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%I #13 Jan 29 2016 17:13:51
%S 1,2,1,3,2,1,1,4,3,2,1,2,1,1,1,5,4,3,2,1,3,2,1,2,1,1,2,1,1,1,1,6,5,4,
%T 3,2,1,4,3,2,1,3,2,1,2,1,1,3,2,1,2,1,1,2,1,1,1,2,1,1,1,1,1,7,6,5,4,3,
%U 2,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,4,3,2,1,3,2,1,2,1,1,3,2,1,2,1,1,2,1,1,1
%N Reversed reduced frequency counts for A004001: a(n) = A265754(A054429(n)).
%C Deleting all 1's and decrementing the remaining terms by one gives the sequence back.
%H Antti Karttunen, <a href="/A266640/b266640.txt">Table of n, a(n) for n = 1..8191</a>
%H T. Kubo and R. Vakil, <a href="http://dx.doi.org/10.1016/0012-365X(94)00303-Z">On Conway's recursive sequence</a>, Discr. Math. 152 (1996), 225-252.
%F a(n) = A265754(A054429(n)).
%F Other identities. For all n >= 0:
%F a(2^n) = n+1.
%e Illustration how the sequence can be constructed by concatenating the reversed reduced frequency counts R_n of each successive level n of A004001-tree:
%e 1
%e / \
%e 2 1
%e /|\ \
%e ____________3 2 1 1
%e / / / | |\ \ \
%e ________4 __3 2 1 2 1 1 1
%e / / / / / / /| /| | |\ \ \ \
%e 5 4 3 2 1 3 2 1 2 1 1 2 1 1 1 1
%e etc.
%o (Scheme) (define (A266640 n) (A265754 (A054429 n)))
%Y Cf. A004001, A054429.
%Y Cf. A000079 (positions of records, where n appears for the first time).
%Y Cf. A265754 (obtained from the mirror image of the same tree).
%K nonn,tabf
%O 1,2
%A _Antti Karttunen_, Jan 22 2016