User talk:Thomas Ordowski

A005234

I am curious how you came by your conjecture, and how large of a number set you have checked against so far.--Bill McEachen 03:38, 11 November 2013 (UTC)

A182514

I am looking this:

The inequality in the definition is asymptotically close to the inequality prime(n+1)-prime(n) > log(n)*log(prime(n)) for sufficiently large n. - Thomas Ordowski, Mar 16 2015

and wondering what does this mean?

((prime(n+1)/prime(n))^n - n) ~ (prime(n+1)-prime(n) - log(n)*log(prime(n)))?

((prime(n+1)/prime(n))^n / n) ~ (prime(n+1)-prime(n)) / (log(n)*log(prime(n)))?

Or something else?

A001223

I see the following two comments:

There exists a constant C such that for n -> infinity, Cramer conjecture a(n) < C log^2 prime(n) is equivalent to (log prime(n+1)/log prime(n))^n < e^C. - Thomas Ordowski, Oct 11 2014

lim sup_{n -> infinity}a(n)/log^2 prime(n) = C <==> lim sup_{n -> infinity}(log prime(n+1)/log prime(n))^n = e^C. - Thomas Ordowski, Mar 09 2015

What is the difference between these statements?

Proof?

John W. Nicholson 02:40, 8 May 2015 (UTC)

A024675

I found your conjecture of interest. Unless I'm mistaken, the conjecture is a bit broader - it applies to every pair of consecutive odd numbers as well. I can supply the Pari code I used.--Bill McEachen (talk) 21:19, 22 February 2018 (EST)

Cf. Andrica's conjecture.