A236859 The length of the initial ascent 123... in the n-th Catalan numeral, A239903(n).

0, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 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, 1, 1, 2, 2, 2, 2

Offset: 0

Author

Antti Karttunen, Apr 18 2014

Keywords

• nonn

Examples

			A239903(1) = 1, thus a(1) = 1.
A239903(2) = 10, thus a(2) = 1.
A239903(4) = 12, thus a(4) = 2.
A239903(39) = 1232, thus a(39) = 3.
A239903(58784) = 1234567899, thus a(58784) = 9.
Note that although the range of validity of A239903 is inherently limited by the decimal representation employed, it doesn't matter here: We have a(58785) = 10, as the corresponding 58785th Catalan String is [1,2,3,4,5,6,7,8,9,10], even though A239903 cannot represent that unambiguously.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16796

Crossrefs

Cf. A239903, A236855, A081291, A126307.

Programs

Scheme

(define (A236859 n) (if (zero? n) n (- (A126307 (A081291 n)) 1)))

Formula

a(0) = 0, and for n>=1, a(n) = A126307(A081291(n))-1.

Each n occurs for the first time (as a record) at the position (C_{n+1})-1, so we have a(A001453(n+1)) = n for all n.