OFFSET
0,3
COMMENTS
The only odd terms are those with indices n = 2^k - 1 (k = 0, 1, 2, 3, ...); see also A038003.
It is conjectured that the only k which yield a(2^k-1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k-1) = 5? Otherwise said, is a(2^k-1) = 3 for all k > 8?
The question is equivalent to: does 2^k - 1 always contain a digit 2 when converted into base 3 for all k > 8? Similar conjecture has been proposed for 2^k, see A004642. - Jianing Song, Sep 04 2018
LINKS
M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (v.1.4, 11/2015).
V. Reshetnikov, A000108(n) ≡ 1 (mod 6), SeqFan list, Nov. 8, 2015.
FORMULA
a(n) = A000108(n) mod 6.
MATHEMATICA
Mod[CatalanNumber[Range[0, 120]], 6] (* Harvey P. Dale, Oct 24 2020 *)
PROG
(PARI) a(n)=binomial(2*n, n)/(n+1)%6
(PARI) A259667(n)=lift(if(n%3!=1, binomod(2*n+1, n, 6)/(2*n+1), if(bittest(n, 0), binomod(2*n, n-1, 6)/n, binomod(2*n, n, 6)/(n+1)))) \\ using binomod.gp by M. Alekseyev, cf. Links.
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 08 2015
STATUS
approved