Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the millionth decimal digit of the $10^{10^{10^{10}}}$th prime?

(This prime is, of course, far larger than the largest currently "known" prime, the latter having nearly 13 million digits.) The answer should include a proof of correctness. I'm posting this question in the spirit of this advice, and will eventually post an answer (with proof of a more general result) if no one else does so.

NB: The notations 10^10^10^10 and $10^{10^{10^{10}}}$ mean the same as 10^(10^(10^10)), and the "millionth digit" means the millionth digit from the left, as usually written (i.e., the most significant digit is leftmost and is called the 1st digit).

Afterthought: It might have been more impressive to have asked for the $10^{10}$th digit of, say, the $10^{10^{10^{10^{10}}}}$th prime (that digit being 8), since perhaps no one has ever before computed the ten-billionth digit of any prime -- the currently "largest known" prime having only about 13 million digits.

share|improve this question
3  
Does $(10^{1000})$th prime number has million digits ? –  pedja Oct 28 '11 at 15:11
1  
I hope this has nothing to do with Benford's Law. –  user13838 Oct 28 '11 at 15:43
3  
The millionth most significant digit, or millionth least significant digit? –  user7530 Oct 28 '11 at 15:53
8  
Known upper and lower bounds for $p_n$ are in principle sharp enough to pin down the millionth digit. However, finding $\log n$ to sufficient accuracy may be difficult. –  André Nicolas Oct 28 '11 at 16:03
1  
@user7530: I've added a note that should make this clear. The intended digit is the millionth most-significant digit (i.e., counting from the left, as usually written, the leftmost digit being considered the 1st digit). –  r.e.s. Oct 28 '11 at 16:13
show 6 more comments

2 Answers

up vote 30 down vote accepted

The following bound on the $n$-th prime ($p_n$) is known: for $n > 6$, $$ n\left(\log{n} + \log\log{n} - 1\right) < p_n < n\left(\log{n} + \log\log{n}\right). $$ For $n=10^{10^{10^{10}}}$, we have $$\log{n} = 10^{10^{10}}\log{10},$$ a number with ten billion and one digits before the decimal point, and $$\log\log{n} = 10^{10}\log{10} + \log\log{10} \approx 23 025 850 931,$$ a number with eleven digits before the decimal point. Since the latter number is about ten billion digits shorter than the former, the millionth digit of $p_n$ is the same as the millionth digit of $n\log{n}$; that is, we can ignore the $n\log\log{n}$ correction$^{\dagger}$. But $$ n\log{n} = 10^{10^{10^{10}}}\cdot 10^{10^{10}}\log{10} = 10^{\left(10^{10^{10}} + 10^{10}\right)}\log{10} $$ is a large power of ten multiplied by $\log{10}$, and so its millionth digit is the same as the millionth digit of $\log{10}$ (i.e., the 999999-th digit after the decimal). This digit can be found in a number of places (e.g., at [numberworld.org]), and is equal to $5$.


$^\dagger$ This relies on our knowing that the digits after the millionth digit of $n\log{n}$ are not an enormous string of $9$'s. In fact, the next digit is $0$ (since that is the millionth digit of $\log{10}$ after the decimal), justifying this step.

share|improve this answer
add comment

A similar approach as in the accepted answer, but using a tighter bound from Dusart ...

Theorem: If $m$ is a positive integer, then the first $k-1$ digits of the $10^{10^m}$th prime are just those of $\log 10$, where $k$ is the greatest integer such that $(k \le \lfloor m - \log_{10} m \rfloor \wedge d_k \lt 9)$, and $d_k$ is the $k$th digit of $\log 10$.

Proof: Rosser proved that $p_n \gt n \log n$, and Dusart proved that $p_n \le (n \log n)(1 + r_n)$ for all $n \ge 39017$, where $r_n = \frac{\log \log n - 0.9484}{\log n} \gt 0$. Therefore,

$p_n = n \log n + (n \log n) \ \epsilon_n $ for all $n \ge 39017$, where $0 \lt \epsilon_n \le r_n$.

Now suppose $n = 10^{10^m}$, where $m$ is a positive integer ($m \ge 1$ ensures $n \ge 39017$, so Dusart's bound applies). Then

$n \log n = 10^{10^m + m} \log 10$

which, in base $10$, is just $\log 10$ with the decimal point shifted $10^m + m$ places to the right.

Now $\log \log 10 < 0.9484$, so

$r_n = 10^{-m} \ (m + \frac{\log \log 10 - 0.9484}{\log 10}) < 10^{-m} m \le 10^{-\lfloor m - \log_{10} m \rfloor}$,

and hence

$(n \log n) \ \epsilon_n < (n \log n) \ 10^{-\lfloor m - \log_{10} m \rfloor}$

where the right-hand side is seen to be, in base $10$, just $n \log n$ with the decimal point shifted $\lfloor m - \log_{10} m \rfloor$ places to the left.

Thus, if $k$ is the greatest integer satisfying both (1) $k \le \lfloor m - \log_{10} m \rfloor$, and (2) the $k$th digit is less than 9 (ensuring that no carry from the right can affect the $(k-1)$th digit), then adding $(n \log n) \ \epsilon_n$ to $n \log n$ does not affect the first $k-1$ digits of the latter term. QED

NB: This result generalizes to the $b^{b^{m}}$th prime, for any integer base $b$ such that $2 \le b \le 13$, and for integer $m \ge \log_b \log_b 39017$. The restrictions are to ensure that Dusart's bound can be applied.

Example: For any $m \ge 10^6 + 8$ (e.g., $m = 10^{10}$, as in the posted question), the first million digits of the $10^{10^m}$th prime are just those of $\log 10$. This is because for $m = 10^6 + 8$, $\lfloor m - \log_{10} m \rfloor = 10^6 + 1$, giving $k$ as the greatest integer such that $(k \le 10^6 + 1 \wedge d_k \lt 9)$. Direct computation (e.g., using Sage, which took less than a minute) shows that $k-1 = 10^6$, $d_{k-1} = 5$, $d_k = 0 \ (\lt 9)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.