1,2

a(n) is the minimal worst-case number of weighings needed by an algorithm to sort a set of n coins of two possible weights into heavy vs. light coins by using a balance; additional known good (heavy, say) coins are available (esp. to distinguish "all heavy" from "all light")

It is known that ceiling(n*log_3(2) + log_3(135/128)) <= a(n) <= ceiling(7n/11) for all n except for n=3. This implies 19 <= a(30) <= 20 and in fact a(n) = ceil(7n/11) for several n <= 187, esp. for all n with n<=50 except n=3, n=30, n=41, n=49.

An-Ping Li, A note on the counterfeit coins problem, arXiv:0902.0841v8 [math.CO]

For n=1, compare the given coin A with a known heavy coin X. If A<X, then A is a light coin; if A=X, then A is a heavy coin; the outcome A>X is not possible. Since one weighing was needed, we have a(1)=1.

For n=3, to sort coins A,B,C, one optimal algorithm is: First compare A:B. If A<B, we know that A is light and B is heavy and can find out about C by comparing e.g. B:C in a second weighing. The case A>B is symmetric. If however A=B, compare A:C. If A<C, we know that A,B are light and C is heavy and vice versa for A>C. The worst case is A=C, which requires a third weighing, e.g. A:X, against a known heavy coin X. Since no algorithm exists that never uses more than 2 weighings, we have a(3) = 3.

Cf. A156301. [From Jonathan Vos Post (jvospost3(AT)gmail.com), May 18 2009]

Sequence in context: A071823 A139338 A057355 this_sequence A055930 A079952 A090638

Adjacent sequences: A160508 A160509 A160510 this_sequence A160512 A160513 A160514

hard,more,nonn

Hagen von Eitzen (math(AT)von-eitzen.de), May 16 2009