A137993 A014138 (= partial sums of Catalan numbers starting with 1,2,5) mod 3.

1, 0, 2, 1, 1, 1, 1, 0, 2, 1, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 1, 2, 0, 1, 1, 1, 1, 2, 0, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0

Offset

0,3

Comments

As usual, "mod 3" means to choose the unique representative in { 0,1,2 } of the equivalence class modulo 3Z.

Here the conventions of A014138 are used, but it seems somehow unnatural to start with offset 0 corresponding to the Catalan number A000108(1).

For m>1, the length of the m-th block of nonzero elements (and thus the approximate length of the m-th string of consecutive 1's) is given by 2 A137822(m)-1.

Links

Table of n, a(n) for n=0..79.

Formula

a(n) = sum( k=1..n+1, C(k) ) (mod 3), where C(k) = binomial(2k,k)/(k+1) = A000108(k).

a(n) = 0 <=> n+1 = 2 A137821(m) for some m.

Prog

(PARI) A137993(n) = lift( sum( k=1, n+1, binomial( 2*k, k )/(k+1), Mod(0, 3) ))

Crossrefs

Cf. A014138, A000108, A137821-A137824, A107755, A137992, A014137(n+1) = a(n)+1 (mod 3).

Sequence in context: A135694 A025924 A025904 * A353329 A282778 A342788

Adjacent sequences: A137990 A137991 A137992 * A137994 A137995 A137996

Keyword

easy,nonn

Author

M. F. Hasler, Mar 16 2008

Status

approved