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

franktaw at netscape.net franktaw at netscape.net
Mon Jul 24 23:05:07 CEST 2006

b^(2/(b-1)) = 4 for b=2, 3 for b=3, and is not an integer for any 
larger b.  Obviously, these two cases do not produce infinitely many 
equalities, since the sum of the digits is 1 in every case.  So there 
are probably no integer values of b and k that produce infinitely many 

One caveat here: this analysis doesn't properly cover the case where 
every prime divisor of b also divides k.  E.g., b=12, k=6 (which a 
quick look suggests has no solutions).

But let's look at 10^(2/9).  For which values of n does floor(10^(2/9 
n)) have digit sum equal to n?  Similarly for ceiling and round.  A 
quick check check up to n = 50 finds only 1,2 for floor, 14,21,23,47,50 
for ceiling, and 14,21,23 for round.

(Obviously there are no solutions for n a multiple of 9, since the 
digit sum for such cases is 1.)

Franklin T. Adams-Watters

-----Original Message-----
From: David Wilson <davidwwilson at comcast.net>

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).
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+.

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