A279205 Length of second run of 1's in binary representation of Catalan(n).

0, 0, 0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 3, 4, 1, 3, 2, 1, 6, 1, 2, 1, 4, 7, 5, 2, 3, 1, 4, 2, 1, 1, 5, 2, 1, 3, 1, 1, 3, 3, 3, 3, 8, 2, 1, 2, 2, 1, 3, 2, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 4, 1, 2, 4, 1, 2, 3, 1, 1, 1, 2, 1, 1, 5, 1, 1, 1, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 3, 2, 2, 1, 1

Offset

0,9

Comments

Suggested by A279026.

What combinatorial problem is this the answer to?

Links

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Example

A000108(13) = 742900_10 = A264663(13) = 10110101010111110100_2, so a(13) = 2.

Mathematica

Q = {};
Num = 100;
T = Table[IntegerDigits[CatalanNumber[n], 2], {n, 0, Num}];
For[i = 1, i <= Num, i++,
c = 0; j = 1;
While[T[[i]][[j]] == 1, j++];
While[T[[i]][[j]] == 0, j++];
c = j;
While[T[[i]][[j]] == 1, j++];
c = j - c;
AppendTo[Q, c]
];
Q (* Benedict W. J. Irwin, Dec 21 2016 *)
Join[{0, 0, 0, 1, 0}, Length[Split[IntegerDigits[#, 2]][[3]]]&/@ CatalanNumber[ Range[5, 100]]] (* Harvey P. Dale, Aug 20 2021 *)

Crossrefs

Cf. A000108, A264663, A279026, A279206.

Sequence in context: A022300 A347552 A300983 * A105690 A214364 A175922

Adjacent sequences: A279202 A279203 A279204 * A279206 A279207 A279208

Keyword

nonn,base

Author

N. J. A. Sloane, Dec 21 2016

Extensions

a(19) to a(99) from Benedict W. J. Irwin, Dec 21 2016

Status

approved