A126306 a(n) = number of double-rises (UU-subsequences) in the n-th Dyck path encoded by A014486(n).

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

Offset

0,9

Links

Table of n, a(n) for n=0..101.

Formula

a(n) = A014081(A014486(n)).

a(n) = A000120(A048735(A014486(n))).

a(A125976(n)) = A057514(n)-1, for all n >= 1.

Example

A014486(20) = 228 (11100100 in binary), encodes the following Dyck path:
    /\
   /..\/\
  /......\
and there is one rising (left-hand side) slope with length 3 and one with length 1, so in the first slope, consisting of 3 U-steps, there are two cases with two consecutive U-steps (overlapping is allowed), thus a(20)=2.

Prog

(Python)
def ok(n):
    if n==0: return True
    B=bin(n)[2:] if n!=0 else '0'
    s=0
    for b in B:
        s+=1 if b=='1' else -1
        if s<0: return False
    return s==0
def a014081(n): return sum(((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1))
print([a014081(n) for n in range(4001) if ok(n)]) # Indranil Ghosh, Jun 13 2017

Crossrefs

Cf. A014081, A014486, A000120, A048735, A125976, A057514.

Sequence in context: A153246 A358006 A025889 * A287356 A029402 A330443

Adjacent sequences: A126303 A126304 A126305 * A126307 A126308 A126309

Keyword

nonn

Author

Antti Karttunen, Jan 02 2007

Status

approved