

A259667


Catalan numbers mod 6.


5



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
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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^k1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k1) = 5? Otherwise said, is a(2^k1) = 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.


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, n1, 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: A222637 A190950 A159985 * A193083 A146103 A245172
Adjacent sequences: A259664 A259665 A259666 * A259668 A259669 A259670


KEYWORD

nonn


AUTHOR

M. F. Hasler, Nov 08 2015


STATUS

approved



