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A137993 A014138 (= partial sums of Catalan numbers starting with 1,2,5) mod 3. 2
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 (list; graph; refs; listen; history; text; internal format)
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 * A282778 A059883 A086967

Adjacent sequences:  A137990 A137991 A137992 * A137994 A137995 A137996

KEYWORD

easy,nonn

AUTHOR

M. F. Hasler, Mar 16 2008

STATUS

approved

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Last modified August 20 18:56 EDT 2019. Contains 326154 sequences. (Running on oeis4.)