



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, 13, 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, 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
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OFFSET

1,1


COMMENTS

If the terms (n>0) are written as an array (leftaligned fashion) with rows of length 2^m, m >= 0:
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, 9,11,13,
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,...
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:
a(2^m + k) = a(2^(m+1)  1  k), m >= 0, 0 <= k < 2^m.
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).
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).
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).
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. AdamsWatters' comment), that is the sequence obtained by adding numerator and denominator in the CalkinWilf 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.
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).


LINKS

Table of n, a(n) for n=1..88.


FORMULA

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).
a(A059893(n)) = a(n) for n > 0.  Yosu Yurramendi, May 30 2017
From Yosu Yurramendi, May 14 2019: (Start)
Take the smallest m > 0 such that 0 <= k < 2^(m1), and choose any M >= m,
a((1/3)*( A016921(2^(m1)+k)*4^(Mm)1)) = 2*a(2^(m1)+k)*(Mm) + a(2^m+2*k ).
a((1/3)*(2*A016921(2^(m1)+k)*4^(Mm)2)) = 2*a(2^(m1)+k)*(Mm) + a(2^m+2*k ) + a(2^(m1)+k).
a((1/3)*( A016969(2^(m1)+k)*4^(Mm)2)) = 2*a(2^(m1)+k)*(Mm) + a(2^m+2*k+1).
a((1/3)*(2*A016969(2^(m1)+k)*4^(Mm)1)) = 2*a(2^(m1)+k)*(Mm) + a(2^m+2*k+1) + a(2^(m1)+k). (End)


EXAMPLE

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


CROSSREFS

Cf. A162909, A162910, A162911, A162912, A007306, A071585, A086592.
Sequence in context: A091238 A178047 A122954 * A257004 A126571 A210874
Adjacent sequences: A268084 A268085 A268086 * A268088 A268089 A268090


KEYWORD

nonn,easy


AUTHOR

Yosu Yurramendi, Jan 26 2016


STATUS

approved



