69923.lean`

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

6 declarations, 1 contain sorry, 0 declared as axioms

Number of primes $p$ such that $2^n \le p \le 2^n + \mathrm{prime}(n)$

The sequence $a(n)$ counts the number of primes $p$ in the interval $[2^n, 2^n + p_n]$, where $p_n$ is the $n$-th prime ($p_1 = 2, p_2 = 3, \dots$): $$a(n) = |\{p \text{ prime} \mid 2^n \le p \le 2^n + p_n\}|$$ for $n \ge 1$, and $a(0) = 0$.

References:

A069923

Imports (1)

FormalConjecturesUtil

def OeisA69923.a Number of primes $p$ such that $2^n \le p \le 2^n + \mathrm{prime}(n)$ line 38 source
Source documentation
Number of primes $p$ such that $2^n \le p \le 2^n + \mathrm{prime}(n)$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat
Proof uses: ite, Eq, OfNat.ofNat, Nat.nth, Nat.Prime, HSub.hSub, Finset.card, Finset.filter, Finset.Icc, HPow.hPow, HAdd.hAdd
13 instances, coercions and auxiliary constants left out.
```
noncomputable def a (n : ℕ) : ℕ :=
…
```
theorem OeisA69923.a_1 Value of the sequence a at 1 line 46 source
Source documentation
Value of the sequence a at 1.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Nat, OeisA69923.a, OfNat.ofNat
Proof uses: id, Eq.mpr, Finset.card, Finset.filter, Nat.Prime, Finset.Icc, HPow.hPow, HAdd.hAdd, Nat.nth, congrArg, Nat.nth_prime_zero_eq_two, of_decide_eq_true, Bool, Decidable.decide, Bool.true, Eq.refl
12 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem a_1 : a 1 = 2 := by
…
```
Mentions: OeisA69923.a
theorem OeisA69923.a_2 Value of the sequence a at 2 line 53 source
Source documentation
Value of the sequence a at 2.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Nat, OeisA69923.a, OfNat.ofNat
Proof uses: id, Eq.mpr, Finset.card, Finset.filter, Nat.Prime, Finset.Icc, HPow.hPow, HAdd.hAdd, Nat.nth, congrArg, Nat.nth_prime_one_eq_three, of_decide_eq_true, Bool, Decidable.decide, Bool.true, Eq.refl
12 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem a_2 : a 2 = 2 := by
…
```
Mentions: OeisA69923.a
theorem OeisA69923.a_3 Value of the sequence a at 3 line 60 source
Source documentation
Value of the sequence a at 3.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Nat, OeisA69923.a, OfNat.ofNat
Proof uses: id, Eq.mpr, Finset.card, Finset.filter, Nat.Prime, Finset.Icc, HPow.hPow, HAdd.hAdd, Nat.nth, congrArg, Nat.nth_prime_two_eq_five, of_decide_eq_true, Bool, Decidable.decide, Bool.true, Eq.refl
12 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem a_3 : a 3 = 2 := by
…
```
Mentions: OeisA69923.a
theorem OeisA69923.a_4 Value of the sequence a at 4 line 67 source
Source documentation
Value of the sequence a at 4.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Eq, Nat, OeisA69923.a, OfNat.ofNat
Proof uses: id, Eq.mpr, Finset.card, Finset.filter, Nat.Prime, Finset.Icc, HPow.hPow, HAdd.hAdd, Nat.nth, congrArg, Nat.nth_prime_three_eq_seven, of_decide_eq_true, Bool, Decidable.decide, Bool.true, Eq.refl
12 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem a_4 : a 4 = 3 := by
…
```
Mentions: OeisA69923.a
theorem OeisA69923.conjecture For any $n > 0$, is there always at least one prime $p$ such that $2^n \le p \le 2^n + \mathrm{prime}(n)$? sorry line 79 source
Source documentation
For any $n > 0$, is there always at least one prime $p$ such that $2^n \le p \le 2^n + \mathrm{prime}(n)$? (checked up to $n = 250$). In this case, that would be stronger than the Schinzel conjecture: "for $m > 1$ there's at least one prime $p$ such that $m \le p \le m + \log(m)^2$" since, for $n > 2$, $\mathrm{prime}(n) < \log(2^n)^2 = n^2 \log(2)$.

depends on sorry what this means
Statement uses: Nat, LT.lt, OfNat.ofNat, LE.le, OeisA69923.a
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
6 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture (n : ℕ) (hn : 0 < n) : 1 ≤ a n := by
…
```
Mentions: OeisA69923.a