A126306 - OEIS

%I #16 May 08 2021 08:35:36

%S 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,

%T 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,

%U 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

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

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

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

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

%e A014486(20) = 228 (11100100 in binary), encodes the following Dyck path:

%e     /\

%e    /..\/\

%e   /......\

%e 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.

%o (Python)

%o def ok(n):

%o     if n==0: return True

%o     B=bin(n)[2:] if n!=0 else '0'

%o     s=0

%o     for b in B:

%o         s+=1 if b=='1' else -1

%o         if s<0: return False

%o     return s==0

%o def a014081(n): return sum(((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1))

%o print([a014081(n) for n in range(4001) if ok(n)]) # _Indranil Ghosh_, Jun 13 2017

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

%K nonn

%O 0,9

%A _Antti Karttunen_, Jan 02 2007