A286389 a(0) = 0; a(n) = n - a(floor(a(n-1)/2)).
0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 8, 8, 9, 10, 10, 11, 12, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 29, 30, 30, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 45, 46, 46, 47, 48, 48, 49, 50, 51, 52, 52, 53, 54, 54, 55, 56, 57, 58, 58
Offset: 0
Links
- Anton Mosunov, Table of n, a(n) for n = 0..400
- Jeffrey Shallit, Proof of Irvine's conjecture via mechanized guessing, arXiv preprint arXiv:2310.14252 [math.CO], October 22 2023.
- Index entries for Hofstadter-type sequences
Programs
-
Mathematica
a[0] = 0; a[n_] := a[n] = n - a[Floor[a[n - 1]/2]]; Array[a, 80, 0]
-
PARI
a(n)=if(n>0,return(n-a(floor(a(n-1)/2))));return(0); \\ Anton Mosunov, May 26 2017
Formula
Conjecture: a(n) ~ c*n, where c = sqrt(3) - 1 = 0.732050807...
From Michel Dekking, Jul 06 2023: (Start)
This conjecture is implied by the conjecture in the COMMENTS, by a simple application of the Perron-Frobenius Theorem.
The vector (1, 1 + sqrt(3)) is a right eigenvector of the incidence matrix of the morphism 0->11, 1->110. Therefore the frequency of 1 in A285431 is equal to sqrt(3) - 1. So if the conjecture in the COMMENTS is true, then this implies that a(n)/n converges to sqrt(3) - 1. (End)
Comments