A286299 -id:A286299 - cp's OEIS frontend

A286298 a(0) = 0, a(1) = 1; thereafter, a(2n) = a(n) + 1 + (n mod 2), a(2n+1) = a(n) + 2 - (n mod 2).

0, 1, 3, 2, 4, 5, 4, 3, 5, 6, 7, 6, 5, 6, 5, 4, 6, 7, 8, 7, 8, 9, 8, 7, 6, 7, 8, 7, 6, 7, 6, 5, 7, 8, 9, 8, 9, 10, 9, 8, 9, 10, 11, 10, 9, 10, 9, 8, 7, 8, 9, 8, 9, 10, 9, 8, 7, 8, 9, 8, 7, 8, 7, 6, 8, 9, 10, 9, 10, 11, 10, 9, 10, 11, 12, 11, 10, 11, 10, 9, 10, 11, 12, 11, 12, 13, 12, 11, 10, 11, 12

Offset: 0

N. J. A. Sloane, Jun 02 2017

Let S be a set containing 0 and let S grow in generations G(i), defined by these rules: If x is in S then 2x is in S and 1 - x is in S. So G(0) = 0, G(1) = {1}, G(2) = {2}, G(3) = {-1,4}, ... Then a(n) is the generation containing the integer k where n = 2k - 1 for k>0 and -2k for k <= 0. The question posed by Clark Kimberling was "Does each generation contain a Fibonacci number or its negative?" It has been proved that every integer occurs in some G(i). - Karyn McLellan, Aug 16 2018

			For k = -5, n = 10 and f(10) = 7, so -5 first appears in G(7). - _Karyn McLellan_, Aug 16 2018

C. Kimberling, Problem proposals, Proceedings of the Sixteenth International Conference on Fibonacci Numbers and Their Applications, P. Anderson, C. Ballot and W. Webb, eds. The Fibonacci Quarterly, 52.5(2014), 5-14.

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Danielle Cox and K. McLellan, A problem on generation sets containing Fibonacci numbers, Fib. Quart., 55 (No. 2, 2017), 105-113.

Cf. A000523, A005811, A286299 (first differences).

f:=proc(n) option remember;
   if n <= 1 then n
   elif (n mod 2) = 0 then f(n/2)+1+((n/2) mod 2);
   else f((n-1)/2) + 2 - ((n-1)/2 mod 2); end if;
end proc;
[seq(f(n),n=0..120)];

a(n) = if (n==0, 0, logint(n, 2) + hammingweight(bitxor(n, n>>1))); \\ Michel Marcus, Aug 21 2018

def A286298(n):
    if n <= 1:
        return n
    a, b = divmod(n, 2)
    return A286298(a) + 1 + b + (-1)**b*(a % 2) # Chai Wah Wu, Jun 02 2017

a(n) = A005811(n) + A000523(n) for n >= 1. - Karyn McLellan, Aug 16 2018

cp's OEIS Frontend

A286298 a(0) = 0, a(1) = 1; thereafter, a(2n) = a(n) + 1 + (n mod 2), a(2n+1) = a(n) + 2 - (n mod 2).

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