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I found a question about the first digits of Graham's number and would like to generalize it :

We want the first n digits of the number $a\uparrow^b c$.

  • Which method is the most effective to do that ?
  • For which magnitude of a,b,c is the calculation feasible ?
  • Is there a hope to calculate at least the first digit of Graham's number ?

If the only way to find the first digits is to simply calculate the number, then already $3\uparrow \uparrow \uparrow 3$ would be out of reach.

I tried to find out some patterns of the leading digits of powers, but there seems to be no structure.

I know, that this question is somewhat broad, nevertheless I hope for some useful hints.

Community
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Peter
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  • If someone can make the question more precise without changing my intention, he/she is invited to edit it. – Peter Jan 09 '14 at 17:03

1 Answers1

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The first decimal digits of number "n" are dependent on fractional part of "x" in its representation as n = 10^x. If "x" is irrational and has too much digits before decimal point, we can't access to its digits which are past decimal point. So we can't calculate leading decimal digits of Graham's number and even $3\uparrow \uparrow \uparrow 3$ in any reasonable time. More briefly:

*The best known method for computing the first digits of a^b is representing it as $10^{b*log_{10}(a)}$, finding fractional part of $b*log_{10}(a) = c$ and computing $10^c$. That should be faster than direct computing.

*We can compute $a \uparrow^{b} c$ for b = 1 (exponentiation) for quite large bases and exponents. Say, we can calculate that $3 \uparrow 7625597484987 = 1258014...$

For b = 2 (tetration), it can be computed only for small c ($2 \uparrow\uparrow 7$'s first digits are already inaccessible).

*No hope. Explained above.

Ikosarakt
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  • Could you quantify a little more what you mean in the second sentence, and perhaps generalized the discussion to the more general case of $a \uparrow^b c$? – Hayden Sep 30 '14 at 10:24
  • By too much I mean that the number of digits overflows computer's memory. Since, say, $3 \uparrow^3 3$ has non-representable number of digits, it's leading digits are inaccessible too. – Ikosarakt Sep 30 '14 at 10:57