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Here, a lower and upper bound for the $n$-th prime are given.

Applying the given bounds

$$n(\ln(n\cdot\ln(n))-1)<p_n<n\cdot\ln(n\cdot\ln(n))$$

and the approximation

$$p_n\approx n(\ln(n\cdot\ln(n))-1)+\frac{n(\ln(\ln(n))-2)}{\ln(n)}$$

we get that $p_{10^{100}}$ is somewhere between $2.346977\cdot 10^{102}$ and $2.35698\cdot 10^{102}$ and approximately $2.3471\cdot 10^{102}$ , so it has $103$ digits.

How many digits can we determine of the googol-th prime with the known methods ?

It is likely that this calculation was already done. In this case, a reference would be nice. (Please present also the result, not only the link).

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  • $\begingroup$ Suggestion: take base 2, as you can add the last one (1). In base 10 related to given approximations at least 2 but not much more than 3 using some other. $\endgroup$ – z100 Jul 4 '16 at 16:28
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    $\begingroup$ A closely related thread $\endgroup$ – Jyrki Lahtonen Jul 4 '16 at 21:07
  • $\begingroup$ $$23471257358657641780361359099363020719654224259786132807$$ $$26748270447868572569974505143186051873726070114$$ is the correct rounded solution of $Li(x)=10^{100}$ $\endgroup$ – Peter Jul 5 '16 at 6:43
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From this paper we have the inequality

$\frac{x}{\log(x)}(1 + \frac{1}{\log(x)} + \frac{2}{\log^2(x)}) < \pi(x) < \frac{x}{\log(x)}(1 + \frac{1}{\log(x)} + \frac{2.334}{\log^2(x)})$ for $x \ge 2 953 652 287$

which leads to $2.347112 \cdot 10^{102} < p_{10^{100}} < 2.347127 \cdot 10^{102}$.


EDIT: I just noticed that the same paper gives

$$p_k < k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right)$$ for $k \ge 688383$ and

$$p_k > k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2.1}{\log k}\right)$$ for $k \ge 3$ which yields $2.3471221 \cdot 10^{102} < p_{10^{100}} < 2.3471265 \cdot 10^{102}$, so six digits are determined.


EDIT 2: Thanks to DanaJ, I see that this 2013 paper by Axler gives the following bounds:

$p_k < k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right) - \frac{(\log(\log(k)))^2 - 6 \log(\log(k)) + 11.847}{(\log (k))^2}$ for $k \ge 2$ and

$p_k > k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right) - \frac{(\log(\log(k)))^2 - 6 \log(\log(k)) + 10.273}{(\log (k))^2}$ for $k \ge 8009824$

which yields $2.347125652 \cdot 10^{102} < p_{10^{100}} < 2.347125801 \cdot 10^{102}$, determining the first seven digits.


Note that, while this paper gives the better asymptotic bound

$|\pi(x) - \mathrm{li}(x)| < 0.2795\frac{x}{(\log (x))^{3/4}}\exp(-\sqrt{\frac{\log (x)}{6.455}})$ for $x \ge 229$

it only determines the first three digits of $p_{10^{100}}$.


Of course, if we assume the Riemann Hypothesis we can get many more digits. The bound

$|\pi(x) - \mathrm{li}(x)| < \frac{\sqrt{x} \log(x)}{8\pi}$ for $x \ge 2657$

will give

$$2.347125735865764178036135909936302071965422425975\cdot10^{102}<p_{10^{100}}<2.347125735865764178036135909936302071965422425983*10^{102}$$

so $47$ digits of $p_{10^{100}}$ are determined.

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    $\begingroup$ Axler 2013 (page viii Korollare G und I) gives tighter bounds vs. Dusart 2010. I believe this gives a seventh digit. $\endgroup$ – DanaJ Jul 5 '16 at 1:15
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    $\begingroup$ Yes, it certainly does! $\endgroup$ – Deedlit Jul 5 '16 at 1:34
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    $\begingroup$ That jump from the first $7$ digits to the first $47$ digits is quite spectacular and really shows the importance of the Riemann Hypothesis. $\endgroup$ – Simply Beautiful Art Aug 11 '16 at 0:57

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