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Search: id:A078415
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| A078415 |
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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. |
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+0
1
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| 1, 6, 9, 18, 162, 261, 4376, 19712, 32805, 65610, 131220, 4785156, 9570312, 272629962, 1208614932, 2542645806624, 154206526918656, 2348694485729280, 9341451062288388, 18049789376104448, 451521135633235968
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OFFSET
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1,2
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COMMENT
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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.
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FORMULA
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Heuristically one might expect (except possibly for some small examples) such an eps for bases a and b at the solution of: (2*eps-1)*log(a)*log(b) = (eps*log(eps)+(1-eps)*log(1-eps))*(log(a)+log(b)). For bases 2 and 3 this is about 0.104939...
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EXAMPLE
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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.
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CROSSREFS
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Sequence in context: A039280 A045091 A086955 this_sequence A023041 A118277 A103186
Adjacent sequences: A078412 A078413 A078414 this_sequence A078416 A078417 A078418
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KEYWORD
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nonn
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AUTHOR
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Robert Harley, Dec 28 2002
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