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

%I A126306 #16 May 08 2021 08:35:36
%S A126306 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 A126306 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 A126306 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 A126306 a(n) = number of double-rises (UU-subsequences) in the n-th Dyck path encoded by A014486(n).
%F A126306 a(n) = A014081(A014486(n)).
%F A126306 a(n) = A000120(A048735(A014486(n))).
%F A126306 a(A125976(n)) = A057514(n)-1, for all n >= 1.
%e A126306 A014486(20) = 228 (11100100 in binary), encodes the following Dyck path:
%e A126306     /\
%e A126306    /..\/\
%e A126306   /......\
%e A126306 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 A126306 (Python)
%o A126306 def ok(n):
%o A126306     if n==0: return True
%o A126306     B=bin(n)[2:] if n!=0 else '0'
%o A126306     s=0
%o A126306     for b in B:
%o A126306         s+=1 if b=='1' else -1
%o A126306         if s<0: return False
%o A126306     return s==0
%o A126306 def a014081(n): return sum(((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1))
%o A126306 print([a014081(n) for n in range(4001) if ok(n)]) # _Indranil Ghosh_, Jun 13 2017
%Y A126306 Cf. A014081, A014486, A000120, A048735, A125976, A057514.
%K A126306 nonn
%O A126306 0,9
%A A126306 _Antti Karttunen_, Jan 02 2007