(sum-of-digits of 5^n)/n approximates pi

[David Wilson]

Let b >= 2, k >= 2. For x >= 1, Let r_b(x) be the standard base-b representation of the integer part of x. Let d_b(x) and s_b(x) be the number and sum, respectively, of the digits of r_b(x).

d_b(x) = floor(log(x)/log(b))+1.  If gcd(k, b) = 1, empirical evidence suggests that the digits of r_b(k^n) for large n are equidistributed, which, if true, implies that s_b(k^n)/d_b(k^n) approaches (b-1)/2, the average of the base-b digits, as n grows without bound. This in turn implies that

          lim  s_b(k^n)/n = ((b-1)/2)(log(k)/log(b)).
        n->inf

When b = 10 and k = 5, this value is (9/2)(log(5)/log(10)) = 3.14536...  Close, but no pi.

We expect s_b(k^n) = n to have an infinite number of solutions exactly when ((b-1)/2)(log(k)/log(b)) = 1, that is, k = b^(2/(b-1)), and modular considerations do not preclude s_b(k^n) = n.  For all other k, we expect s_b(k^n) = n to have a finite number of solutions (e.g, when b = 10 and k = 2, which was Tanya's original problem).  For base b = 10, we expect s_10(k^n) = n to have an infinite number of solutions only when k = 10^(2/9) = 1.6681+.

----- Original Message ----- 
  From: Hans Havermann 
  To: seqfan at ext.jussieu.fr 
  Sent: Monday, July 24, 2006 7:09 AM
  Subject: (sum-of-digits of 5^n)/n approximates pi


    Each n provides a value of (sum-of-digits of 5^n)/n that is closer to pi than the previous value. I'm guessing there aren't many more terms.


  I let this run overnight and did get two more terms: {1, 2, 4, 6, 8, 139, 309, 390, 819, 2868, 6751, 8045, 9414, 15008, 15375, 56839, 84383, 151286, ...}


  (sum-of-digits of 5^151286)/151286 is about 3.141592745.


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