[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[L. Edson Jeffery]

>I think it should be
>binomial(2n-2, n-1) = binomial(2^(m+1)-4, 2^m-2)
>instead of
>binomial(2n-2, n-1) = binomial(2^(m+1)-4, 2^m-1)

Emmanuel,

C(n) = (1/(n+1))*binomial(2*n, n),

so

C(2^m-1) = (1/2^m)*binomial(2^(m+1)-2, 2^m-1).

Ed Jeffery