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import FormalConjecturesUtilNumber of primes $p$ such that $n^n \le p \le n^n + n^2$
The sequence $a(n)$ counts the number of prime numbers in the interval $[n^n, n^n + n^2]$: $$a(n) = |{p \text{ prime} \mid n^n \le p \le n^n + n^2}|$$
References:
namespace OeisA69922open FinsetNumber of primes $p$ such that $n^n \le p \le n^n + n^2$.
def a (n : ℕ) : ℕ :=
((Icc (n ^ n) (n ^ n + n ^ 2)).filter Nat.Prime).card
Value of the sequence a at 1.
@[category test, AMS 11]
theorem a_1 : a 1 = 1 := ⊢ a 1 = 1
All goals completed! 🐙
Value of the sequence a at 2.
@[category test, AMS 11]
theorem a_2 : a 2 = 2 := ⊢ a 2 = 2
All goals completed! 🐙
Value of the sequence a at 3.
@[category test, AMS 11]
theorem a_3 : a 3 = 2 := ⊢ a 3 = 2
All goals completed! 🐙
Value of the sequence a at 4.
@[category test, AMS 11]
theorem a_4 : a 4 = 4 := ⊢ a 4 = 4
All goals completed! 🐙
Value of the sequence a at 5.
@[category test, AMS 11]
theorem a_5 : a 5 = 1 := ⊢ a 5 = 1
All goals completed! 🐙Question: for any $n > 0$, is there at least one prime $p$ such that $n^n \le p \le n^n + n^2$? 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 $n^2 < \log(n^n)^2 = n^2 \log(n)^2$.
@[category research open, AMS 11]
theorem conjecture (n : ℕ) (hn : 0 < n) : 1 ≤ a n := n:ℕhn:0 < n⊢ 1 ≤ a n
All goals completed! 🐙end OeisA69922