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%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
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