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

Author

M. F. Hasler, Mar 16 2008

Keywords

• easy
• nonn

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.

Crossrefs

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

Programs

PARI

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

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.