[seqfan] Re: A000108(n) ≡ 1 (mod 6)

L. Edson Jeffery lejeffery2 at gmail.com
Wed Dec 2 22:03:26 CET 2015


Let C(n) denote the n-th Catalan number. I think it was Deutsch and Sagan
who proved that binomial(2n,n) == 0 (mod 3) if and only if the base 3
representation of n contains at least one 2. Can this result be extended in
some way to prove the following conjecture?

Conjecture 1: C(n) == 0 (mod 3) if and only if the base 3 representations
of both n and n+1 contain at least one 2.

If Conjecture 1 is true, then it should be enough to prove, for all m>8,
that the base 3 representations of both 2^m - 1 and 2^m contain at least
one 2, from which William Keith's conjecture that C(2^m-1) == 0 (mod 3)
would then follow.

Ed Jeffery


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