1,1

A set S is strongly triple-free if x in S implies 2x not in S and 3x not in S.

Conjecture: for k=1,2,3,..., a(6k+1)=2a(6k) and a(6k+5)=2a(6k+4) (these relations hold through a(35)). - John W. Layman (layman(AT)math.vt.edu), Jun 22 2002

Comment from Pradhan Prashanth Kumar (pradhan.ptr(AT)gmail.com), Feb 03 2008: The conjecture is true. Proof:

Let b(6k+1) = Number of strongly triple-free subsets of {1,2,...,6k+1} which do not contain 6k+1 and c(6k+1) = Number of strongly triple-free subsets of {1,2,...,6k+1} which contain 6k+1. Now a(6k+1) = b(6k+1) + c(6k+1) and b(6k+1) = a(6k).

1) c(6k+1)<=a(6k) : Take any strongly triple-free subset of {1,2,..,6k+1}, which contains 6k+1 and delete 6k+1. The new set is a subset of {1,2,...,6k} and is trongly triple-free. Hence c(6k+1)<=a(6k).

2) c(6k+1)>=a(6k) : Take any strongly triple-free subset of {1,2,...,6k}. Add 6k+1 to it. Since 6k+1 is not divisible by 2 or 3, this new set is still strongly triple-free. Hence c(6k+1)>=a(6k).

This shows that c(6k+1) = a(6k) and therefore a(6k+1) = b(6k+1)+c(6k+1) = 2a(6k). QED

Another proof for the conjecture: a(6k+r) = 2a(6k+r-1) when r={1,5} (with a(0)=1) would be: Any positive integer of form (6k+1) or (6k+5) is neither divisible by 2 nor by 3. Hence adding the number (6k+1) or (6k+5) to the each strongly triple-free subset of {1, ..., 6k} or {1, ..., 6k+4} does not violate the property and hence we would have 2a(6k) or 2a(6k+4) such subsets for a(6k+1) or a(6k+5). [From Ramasamy Chandramouli (thedavinci(AT)gmail.com), Aug 30 2008]

A068060 is the weakly triple-free analog of this sequence. [From S. R. Finch (Steven.Finch(AT)inria.fr), Mar 02 2009]

S. R. Finch, Triple-Free Sets of Integers

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Cf. A050291-A050296.

Sequence in context: A049861 A130476 A096592 this_sequence A121649 A030034 A093000

Adjacent sequences: A050292 A050293 A050294 this_sequence A050296 A050297 A050298

nonn

Eric Weisstein (eric(AT)weisstein.com)

More terms from John W. Layman (layman(AT)math.vt.edu), Jun 22 2002