 ^ (a n) [MOD n] := by
…
```
Mentions: OeisA51903.a