A sufficient condition for full density of V(1,1)V(1,1)

Let V(1,1)V(1,1) be the visibility set considered in the paper. For 1a,bN1\leqslant a,b\leqslant N, let b1<b2<<bkab_1<b_2<\dots<b_{k_a} be the values of bb such that (a,lbi)V(1,1)(a,lb_i)\notin V(1,1) for every lNl\in\mathbb{N} with 1lN/bi1\leqslant l\leqslant N/b_i, and define

c(a):=n(a,N)N,c(a):=\frac{n(a,N)}{N},

where n(a,N)n(a,N) is the number of bb with (a,b)V(1,1)(a,b)\notin V(1,1). Set

c1(a):=i=1ka1bi.c_1(a):=\sum_{i=1}^{k_a}\frac{1}{b_i}.

Full-density criterion for V(1,1)V(1,1). If

a=1Nc1(a)=o(N),\sum_{a=1}^N c_1(a)=o(N),

then dens(V(1,1))=1\operatorname{dens}(V(1,1))=1.

Sources & referencesView supporting material

Primary source

Sneha Chaubey and Ashish Kumar Pandey, “On the density of visible lattice points along polynomials”, arXiv:2109.08431 (2021).

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