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a(n) = A162909(n) + A162910(n).
8

%I #40 Apr 25 2024 11:06:03

%S 2,3,3,5,4,4,5,8,7,5,7,7,5,7,8,13,11,9,12,9,6,10,11,11,10,6,9,12,9,11,

%T 13,21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,

%U 13,14,19,17,11,16,19,14,18,21,34,29,23,31,25,17,27,30,25,23,13,20,29,22,26,31,23,19,17,22,13,8,16,17,27

%N a(n) = A162909(n) + A162910(n).

%C If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m >= 0:

%C 2,

%C 3, 3,

%C 5, 4, 4, 5,

%C 8, 7, 5, 7, 7, 5, 7, 8,

%C 13,11, 9,12, 9, 6,10,11,11,10,6, 9,12, 9,11,13,

%C 21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,13,14,19,17,11,16, ...

%C a(n) is palindromic in each level m >= 0 (ranks between 2^m and 2^(m+1)-1), because in each level m >= 0 A162910 is the reverse of A162909:

%C a(2^m + k) = a(2^(m+1) - 1 - k), m >= 0, 0 <= k < 2^m.

%C All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).

%C a(2^m + k) = A162909(2^(m+2) + k), a(2^m + k) = A162909(2^(m+1)+ 2^m + k), a(2^m + k) = A162910(2^(m+1) + k), m >= 0, 0 <= k < 2^m (empirical observations).

%C a(n) = A162911(n) + A162912(n), where A162911(n)/A162912(n) is the bit reversal permutation of A162909(n)/A162910(n) in each level m >= 0 (empirical observations).

%C a(n) = A162911(2n+1), a(n) = A162912(2n) for n > 0 (empirical observations). n > 1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), which is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A162909(n)/A162910(n) is also an enumeration system of all positive rationals (Bird system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306.

%C The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), and A086592 (A020650+A020651).

%F a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k) with m = 0, 1, 2, ... and 0 <= k < 2^m (empirical observation).

%F a(A059893(n)) = a(n) for n > 0. - _Yosu Yurramendi_, May 30 2017

%F From _Yosu Yurramendi_, May 14 2019: (Start)

%F Take the smallest m > 0 such that 0 <= k < 2^(m-1), and choose any M >= m,

%F a((1/3)*( A016921(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k ).

%F a((1/3)*(2*A016921(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k ) + a(2^(m-1)+k).

%F a((1/3)*( A016969(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1).

%F a((1/3)*(2*A016969(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1) + a(2^(m-1)+k). (End)

%F a(n) = A007306(A258996(n)), n > 0. - _Yosu Yurramendi_, Jun 23 2021

%e m = 3, k = 6: a(38) = 17, a(22) = 10, a(14) = 7.

%o (PARI) a(n) = my(x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, x], [x, y]=[y, x+y])); x \\ _Mikhail Kurkov_, Mar 10 2023

%Y Cf. A162909, A162910, A162911, A162912, A007306, A071585, A086592.

%K nonn,easy

%O 1,1

%A _Yosu Yurramendi_, Jan 26 2016