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

0,3

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

Table of n, a(n) for n=0..119.

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

Cf. A000108, A004642, A038003.

Sequence in context: A190950 A355676 A159985 * A321216 A193083 A146103

Adjacent sequences: A259664 A259665 A259666 * A259668 A259669 A259670

Keyword

nonn

Author

M. F. Hasler, Nov 08 2015

Status

approved