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A064099 a(n) = ceiling(log(3 + 2*n)/log(3)).

%I A064099 #30 Oct 26 2015 18:01:18
%S A064099 1,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,
%T A064099 4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,
%U A064099 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5
%N A064099 a(n) = ceiling(log(3 + 2*n)/log(3)).
%C A064099 Minimal number of weighings to detect a heavier or lighter counterfeit coin among n coins.
%C A064099 The relation is given via the inverse (A003462) and the comments in A029858. - _R. J. Mathar_, Sep 10 2015
%D A064099 J. G. Mauldon, Strong solutions for the counterfeit coin problem. IBM Research Report RC 7476 (#31437) 9/15/78, IBM Thomas J. Watson Research Center, P. O. Box 218, Yorktown Heights, N. Y. 10598
%H A064099 Harry J. Smith, <a href="/A064099/b064099.txt">Table of n, a(n) for n=0,...,1000</a>
%H A064099 Gary Darby, <a href="http://delphiforfun.org/programs/counterfeitcoin.htm">The Counterfeit Coin</a>
%H A064099 Gary Darby, <a href="http://delphiforfun.org/programs/CounterfeitCoin_Gardner.htm">Martin Gardner and The Counterfeit Coin Problem</a>
%H A064099 M. Gardner, <a href="http://www.faqs.org/faqs/puzzles/archive/logic/part5/">logic/weighing/balance.s</a> on the counterfeit coin weighing.
%F A064099 a(n) = A134021(n+1). - _Reinhard Zumkeller_, Oct 19 2007
%e A064099 It would be nice to have some examples showing how the sequence is related to the coin problem! - _N. J. A. Sloane_, Jun 25 2002
%p A064099 A064099 := n->ceil(evalf(log(3+2*n)/log(3)));
%t A064099 Table[Ceiling[Log[3,3+2n]],{n,0,100}] (* _Harvey P. Dale_, Oct 26 2015 *)
%o A064099 (PARI) { for (n=0, 1000, write("b064099.txt", n, " ", ceil(log(3 + 2*n)/log(3))) ) } \\ _Harry J. Smith_, Sep 07 2009
%Y A064099 Cf. A003462 ((3^n-1)/2, the inverse).
%K A064099 nice,easy,nonn
%O A064099 0,2
%A A064099 Eugene McDonnell (EEMcD(AT)AOL.com), Sep 16 2001