OFFSET
1,1
COMMENTS
If the terms (n>0) are written as an array (in a left-aligned 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. 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.
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^(m-1), and choose any M >= m,
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 ).
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).
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).
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)
EXAMPLE
m = 3, k = 6: a(38) = 17, a(22) = 10, a(14) = 7.
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Yosu Yurramendi, Jan 26 2016
STATUS
approved