Shouldn't that be 4.5 log_10(2) n? Admittedly, 1.354635 is close to 1.359141. Rich ----- Quoting "W. Edwin Clark" <wclark@mail.usf.edu>:
S(n) is the sequence http://oeis.org/A001370 where it is asserted that S(n) ~ ne/2 Maple calculates the first 5000 terms in just over 9 seconds -- so experiments should be easy.
On Wed, Apr 17, 2013 at 3:46 PM, James Propp <jamespropp@gmail.com> wrote:
Let E(n) and V(n) be the expected value and variance, respectively, of the sum of n independent draws from {0,1,2,...,9}; we can easily write E(n) and V(n) in closed form, and thereby extend to E(x) and V(x) where x is any positive real number.
Let S(n) be the sum of the digits of 2^n, and let c = log_10 2, so that 2^n has about cn digits.
What if anything is known (empirically or rigorously) about the distribution and autocorrelations of the sequence D_n whose nth term is (S(n)-E(cn))/sqrt(V(cn))?
If that's obscure, just think of it as the sum of the digits of 2^n, adjusted to make it look like a Gaussian of mean 0 and variance 1. (The "D" is for discrepancy.)
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