[seqfan] Re: A000108(n) ≡ 1 (mod 6)
israel at math.ubc.ca
israel at math.ubc.ca
Thu Dec 3 01:19:53 CET 2015
The fact that binomial(2n,n) == 0 (mod 3) iff the base 3 representation of
n contains at least one 2 follows directly from Kummer's theorem on
binomial coefficients.
Since C(n) = binomial(2n,n) - binomial(2n,n-1), Kummer's theorem says that
a necessary condition for C(n) == 0 (mod 3) is that either both or neither
of the additions 2n = n + n and 2n = (n+1) + (n-1) in base 3 have carries.
A sufficient condition is that both have carries and they have a different
number of carries.
Cheers,
Robert
On Dec 2 2015, L. Edson Jeffery wrote:
>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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