nines

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)