A078415 - OEIS

%I #21 Oct 18 2014 05:03:07

%S 1,6,9,18,162,261,4376,19712,32805,65610,131220,4785156,9570312,

%T 272629962,1208614932,2542645806624,154206526918656,2348694485729280,

%U 9341451062288388,18049789376104448,451521135633235968

%N Let f(n) = fraction of digits that are nonzero when n is written in base 2 and g(n) the same fraction for base 3. Let h(n) = max {f(n), g(n)}. Sequence gives n for which h(n) sets a new low record.

%C Suggested by the following question from Andreas Weingartner. Prove or disprove: there exists an epsilon>0 such that no natural number has the property that in base 2 as well as in base 3 at most (epsilon)*100% of the digits are nonzero.

%H Robert Harley and Benne de Weger, <a href="/A078415/b078415.txt">Table of n, a(n) for n = 1..23</a> (terms 1 through 21 were computed by Robert Harley; terms 22 and 23 by Benne de Weger, Oct 16 2014)

%F Heuristically one might expect (except possibly for some small examples) such an eps for bases a and b at the solution of: (log(a-1)*log(b)+log(b-1)*log(a))*eps - log(a)*log(b) = (eps*log(eps)+(1-eps)*log(1-eps))*(log(a)+log(b)). For bases 2 and 3 this is about 0.131737...  (Formula and value for eps corrected by _Benne de Weger_, Oct 13 2014)

%e 18 = 10010 in base 2, f(18) = 2/5, 18 = 200 in base 3, g(18) = 1/3, h(n) = max {2/5, 1/3} = 2/5 and this is a new (low) record.

%K nonn,base

%O 1,2

%A Robert Harley, Dec 28 2002