Created
December 3, 2016 18:05
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The complete works of Shakespeare in binary have length | |
s = 8 * 5.3M = 8 * 5458199 = 43665592 | |
Let S be the set of numbers that contain the complete works of Shakespeare in binary. | |
If N is the number of occurrences of the complete works of Shakespeare, N is approximately | |
Poisson distributed with mean | |
E(N) = (lg n - s) 2^-s = mu | |
where n is viewed as a random number with the given number of digits. We have | |
Pr(n not in S) = Pr(N = 0) | |
= e^-mu | |
= exp(-(log n / log 2 - s) 2^-s) | |
= exp(-log n * 2^-s / log 2 ) exp(s 2^-s / log 2) | |
= exp(-A log n) B | |
= n^-A B | |
where | |
A = 2^-s / log 2 | |
B = exp(s 2^-s / log 2) ~ 1 | |
The harmonic sum of -S is about | |
H = sum_{n not in S} 1 / n | |
~ sum_n Pr(n not in S) / n | |
~ sum_n n^-A / n | |
~ sum_n n^-(1 + A) | |
~ int_t t^-(1 + A) | |
~ t^-A / A ]_1^inf | |
~ 1 / A | |
= 1 / (2^-s / log 2) | |
~ 2^s log 2 | |
log10 H ~ log10 (2^s) + log10 (log 2) | |
= s log10 2 + log10 (log 2) | |
= 13144652.811250633 | |
H ~ 6e13144652 | |
~ 10^(10^7.1) |
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