Does anyone have any ideas how to compute the first and last (before decimal point, if it is transcendental) digit of Skewes' number? Ikosarakt1 (talk) 13:15, December 20, 2012 (UTC)

This article is in large part false. There is infinitely many n's such that \(\pi(n)<li(n)\), but also infinitely many for which \(\pi(n)>li(n)\). Exact formulation is that this inequality switches direction infinitely many times. Say largest number before first swap is \(a\). Assuming Riemann hypothesis \(a<Sk_1\). Inequality \(a<Sk_2\) holds independently of Riemann hypothesis. LittlePeng9 (talk) 17:59, May 5, 2013 (UTC)

Transcendence

Schanuel's conjecture would imply the transcendence of both Skewes Numbers:

Let a be equal to either 79 or 7.705.

Firstly, a is a nonzero algebraic number, so the Lindemann–Weierstrass theorem does imply the transcendence of e^a.

Secondly, a and e^a are linearly independent over Q, so Q(a, e^a, e^a, e^(e^a)) has transcendence degree of at least 2 over Q. Therefore, e^a and e^(e^a) are algebraically independent over Q.

Thirdly, a, e^a, and e^(e^a) are linearly independent over Q, so Q(a, e^a, e^(e^a), e^a, e^(e^a), e^(e^(e^a))) has transcendence degree of at least 3 over Q. Therefore, e^a, e^(e^a), and e^(e^(e^a)) are algebraically independent over Q. With a = 79, the transcendence of the first Skewes Number follows.

And fourthly, a, e^a, e^(e^a), and e^(e^(e^a)) are linearly independent over Q, so Q(a, e^a, e^(e^a), e^(e^(e^a)), e^a, e^(e^a), e^(e^(e^a)), e^(e^(e^(e^a)))) has transcendence degree of at least 4 over Q. Therefore, e^a, e^(e^a), e^(e^(e^a)), and e^(e^(e^(e^a))) are algebraically independent over Q. With a = 7.705, the transcendence of the second Skewes Number follows. --84.61.186.220 18:50, September 29, 2014 (UTC)

Is the given lower bound still accurate?

According to this webpage , values of the prime counting function up to 10^25 (also known as ten septillion, or a "minnow" in the guppy regiment) have been found.

Allegedly, π(10^25) has been proven to be 176,846,309,399,143,769,411,680.

whereas li(10^25) = 176,846,309,399,198,930,392,619.378663572790738111035705... (calculated here )

As you can see, li(10^25) > π(10^25) still.

If the information presented in the website given is true, either I just improved the lower bound or this article hasn't been updated in ages.

ArtismScrub (talk) 02:04, November 6, 2017 (UTC)

Even if we know \(li(10^{25}) > \pi(10^{25})\), we don't know whether the inequality isn't reversed for some number below it. The inequality has been verified for all numbers below \(10^{14}\), but only for very few numbers above it. LittlePeng9 (talk) 08:02, November 6, 2017 (UTC)
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