Extrema of a(n)/n^(log_phi 2) for the Fibbinary Numbers on Intervals Between Consecutive Fibonacci Numbers

The Fibbinary numbers, OEIS A003714, may be obtained by reading the Zeckendorf representation of an integer as a binary expansion. More precisely, let F1=F2=1F_1=F_2=1 and Fi+1=Fi+Fi−1F_{i+1}=F_i+F_{i-1} for i≥2i\ge2. If n=∑i≥2εiFin=\sum_{i\ge2}\varepsilon_iF_i is the unique Zeckendorf representation of nn, with εi∈{0,1}\varepsilon_i\in\{0,1\} and no two adjacent coefficients equal to 11, define a(n)=∑i≥2εi2i−2a(n)=\sum_{i\ge2}\varepsilon_i2^{i-2}, with a(0)=0a(0)=0. Thus a(n)a(n) is the nn-th nonnegative integer whose binary expansion contains no two adjacent 1s. Writing φ=(1+5)/2\varphi=(1+\sqrt5)/2 and κ=log⁡φ2\kappa=\log_\varphi2, this note determines the exact extrema of the normalized sequence a(n)/nκa(n)/n^\kappa between consecutive Fibonacci numbers. For every k≥2k\ge2 and Fk≤n<Fk+1F_k\le n<F_{k+1}, one has ⌊2k/3⌋/(Fk+1−1)κ≤a(n)/nκ≤2k−2/Fkκ\lfloor2^k/3\rfloor/(F_{k+1}-1)^\kappa\le a(n)/n^\kappa\le2^{k-2}/F_k^\kappa, with a(Fk)=2k−2a(F_k)=2^{k-2} and a(Fk+1−1)=⌊2k/3⌋a(F_{k+1}-1)=\lfloor2^k/3\rfloor. For k≥4k\ge4, the upper bound is attained only at n=Fkn=F_k, while the lower bound is attained only at n=Fk+1−1n=F_{k+1}-1. These interval extrema determine the global asymptotic bounds: lim inf⁡n→∞a(n)/nκ=(5)κ/6=0.5311949872…\liminf_{n\to\infty}a(n)/n^\kappa=(\sqrt5)^\kappa/6=0.5311949872\ldots and lim sup⁡n→∞a(n)/nκ=(5)κ/4=0.7967924808…\limsup_{n\to\infty}a(n)/n^\kappa=(\sqrt5)^\kappa/4=0.7967924808\ldots. In particular, (5)κ/6<a(n)/nκ≤1(\sqrt5)^\kappa/6