# Find $N$, in the decimal expansion of the large number $N=4^{4^{4^4}}$

Find $N$, in the decimal expansion of the large number

$$N=4^{4^{4^4}}$$

Following on from the question I posted yesterday about finding the number of digits

I now wanted to find the $N$ (decimal expansion itself).

could I use this formula possibly?

$$\sum_{i=1}^\infty 10^{-i} d_i$$

I wanted to work out $N$ because I was faced with the next part of the question

Say a robot could type $10$ billion digits a second! Find the time $T_n$, in years to type out the number $N$ in the previous part of this question.

I don't know how to go about calculating this..

any help is appreciated :)

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What exactly is the question? Not the number of digits? Because the "next part of the question" seems to be related only to the number of digits. – TMM Dec 28 '12 at 17:54
the question now is ''Say a robot could type 10 billion digits a second! Find the time Tn, in years to type out the number N in the previous part of this question.'' but obviously to solve this problem N needs to be found which is the decimal expansion of 4^4^4^4. The number of digits was the first part of this entire question which is now done with. :) – Anona anon Dec 28 '12 at 17:57
I took a shot at tex-ing the formula, but I was reluctant to guess at the weird characters that are rendering on my screen. Hopefully someone can finish it off! – rschwieb Dec 28 '12 at 18:09
@Anonaanon Huh? Can't you answer that question by knowing how many digits it has? Divide by 10 billion to get the number of seconds, divide by 60 to get the number of minutes, etc. – Mike Dec 28 '12 at 21:55
yep, done! thank you :) – Anona anon Dec 28 '12 at 21:56

Given that you know approximately the number of digits $D$, and that there are about $3.16 \cdot 10^7$ seconds in a year, it would take $\frac D{3.16 \cdot 10^{17}}$ years
@Anonaanon: No, the robot is writing digits at the given rate so we need to know how many digits it needs to write. If the number had $10^9$ digits, it would take $0.1$ seconds. – Ross Millikan Dec 28 '12 at 18:14