|
Search: id:A137993
|
|
|
| A137993 |
|
A014138 (= partial sums of Catalan numbers starting with 1,2,5) mod 3. |
|
+0
1
|
|
| 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; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
As usual, "mod 3" means to chose 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.
|
|
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.
|
|
PROGRAM
|
(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 this_sequence A059883 A086967 A098490
Adjacent sequences: A137990 A137991 A137992 this_sequence A137994 A137995 A137996
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
M. F. Hasler (MHasler(AT)univ-ag.fr), Mar 16 2008
|
|
|
Search completed in 0.002 seconds
|
