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Erdős Problem 933

References:

    erdosproblems.com/933

    [Er76d] Erdős, P, Problems and results on number theoretic properties of consecutive integers and related questions. Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1975) (1976), 25-44.

open Filternamespace Erdos933

The 2-adic valuation of $n(n+1)$.

def k (n : ) : := padicValNat 2 (n * (n + 1))

The 3-adic valuation of $n(n+1)$.

def l (n : ) : := padicValNat 3 (n * (n + 1))

If $n(n+1)=2^k3^lm$, where $(m,6)=1$, then is it true that $\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty$?

@[category research open, AMS 11] theorem erdos_933 : answer(sorry) atTop.limsup (fun n : ((((2 ^ k n * 3 ^ l n : ) : ) / ((n : ) * Real.log (n : ))) : EReal)) = := True limsup (fun n (2 ^ k n * 3 ^ l n) / (n * (Real.log n))) atTop = All goals completed! 🐙

Mahler proved (a more general result that implies in particular) that $2^k3^l<n^{1+o(1)}$.

@[category research solved, AMS 11] theorem erdos_933.variants.mahler : c : , (c =o[atTop] (1 : )) ∀ᶠ n in atTop, ((2 ^ k n * 3 ^ l n : ) : ) < (n : ) ^ (1 + c n) := c, c =o[atTop] 1 ∀ᶠ (n : ) in atTop, (2 ^ k n * 3 ^ l n) < n ^ (1 + c n) All goals completed! 🐙

Erdős [Er76d] wrote 'it is easy to see' that for infinitely many $n$, $2^k 3^l > n\log n$.

Steinerberger has noted a simple proof of this fact follows from taking $n=2^{3^r}$ for any integer $r\geq 1$, when $k=3^r$ and $l=r+1$.

r:m: := 3 ^ (r + 1)hk:k (2 ^ m) = mhl:l (2 ^ m) = r + 2hlog:m * Real.log 2 < 3 ^ (r + 2)2 ^ m * 3 ^ (r + 2) > 2 ^ m * (m * Real.log 2) exact mul_lt_mul_of_pos_left hlog (r:m: := 3 ^ (r + 1)hk:k (2 ^ m) = mhl:l (2 ^ m) = r + 2hlog:m * Real.log 2 < 3 ^ (r + 2)0 < 2 ^ m All goals completed! 🐙)end Erdos933