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%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
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