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%I #11 Jun 06 2017 12:00:45
%S 0,1,2,1,3,2,4,2,4,1,3,3,5,3,5,2,5,4,6,3,4,2,4,2,6,2,4,3,6,1,3,4,6,3,
%T 5,4,7,4,6,4,5,5,7,2,5,3,5,3,7,5,7,3,5,4,6,3,7,6,8,4,4,3,5,5,7,6,8,4,
%U 6,3,5,6,8,3,5,5,7,5,7,3,6,4,6,5,8,1,3,5,6,2,4,4,6,2,4,2,8,4,6,6,8,2,4,2,6,3,5,4,7,4,6,5,8,5,7,5,9,5,7,4,5
%N Base-3 digit sum of A254103: a(n) = A053735(A254103(n)).
%C Reflecting the structure of A254103 also this sequence can be represented as a binary tree:
%C 0
%C |
%C ...................1...................
%C 2 1
%C 3......../ \........2 4......../ \........2
%C / \ / \ / \ / \
%C / \ / \ / \ / \
%C / \ / \ / \ / \
%C 4 1 3 3 5 3 5 2
%C 5 4 6 3 4 2 4 2 6 2 4 3 6 1 3 4
%C etc.
%H Antti Karttunen, <a href="/A286632/b286632.txt">Table of n, a(n) for n = 0..8191</a>
%F a(n) = A053735(A254103(n)).
%F a(n) = A056239(A286633(n)).
%F For all n >= 0, a(A000079(n)) = n+1.
%o (Scheme) (define (A286632 n) (A053735 (A254103 n)))
%o (Python)
%o from sympy.ntheory.factor_ import digits
%o def a254103(n):
%o if n==0: return 0
%o if n%2==0: return 3*a254103(n/2) - 1
%o else: return floor((3*(1 + a254103((n - 1)/2)))/2)
%o def a(n): return sum(digits(a254103(n), 3)[1:]) # _Indranil Ghosh_, Jun 06 2017
%Y Cf. A053735, A254103, A286585, A286633.
%K nonn,base
%O 0,3
%A _Antti Karttunen_, Jun 03 2017
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