OFFSET
0,4
COMMENTS
a(n) = A000120(A010078(n)), n>0; a(n) = A023416(A004754(n-1)), n>1. - Reinhard Zumkeller, Dec 04 2015
Conjecture: a(n)+1 is the length of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
Terms a(n); n >= 2 can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, let S(k) = {S(k-1)+1, S(k-1)}, where +1 means +1 on every term of S(k-1); see Example. Each step of the recursion gives the next 2^k terms of the sequence. - David James Sycamore, Jul 15 2024
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Wikipedia, Two's complement
FORMULA
a(n) = A023416(n-1) + 1.
a(n) = if n<=1 then n else (n mod 2) + a((n mod 2) + floor(n/2)). - Reinhard Zumkeller, Feb 05 2007
a(n) = if n<2 then n else a(ceiling(n/2)) + n mod 2. - Reinhard Zumkeller, Jul 25 2006
Min{m: a(m)=n} = if n>0 then A083318(n-1) else 0. - Reinhard Zumkeller, Jul 25 2006
EXAMPLE
Using the above recursion for a(n); n >= 2, we have:
S(0) = {1} so a(2) = 1;
S(1) = {2,1} so a(3,4) = 2,1;
S(2) = {3,2,2,1}, so a(5,6,7,8) = 3,2,2,1;
As irregular table the sequence for n >= 2 begins:
1;
2,1;
3,2,2,1;
4,3,3,2,3,2,2,1;
5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
6,5,5,4,5,4,4,3,5,4,3,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
and so on (the k-th row contains 2^k terms; k>=0). - David James Sycamore, Jul 15 2024
MATHEMATICA
a[0] = 0; a[1] = 1; a[n_] := a[n] = Mod[n, 2] + a[Mod[n, 2] + Floor[n/2]]; Array[a, 96, 0] (* Jean-François Alcover, Aug 12 2017, after Reinhard Zumkeller *)
PROG
(Haskell)
a008687 n = a008687_list !! n
a008687_list = 0 : 1 : c [1] where c (e:es) = e : c (es ++ [e+1, e])
-- Reinhard Zumkeller, Mar 07 2011
(PARI) a(n) = if(n<2, n, n--; logint(n, 2) - hammingweight(n) + 2); \\ Kevin Ryde, Apr 14 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
STATUS
approved