A259667 Catalan numbers mod 6.

1, 1, 2, 5, 2, 0, 0, 3, 2, 2, 2, 4, 4, 4, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 1, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 4, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 2, 2, 2, 4

Offset: 0

Author

M. F. Hasler, Nov 08 2015

Keywords

• nonn

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.

Crossrefs

Cf. A000108, A004642, A038003.

Programs

Mathematica

Mod[CatalanNumber[Range[0,120]],6] (* Harvey P. Dale, Oct 24 2020 *)

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.

Formula

a(n) = A000108(n) mod 6.