%I #7 Jan 05 2025 19:51:42
%S 0,0,2,2,2,4,4,4,4,4,4,4,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,8,8,8,8,
%T 8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
%U 8,8,8,8,8,8,8,8,8,10,10,10,10,10,10,10
%N Length of the "fractional part" of the phi-representation of n.
%C The phi-representation of n is the (essentially) unique way to write n = Sum_{j=L..R} b(j)*phi^j, where b(j) is in {0,1} and -oo < L <= 0 <= R, where phi = (1+sqrt(5))/2, subject to the condition that b(j)b(j+1) != 1. The "fractional" part is the string of bits b(L)b(L+1)...b(-1), and its length is thus L.
%C The gaps between consecutive terms are all either 0 or 2, and a gap of 2 occurs if and only if n = L(2i+1) for i >= 0. This is equivalent to Theorem 2.1 of Sanchis and Sanchis (2001).
%H George Bergman, <a href="https://math.berkeley.edu/~gbergman/papers/base_tau.pdf">A number system with an irrational base</a>, Math. Mag. 31 (1957), 98-110.
%H G. R. Sanchis and L. A. Sanchis, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/39-2/sanchis.pdf">On the frequency of occurrence of α^i in the α-expansions of the positive integers</a>, Fibonacci Quart. 39 (2001), 123-137.
%F There is a linear representation of rank 11 for a(n).
%e The phi-representation of 20 is 1000010.010001, so a(20) = 6.
%Y Cf. A341722, A362692.
%K nonn
%O 0,3
%A _Jeffrey Shallit_, May 07 2023