Let {*a*_{1},*a*_{2},…,*a*_{k}} be a set of integers such that 0 < *a*_{1} < ⋯ < *a*_{k}. We define the circulant graph *C*_{∞}(*a*_{1},…,*a*_{k}) as the infinite graph with vertices ℤ and such that
$\begin{array}{}N(j)=\{j\pm {a}_{i}{\}}_{i=1}^{k}\end{array}$ is the set of neighbors of each vertex *j* ∈ ℤ. Then *C*_{∞}(*a*_{1},…,*a*_{k}) is a regular graph of degree 2*k*. If *k* = 1, then *C*_{∞}(1) is isometric to the Cayley graph of ℤ, which is 0-hyperbolic. Hence, in what follows we just consider circulant graphs with *k* > 1.

Denote by *J*(*G*) the set of vertices and midpoints of edges in *G*, and by ⌊*t*⌋ the lower integer part of *t*.

The following results in [25] will be useful.

#### Theorem 3.1

([25, Theorem 2.6]). *For every hyperbolic graph G*, *δ*(*G*) *is a multiple of* 1/4.

As usual, by *cycle* we mean a simple closed curve, i.e., a path with different vertices, unless the last one, which is equal to the first vertex.

#### Theorem 3.2

([25, Theorem 2.7]). *For any hyperbolic graph G*, *there exists a geodesic triangle T* = {*x*, *y*, *z*} *that is a cycle with x*, *y*, *z* ∈ *J*(*G*) *and δ*(*T*) = *δ*(*G*).

We also need the following technical lemmas.

#### Lemma 3.3

*For any integers k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k}, *consider G* = *C*_{∞}(1,*a*_{2},…,*a*_{k}). *Then the following statements are equivalent*:

*d*_{G}(0,⌊*a*_{k}/2⌋) = ⌊*a*_{k}/2⌋ *and*, *if a*_{k} is odd, *then d*_{G}(0,⌊*a*_{k}/2⌋+1) ≥ ⌊*a*_{k}/2⌋.

*a*_{2} ≥ *a*_{k} − 1.

*We have either k* = 2 *or k* = 3 *and a*_{2} = *a*_{3} − 1.

#### Proof

Assume that (1) holds. If *k* = 2, then (2) holds; hence, we can assume *k* ≥ 3. Define *r* := ⌊*a*_{k}/2⌋. If *a*_{2} ≤ *r*, then *r* = *d*_{G}(0,*r*) by hypothesis and *r* = *d*_{G}(0,*r*) ≤ *d*_{G}(0,*a*_{2}) + *d*_{G}(*a*_{2},*r*) ≤ 1 + *r* − *a*_{2} < *r*, which is a contradiction. Thus we conclude *a*_{2} > *r*. Therefore,
$$\begin{array}{}r={d}_{G}(0,r)\le {d}_{G}(0,{a}_{2})+{d}_{G}({a}_{2},r)\le 1+{a}_{2}-r,\end{array}$$
and *a*_{2} ≥ 2*r* − 1. Hence, *a*_{2} ≥ *a*_{k} − 1 if *a*_{k} is even. Since *a*_{2} > *r*, if *a*_{k} is odd, then
$$\begin{array}{}r\le {d}_{G}(0,r+1)\le {d}_{G}(0,{a}_{2})+{d}_{G}({a}_{2},r+1)\le 1+{a}_{2}-(r+1),\end{array}$$
and *a*_{2} ≥ 2*r* = *a*_{k} − 1. Then (2) holds.

A simple computation provides the converse implication.

The equivalence of (2) and (3) is elementary.□

Let us define the subset 𝓔 of infinite circulant graphs as 𝓔:= {*C*_{∞}(1,2*m* + 1,2*m* + 2,2*m* + 3)}_{m ≥ 1}.

#### Lemma 3.4

*Consider any integers k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k} with a_{2} < *a*_{k} − 1, *and G* = *C*_{∞}(1,*a*_{2},…,*a*_{k}) ∉ 𝓔. *Then*
$$\begin{array}{}{\displaystyle min\{{d}_{G}(0,u),{d}_{G}({a}_{j},u)\}\le \lfloor \frac{{a}_{k}}{2}\rfloor -1,}\end{array}$$(1)
*for every u* ∈ ℤ *with* 0 ≤ *u* ≤ *a*_{j} and 1 ≤ *j* ≤ *k*.

#### Proof

Since *a*_{2} < *a*_{k} − 1, Lemma 3.3 gives:

if *a*_{k} is even, then *d*_{G}(0,⌊*a*_{k}/2⌋) < ⌊*a*_{k}/2⌋,

if *a*_{k} is odd, then *d*_{G}(0,⌊*a*_{k}/2⌋) < ⌊*a*_{k}/2⌋ or *d*_{G}(0,⌊*a*_{k}/2⌋ + 1) < ⌊*a*_{k}/2⌋.

If *d*_{G}(0,⌊*a*_{k}/2⌋) < ⌊*a*_{k}/2⌋, then inequality (1) trivially holds. Hence, we can assume that *a*_{k} is odd, *d*_{G}(0,⌊*a*_{k}/2⌋) = ⌊*a*_{k}/2⌋ and *d*_{G}(0,⌊*a*_{k}/2⌋ + 1) < ⌊*a*_{k}/2⌋. These facts imply that *d*_{G}(0,⌊*a*_{k}/2⌋ + 1) = ⌊*a*_{k}/2⌋ − 1.

If *a*_{2} ≤ ⌊*a*_{k}/2⌋, then *d*_{G}(0,⌊*a*_{k}/2⌋) < ⌊*a*_{k}/2⌋, which is a contradiction. Therefore, *a*_{2} > ⌊*a*_{k}/2⌋ and
$$\begin{array}{}{\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor -1={d}_{G}(0,\lfloor \frac{{a}_{k}}{2}\rfloor +1)\le 1+{a}_{2}-\lfloor \frac{{a}_{k}}{2}\rfloor -1,\phantom{\rule{1em}{0ex}}{a}_{k}-2\le {a}_{2}<{a}_{k}-1.}\end{array}$$

We conclude *a*_{2} = *a*_{k} − 2, and *k* = 3 or *k* = 4. If *k* = 4, then *a*_{2} = *a*_{4} − 2, *a*_{3} = *a*_{4} − 1 and *G* ∈ 𝓔, since *a*_{k} is odd. This is a contradiction, and we conclude *k* = 3 and *a*_{2} = *a*_{3} − 2.

If *j* = 1, then min{*d*_{G}(0,*u*),*d*_{G}(1,*u*)} = 0 for every 0 ≤ *u* ≤ 1.

If *j* = 2, then for every 0 ≤ *u* ≤ *a*_{2}
$$\begin{array}{}{\displaystyle min\{{d}_{G}(0,u),{d}_{G}({a}_{2},u)\}\le min\{u,{a}_{2}-u\}\le \lfloor \frac{{a}_{2}}{2}\rfloor =\lfloor \frac{{a}_{3}}{2}\rfloor -1.}\end{array}$$

If *j* = 3, then *d*_{G}(⌊*a*_{3}/2⌋,*a*_{3}) = *d*_{G}(0,⌊*a*_{3}/2⌋ + 1) = ⌊*a*_{3}/2⌋ − 1, and for every 0 ≤ *u* ≤ *a*_{3} with *u* ≠ ⌊*a*_{3}/2⌋,⌊*a*_{3}/2⌋ + 1,
$$\begin{array}{}{\displaystyle min\{{d}_{G}(0,u),{d}_{G}({a}_{3},u)\}\le min\{u,{a}_{3}-u\}\le \lfloor \frac{{a}_{3}}{2}\rfloor -1.}\end{array}$$

This finishes the proof. □

#### Proposition 3.5

*Consider any integers k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k}. *If we have either* *k* = 2, *or k* = 3 *and* *a*_{2} = *a*_{3}−1, *or* *C*_{∞} (1, *a*_{2}, …, *a*_{k}) ∈ 𝓔, *then*
$$\begin{array}{}\delta ({C}_{\mathrm{\infty}}(1,{a}_{2},\dots ,{a}_{k}))\ge {\displaystyle \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

#### Proof

Define *r* := ⌊ *a*_{k}/2⌋ and *G* := *C*_{∞} (1, *a*_{2}, …, *a*_{k}).

Assume first that we have either *k* = 2, or *k* = 3 and *a*_{2} = *a*_{3}−1. Consider the curves *γ*_{1}*γ*_{2} in *G* joining *x* := 0 and *y* := *r* + (*r* + 1)*a*_{k} given by
$$\begin{array}{}{\gamma}_{1}:=[0,1]\cup [1,2]\cup \cdots \cup [r-1,r]\cup [r,r+{a}_{k}]\cup [r+{a}_{k},r+2{a}_{k}]\cup \cdots \cup [r+r{a}_{k},r+(r+1){a}_{k}],\\ {\gamma}_{2}:=[0,{a}_{k}]\cup [{a}_{k},2{a}_{k}]\cup \cdots \cup [r{a}_{k},(r+1){a}_{k}]\cup [(r+1){a}_{k},(r+1){a}_{k}+1]\cup \cdots \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\cdots \cup [(r+1){a}_{k}+1,(r+1){a}_{k}+2]\cup [r+(r+1){a}_{k}-1,r+(r+1){a}_{k}].\end{array}$$

Lemma 3.3 gives that *γ*_{1} and *γ*_{2} are geodesics; then *d*_{G}(*x*, *y*) = *L*(*γ*_{1}) = *L*(*γ*_{2}) = 2*r*+1. Let *T* be the geodesic bigon *T* = {*γ*_{1}, *γ*_{2}} in *G*. If *p* is the midpoint of [*r*, *r* + *a*_{k}], then *d*_{G}(*p*, *x*) = *d*_{G}(*p*, *y*) = *r* + 1/2 and Lemma 3.3 gives that *d*_{G}(*p*, *γ*_{2}) = *r* + 1/2. Hence,
$$\begin{array}{}\delta ({C}_{\mathrm{\infty}}(1,{a}_{2},\cdots ,{a}_{k}))\ge {d}_{G}(p,{\gamma}_{2})={\displaystyle \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
Assume now that *C*_{∞} (1, *a*_{2}, …, *a*_{k}) ∈ 𝓔. Note that *r* + *ra*_{k} = (*r*+1) *a*_{k−1}, since *a*_{k} = *a*_{k−1} + 1 is odd. Consider the curves *γ*_{1},*γ*_{2}, *γ*_{3} in *G*
$$\begin{array}{}{\gamma}_{1}:=[-r-r{a}_{k},-r-(r-1){a}_{k}]\cup \cdots \cup [-r-2{a}_{k},-r-{a}_{k}]\cup [-r-{a}_{k},-r]\cup [-r,r]\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\cup [r,r+{a}_{k}]\cup [r+{a}_{k},r+2{a}_{k}]\cup \cdots \cup [r+(r-1){a}_{k},r+r{a}_{k}],\\ {\gamma}_{2}:=[-(r+1){a}_{k-1},-r{a}_{k-1}]\cup \cdots \cup [-2{a}_{k-1},-{a}_{k-1}]\cup [-{a}_{k-1},0],\\ {\gamma}_{3}:=[0,{a}_{k-1}]\cup [{a}_{k-1},2{a}_{k-1}]\cup \cdots \cup [r{a}_{k-1},(r+1){a}_{k-1}],\end{array}$$
joining *x* := −(*r*+1)*a*_{k−1}, *y* := (*r*+1)*a*_{k−1} and *z* := 0. One can check that *γ*_{1}, *γ*_{2} and *γ*_{3} are geodesics in *G*. Let *T* be the geodesic triangle *T* = {*γ*_{1}, *γ*_{2}, *γ*_{3}}. Note that *d*_{G}(0, *r*) = *r*, since *G* ∈ 𝓔. Therefore, if *p* is the midpoint of [−*r*, *r*], then
$$\begin{array}{}\delta (G)\ge {d}_{G}(p,{\gamma}_{2}\cup {\gamma}_{3})={\displaystyle \frac{1}{2}+{d}_{G}(r,{\gamma}_{3})=\frac{1}{2}+{d}_{G}(r,\{0,2r\})=\frac{1}{2}+r=\frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$ □

Let *G* = *C*_{∞} (1, *a*_{2}, …, *a*_{k}). If x∈ *V*(*G*), we define *x*_{1} := *x*_{2} := *x*; if *x* ∈ *G* \ *V*(*G*), we define *x*_{1} and *x*_{2} as the endpoints of the edge containing *x* with *x*_{1} < *x*_{2}. Therefore, 1 ≤ *x*_{2}−*x*_{1} ≤ *a*_{k} if *x* ∈ *G* \ *V*(*G*). If *x*, *y* ∈ *J*(*G*), we say that *xL y* if *x* = *y* or *x*_{2} ≤ *y*_{1}, and *x* and *y* are *related* (and we write *xR y*) if *xL y* or *yL x*. Note that *L* is an order relation on *J*(*G*).

#### Lemma 3.6

*Consider any integers* *k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k}. *If x*, *y* ∈ *J*(*C*_{∞}(1, *a*_{2}, …, *a*_{k})) *and* *x* *and* *y* *are not related*, *then*
$$\begin{array}{}{d}_{G}(x,y)\le 1+{\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
*Furthermore*, *if a*_{2} < *a*_{k}−1 *and* *C*_{∞} (1, *a*_{2}, …, *a*_{k})∉ 𝓔, *then*
$$\begin{array}{}{d}_{G}(x,y)\le {\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

#### Proof

Let *G* = *C*_{∞} (1, *a*_{2}, …, *a*_{k}). If *x*_{1} ≤ *y*_{1} ≤ *y*_{2} ≤ *x*_{2}, then the cycle
$$\begin{array}{}\sigma :=[{x}_{1},{x}_{1}+1]\cup [{x}_{1}+1,{x}_{1}+2]\cup \cdots \cup [{y}_{1}-1,{y}_{1}]\cup [{y}_{1},{y}_{2}]\cup [{y}_{2},{y}_{2}+1]\cup \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\cdots \cup [{x}_{2}-2,{x}_{2}-1]\cup [{x}_{2}-1,{x}_{2}]\cup [{x}_{2},{x}_{1}]\end{array}$$
has length at most 1 + *a*_{k}. Since *σ* is a cycle containing the points *x* and *y*, we have
$$\begin{array}{}{d}_{G}(x,y)\le {\displaystyle \frac{1}{2}L(\sigma )\le \frac{1+{a}_{k}}{2}\le 1+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
If *a*_{2} < *a*_{k} − 1 and *G* ∉ 𝓔, then Lemma 3.4 gives *d*_{G}(*x*, *y*_{1}) ≤ 1/2 + ⌊ *a*_{k}/2⌋−1=⌊*a*_{k}/2⌋−1/2 and *d*_{G}(*x*, *y*) ≤ ⌊ *a*_{k}/2⌋.

If *y*_{1} ≤ *x*_{1} ≤ *x*_{2} ≤ *y*_{2}, then the same argument gives these inequalities.

If *x*_{1} ≤ *y*_{1} < *x*_{2} ≤ *y*_{2}, then *x*_{2} − *y*_{1}+*y*_{1} − *x*_{1} ≤ *a*_{k} and min{*x*_{2}−*y*_{1}, *y*_{1} − *x*_{1}} ≤ ⌊ *a*_{k}/2⌋. If *x*_{2} − *y*_{1} ≤ ⌊ *a*_{k}/2⌋, then *d*_{G}(*x*, *y*) ≤ *d*_{G}(*x*, *x*_{2})+*x*_{2} − *y*_{1}+*d*_{G}(*y*_{1}, *y*) ≤ 1+⌊ *a*_{k}/2⌋. If *y*_{1} − *x*_{1} ≤ ⌊ *a*_{k}/2⌋, then *d*_{G}(*x*, *y*) ≤ *d*_{G}(*x*, *x*_{1})+*y*_{1} − *x*_{1}+*d*_{G}(*y*_{1}, *y*) ≤ 1+⌊ *a*_{k}/2⌋. If *a*_{2} < *a*_{k}−1 and *C*_{∞}(1, *a*_{2}, …, *a*_{k})∉ 𝓔, then Lemma 3.4 also gives *d*_{G}(*x*, *y*_{1}) ≤ ⌊ *a*_{k}/2⌋−1/2 and *d*_{G}(*x*, *y*) ≤ ⌊ *a*_{k}/2⌋.

If *y*_{1} ≤ *x*_{1} < *y*_{2} ≤ *x*_{2}, then the same argument gives these inequalities. □

We can state now the main result of this section, which provides a sharp upper bound for the hyperbolicity constant of infinite circulant graphs.

#### Theorem 3.7

*For any integers* *k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k},
$$\begin{array}{}\delta ({C}_{\mathrm{\infty}}(1,{a}_{2},\cdots ,{a}_{k}))\le {\displaystyle \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
*and the equality is attained if and only if we have either* *k* = 2, *or* *k* = 3 *and* *a*_{2} = *a*_{3}−1, *or k* = 4, *a*_{2} = *a*_{4}−2, *a*_{3} = *a*_{4}−1 *and* *a*_{4} *is odd*.

#### Proof

In order to bound *δ*(*G*) with *G* = *C*_{∞}(1, *a*_{2}, …, *a*_{k}), let us consider a geodesic triangle *T* = {*x*, *y*, *z*} in *G* and *p* ∈ [*xy*]; by Theorem 3.2 we can assume that *T* is a cycle with *x*, *y*, *z* ∈ *J*(*G*). We consider several cases.

(A)

If *x* and *y* are not related, then Lemma 3.6 gives
$$\begin{array}{}{d}_{G}(p,[xz]{\displaystyle \cup [yz])\le {d}_{G}(p,\{x,y\})\le \frac{1}{2}{d}_{G}(x,y)\le \frac{1}{2}+\frac{1}{2}\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$(2)

(B)

Assume that *xR y*. Without loss of generality we can assume that *xL y*. Denote by *w*_{1}, …, *w*_{m} the vertices in [*xy*] such that *w*_{1} ∈ {*x*_{1}, *x*_{2}}, *w*_{m} ∈ {*y*_{1}, *y*_{2}} and *d*_{G}(*w*_{j}, *w*_{j+1}) = 1 for every 1 ≤ *j* < *m*; we define
$$\begin{array}{}{i}_{0}:=min\{1\le i\le m|{w}_{j}\ge {x}_{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}i\le j\le m\},\phantom{\rule{2em}{0ex}}{x}_{0}:={w}_{{i}_{0}}.\end{array}$$

(*B*.1) Assume that either *x* ∈ *V*(*G*) or *x* is the midpoint of an edge [*r*, *r* +1]∈ *E*(*G*) for some *r* ∈ ℤ. If *x*_{0} = *x*_{2}, then *L*([*xx*_{0}]) ≤ 1/2. If *x*_{0}>*x*_{2}, then the cycle
$$\begin{array}{}\sigma :=[{w}_{{i}_{0}},{w}_{{i}_{0}-1}]\cup [{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}+1]\cup [{w}_{{i}_{0}-1}+1,{w}_{{i}_{0}-1}+2]\cup \cdots \cup [{w}_{{i}_{0}}-2,{w}_{{i}_{0}}-1]\cup [{w}_{{i}_{0}}-1,{w}_{{i}_{0}}]\end{array}$$
has length at most 1 + *a*_{k}. Since *x* ∈ *V*(*G*) or *x* is the midpoint of an edge [*r*, *r* + 1], thus *σ* is a cycle containing the points *x* and *x*_{0}, and we have
$$\begin{array}{}L([x{x}_{0}])={d}_{G}(x,{x}_{0})\le {\displaystyle \frac{1}{2}L(\sigma )\le \frac{1+{a}_{k}}{2}<\frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

(*B*.2) Assume now that *x* is the midpoint of an edge [*r*, *r* + *a*_{j}]∈ *E*(*G*) for some *r* ∈ ℤ and 1 < *j* ≤ *k*. If *x*_{0} = *x*_{2}, then *L*([*xx*_{0}]) = 1/2. If *w*_{i0−1} = *x*_{1}, then *L*([*xx*_{0}]) = 3/2. If *x*_{0} > *x*_{2} and *w*_{i0−1} ≠ *x*_{1}, then we have either *w*_{i0−1} < *x*_{1} < *x*_{2} < *w*_{i0} or *x*_{1} < *w*_{i0−1} < *x*_{2} < *w*_{i0}. In the first case the cycle
$$\begin{array}{}\sigma :=[{w}_{{i}_{0}},{w}_{{i}_{0}-1}]\cup [{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}+1]\cup \cdots \cup [{x}_{1}-1,{x}_{1}]\cup [{x}_{1},{x}_{2}]\cup [{x}_{2},{x}_{2}+1]\cup \cdots \cup [{w}_{{i}_{0}}-1,{w}_{{i}_{0}}]\end{array}$$
has length at most *a*_{k}. Since *σ* is a cycle containing the points *x* and *x*_{0}, we have
$$\begin{array}{}L([x{x}_{0}])={d}_{G}(x,{x}_{0})\le {\displaystyle \frac{1}{2}L(\sigma )\le \frac{{a}_{k}}{2}.}\end{array}$$
Assume that *x*_{1} < *w*_{i0−1} < *x*_{2} < *w*_{i0}. Then
$$\begin{array}{}L([x{x}_{0}])={d}_{G}(x,{x}_{0})\le {\displaystyle \frac{1}{2}+min\{{d}_{G}({x}_{2},{w}_{{i}_{0}-1}),{d}_{G}({w}_{{i}_{0}-1},{x}_{1})\}+1}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\le {\displaystyle \frac{3}{2}+min\{{x}_{2}-{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}-{x}_{1}\}\le \frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
Therefore, in Case (*B*), if *p* ∈[*xx*_{0}]\ *B*(*x*_{0}, 3/4), then
$$\begin{array}{}{d}_{G}(p,[xz]{\displaystyle \cup [yz])\le {d}_{G}(p,x)\le L([x{x}_{0}]\mathrm{\setminus}\phantom{\rule{thinmathspace}{0ex}}B({x}_{0},3/4))=L([x{x}_{0}])-\frac{3}{4}\le \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$(3)

Define
$$\begin{array}{}{j}_{0}:=max\{1\le i\le m|{w}_{j}\le {y}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\le j\le i\},\phantom{\rule{2em}{0ex}}{y}_{0}:={w}_{{j}_{0}}.\end{array}$$
A similar argument to the previous one shows that if *p* ∈[*y*_{0}*y*]\ *B*(*y*_{0}, 3/4), then
$$\begin{array}{}{d}_{G}(p,[xz]{\displaystyle \cup [yz])\le \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$(4)

Since *T* is a continuous curve, we have
$$\begin{array}{}([xz]\cup [yz])\cap \{{x}_{1},{x}_{2}\}\ne \mathrm{\varnothing},\phantom{\rule{2em}{0ex}}([xz]\cup [yz])\cap \{{y}_{1},{y}_{2}\}\ne \mathrm{\varnothing}.\end{array}$$

(*a*) Assume that *y*_{0} ∈ [*xx*_{0}]\ {*x*_{0}}.

(*a*.1) If *d*_{G}(*x*_{0}, *y*_{0})≥2, then
$$\begin{array}{}L([xy])={\displaystyle L([x{x}_{0}])+L([{y}_{0}y])-L([{y}_{0}{x}_{0}])\le \frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor +\frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor -2=1+2\lfloor \frac{{a}_{k}}{2}\rfloor ,}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{\displaystyle {d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,\{x,y\})\le \frac{1}{2}L([xy])\le \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

(*a*.2) If *d*_{G}(*x*_{0}, *y*_{0}) = 1, then *w*_{j0} = *y*_{0} = *w*_{i0−1}, *w*_{i0} = *x*_{0} = *w*_{j0+1} and the definitions of *x*_{0} and *y*_{0} give *y*_{0} < *x*_{2} ≤ *x*_{0} and *y*_{0} ≤ *y*_{1} < *x*_{0}. Since [*x*_{0}, *y*_{0}] ∈ *E*(*G*),
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle {d}_{G}({x}_{2},{y}_{1})\le \frac{1+{a}_{k}}{2},}\\ \phantom{\rule{thinmathspace}{0ex}}L([xy])={d}_{G}(x,y)\le {d}_{G}(x,{x}_{2})+{d}_{G}({x}_{2},{y}_{1})+{d}_{G}({y}_{1},y)\le {\displaystyle \frac{1}{2}+\frac{1+{a}_{k}}{2}+\frac{1}{2}=\frac{3+{a}_{k}}{2},}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,\{x,y\})\le {\displaystyle \frac{1}{2}L([xy])\le \frac{3+{a}_{k}}{4}\le \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

(*b*) Assume that *y*_{0}∉[*xx*_{0}]\ {*x*_{0}}. Since *x*_{2} ≤ *x*_{0}, *y*_{0} ≤ *y*_{1} and [*xz*] ∪ [*yz*] is a continuous curve joining *x* and *y*, if *p* ∈ *V*(*G*)∩[*x*_{0}*y*_{0}]⊂[*xy*], then there exist *u*, *v* ∈ *V*(*G*)∩([*xz*]∪[*yz*]) with [*u*, *v*] ∈ *E*(*G*) and *u* ≤ *p* ≤ *v*. Since *T* is a cycle, we have
$$\begin{array}{}{d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,\{u,v\})\le {\displaystyle min\{p-u,v-p\}\le \lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
Therefore, if *p* ∈[*x*_{0}*y*_{0}]∪ *B*(*x*_{0}, 3/4)∪ *B*(*y*_{0}, 3/4), then
$$\begin{array}{}{\displaystyle {d}_{G}(p,[xz]\cup [yz])<\frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$

This inequality, (2), (3) and (4) give
$$\begin{array}{}{\displaystyle \delta ({C}_{\mathrm{\infty}}(1,{a}_{2},\dots ,{a}_{k}))\le \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$(5)

By Theorem 3.1, in order to finish the proof it suffices to show that the equality in (5) is not attained. Seeking for a contradiction, assume that the equality is attained. The proof of (5) gives that we have
$$\begin{array}{}{\displaystyle {d}_{G}(p,[xz]\cup [yz])=\frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
where *p* is the point in [*xx*_{0}] with *d*_{G}(*p*, *x*_{0}) = 3/4 or the point in [*y*_{0}*y*] with *d*_{G}(*p*, *y*_{0}) = 3/4. By symmetry, without loss of generality we can assume that *p* is the point in [*xx*_{0}] with *d*_{G}(*p*, *x*_{0}) = 3/4; thus *d*_{G}(*p*, *w*_{i0−1}) = 1/4. Therefore, we are in Case (*B*.2) with *x*_{1} < *w*_{i0−1} < *x*_{2} < *w*_{i0} and
$$\begin{array}{}{\displaystyle L([x{x}_{0}])=\frac{3}{2}+min\{{d}_{G}({x}_{2},{w}_{{i}_{0}-1}),{d}_{G}({w}_{{i}_{0}-1},{x}_{1})\}=\frac{3}{2}+min\{{x}_{2}-{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}-{x}_{1}\}=\frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
Then
$$\begin{array}{}{\displaystyle min\{{d}_{G}({x}_{2},{w}_{{i}_{0}-1}),{d}_{G}({w}_{{i}_{0}-1},{x}_{1})\}=min\{{x}_{2}-{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}-{x}_{1}\}=\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
and we conclude
$$\begin{array}{}{\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor \le {x}_{2}-{w}_{{i}_{0}-1},{w}_{{i}_{0}-1}-{x}_{1}\le \lfloor \frac{{a}_{k}}{2}\rfloor +1.}\end{array}$$(6)

(*I*) Assume that *x*_{2}∈[*xx*_{0}]. Since *T* is a cycle, then *x*_{1}∈[*xz*]∪[*yz*] and *x*_{1} < *w*_{i0−1}. Hence, since [*xz*]∪[*yz*] is a continuous curve joining *x* and *y*, and *d*_{G}(*p*, *w*_{i0−1}) = 1/4, we obtain as above
$$\begin{array}{}{\displaystyle \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ={d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,{w}_{{i}_{0}-1})+{d}_{G}({w}_{{i}_{0}-1},[xz]\cup [yz])\le \frac{1}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
which is a contradiction.

(*II*) Assume that *x*_{1}∈[*xx*_{0}].

(*II*. 1) If *x*_{2} − *x*_{1} = *x*_{0} − *w*_{i0−1} = *a*_{k}, then *w*_{i0−1} − *x*_{1} = *x*_{0} − *x*_{2}, *d*_{G}(*x*_{1}, *w*_{i0−1}) = *d*_{G}(*x*_{2}, *x*_{0}) and
$$\begin{array}{}{d}_{G}(x,{x}_{0})={d}_{G}(x,{x}_{1})+{d}_{G}({x}_{1},{w}_{{i}_{0}-1})+{d}_{G}({w}_{{i}_{0}-1},{x}_{0})>{d}_{G}(x,{x}_{1})+{d}_{G}({x}_{1},{w}_{{i}_{0}-1})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}={\displaystyle \frac{1}{2}+{d}_{G}({x}_{2},{x}_{0})={d}_{G}(x,{x}_{2})+{d}_{G}({x}_{2},{x}_{0})\ge {d}_{G}(x,{x}_{0}),}\end{array}$$
which is a contradiction.

(*II*.2) If *x*_{2} − *x*_{1} < *a*_{k}, then (6) gives
$$\begin{array}{}{x}_{2}-{w}_{{i}_{0}-1}={w}_{{i}_{0}-1}-{x}_{1}={\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
and
$$\begin{array}{}{\displaystyle \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ={d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,{x}_{2})\le {d}_{G}(p,{w}_{{i}_{0}-1})+{d}_{G}({w}_{{i}_{0}-1},{x}_{2})\le \frac{1}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
which is a contradiction.

(*II*.3) Assume *x*_{2} − *x*_{1} = *a*_{k} and *x*_{0} − *w*_{i0−1} < *a*_{k}. Since *x*_{1}∈[*xx*_{0}], we have *w*_{i0−1} − *x*_{1} = ⌊ *a*_{k}/ 2 ⌋.

(*II*.3.1) If *x*_{2} − *w*_{i0−1} = ⌊*a*_{k}/2⌋, then
$$\begin{array}{}{\displaystyle \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ={d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,{x}_{2})\le {d}_{G}(p,{w}_{{i}_{0}-1})+{d}_{G}({w}_{{i}_{0}-1},{x}_{2})\le \frac{1}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
which is a contradiction.

(*II*.3.2) If *x*_{2} − *w*_{i0−1} = ⌊ *a*_{k}/2⌋+1, then *a*_{k} = *x*_{2} − *x*_{1} = 2⌊ *a*_{k}/2⌋+1 and
$$\begin{array}{}{\displaystyle {d}_{G}({x}_{0},{x}_{2})\le {x}_{0}-{x}_{2}={x}_{0}-{w}_{{i}_{0}-1}-({x}_{2}-{w}_{{i}_{0}-1})<{a}_{k}-\lfloor \frac{{a}_{k}}{2}\rfloor -1=\lfloor \frac{{a}_{k}}{2}\rfloor ,}\\ {\displaystyle \frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ={d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,{x}_{2})\le {d}_{G}(p,{x}_{0})+{d}_{G}({x}_{0},{x}_{2})<\frac{3}{4}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\end{array}$$
which is a contradiction.

This finishes the proof of the inequality.

If we have either *k* = 2, or *k* = 3 and *a*_{2} = *a*_{3}−1, or *k* = 4, *a*_{2} = *a*_{4}−2, *a*_{3} = *a*_{4}−1 and *a*_{4} is odd, then Proposition 3.5 gives
$$\begin{array}{}{\displaystyle \delta ({C}_{\mathrm{\infty}}(1,{a}_{2},\cdots ,{a}_{k}))\ge \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor .}\end{array}$$
Since we have the converse inequality, we conclude that the equality holds.

Assume now that the equality holds. Denote by *G* the circulant graph *C*_{∞} (1, *a*_{2}, …, *a*_{k}). By Theorem 3.2 there exist a geodesic triangle *T* = {*x*, *y*, *z*} in *G* that is a cycle with *x*, *y*, *z* ∈ *J*(*G*), and *p* ∈ [*xy*] with *d*_{G}(*p*, [*xz*] ∪ [*yz*]) = 1/2+⌊ *a*_{k}/2⌋. Hence, *p* ∈ *J*(*G*).

Seeking for a contradiction, assume that *a*_{2} < *a*_{k}−1 and that *G* ∉ 𝓔.

If *x* and *y* are not related, then Lemma 3.6 gives *d*_{G}(*x*, *y*) ≤ ⌊ *a*_{k}/2⌋. Since *d*_{G}(*x*, *y*) = *d*_{G}(*x*, *p*) + *d*_{G}(*p*, *y*)≥ 1 + 2 ⌊ *a*_{k}/2⌋, thus *x* and *y* are related, and without loss of generality we can assume *xL y*. Since *d*_{G}(*p*, *x*), *d*_{G}(*p*, *y*)≥1/2+⌊ *a*_{k}/2⌋, the previous argument gives that *x* and *p* are related, and *p* and *y* are related.

(*i*) Assume first *xL p* and *pL y*.

Since *p*_{1} − *x*_{1}≥*p*_{1} − *x*_{2}≥*d*_{G}(*x*_{2},*p*_{1})≥⌊ *a*_{k}/2⌋−1/2, we have *p*_{1} − *x*_{1}≥*p*_{1} − *x*_{2}≥⌊ *a*_{k}/2⌋, and there is some point *u* ∈([*xz*]∪[*yz*])∩ *V*(*G*) with *u* ≤ *p*_{1}. A similar argument gives that there is some point *v* ∈ ([*xz*]∪[*yz*])∩ *V*(*G*) with *p*_{2} ≤ *v*. Denote by *u* and *v* any couple of vertices satisfying these properties. We are going to prove that *v* − *u* > *a*_{k}.

If *p* ∈ *V*(*G*), then
$$\begin{array}{}{\displaystyle min\{p-u,v-p\}\ge min\{{d}_{G}(u,p),{d}_{G}(p,v)\}\ge \frac{1}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\\ {\displaystyle min\{p-u,v-p\}\ge 1+\lfloor \frac{{a}_{k}}{2}\rfloor ,\phantom{\rule{2em}{0ex}}v-u\ge 2+2\lfloor \frac{{a}_{k}}{2}\rfloor >{a}_{k}.}\end{array}$$
If *p* is the midpoint of some edge [*m*, *m*+*a*_{j}] with *m* ∈ ℤ and 2 ≤ *j* ≤ *k*, then Lemma 3.4 gives *d*_{G}(*p*, *w*) ≤ ⌊ *a*_{k}/2⌋−1/2 < 1/2 + ⌊ *a*_{k}/2⌋ for every *w* ∈ ℤ with *m* ≤ *w* ≤ *m* + *a*_{j}, since *a*_{2} < *a*_{k}−1 and *G* ∉ 𝓔. Furthermore,
$$\begin{array}{}{\displaystyle min\{m-u,v-m-{a}_{j}\}\ge min\{{d}_{G}(u,m),{d}_{G}(m+{a}_{j},v)\}\ge \lfloor \frac{{a}_{k}}{2}\rfloor ,}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}v-u\ge {a}_{j}+2{\displaystyle \lfloor \frac{{a}_{k}}{2}\rfloor \ge 2+2\lfloor \frac{{a}_{k}}{2}\rfloor >{a}_{k}.}\end{array}$$

Finally, assume that *p* is the midpoint of some edge [*m*, *m*+1] with *m* ∈ ℤ. By Lemma 3.3, we have *d*_{G}(0, ⌊ *a*_{k}/2⌋) ≤ ⌊ *a*_{k}/2⌋−1 or *d*_{G}(0, ⌊ *a*_{k}/2⌋+1) ≤ ⌊ *a*_{k}/2⌋−1. Therefore,
$$\begin{array}{}{\displaystyle {d}_{G}(p,p\pm (\lfloor \frac{{a}_{k}}{2}\rfloor +\frac{1}{2}))\le \lfloor \frac{{a}_{k}}{2}\rfloor -\frac{1}{2},\phantom{\rule{2em}{0ex}}min\{p-u,v-p\}\ge \frac{3}{2}+\lfloor \frac{{a}_{k}}{2}\rfloor ,}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}v-u\ge {\displaystyle 3+2\lfloor \frac{{a}_{k}}{2}\rfloor \ge 2+2\lfloor \frac{{a}_{k}}{2}\rfloor >{a}_{k}.}\end{array}$$

Hence, we have proved *v* − *u* >*a*_{k} in every case.

Since if *p* is the midpoint of some edge [*m*, *m* +*a*_{j}] with *m* ∈ ℤ and 2 ≤ *j* ≤ *k* then *d*_{G}(*p*, *w*) < 1/2 + ⌊ *a*_{k}/2 ⌋ for every *w* ∈ ℤ with *m* ≤ *w* ≤ *m* + *a*_{j}, and [*xz*]∪[*yz*] is a continuous curve joining *u* and *v*, there exist *u*_{0}, *v*_{0} ∈([*xz*]∪[*yz*])∩ *V*(*G*) with *u*_{0} ≤ *p*_{1}, *p*_{2} ≤ *v*_{0} and [*u*_{0}, *v*_{0}]∈ *E*(*G*). Hence, *a*_{k}≥*v*_{0} − *u*_{0} > *a*_{k}, which is a contradiction.

(*ii*) Assume now that *pL x* or *yL p*. By symmetry, we can assume that *pL x* and thus *p* ∈[*xx*_{0}] and *d*_{G}(*p*, *x*_{0})≥1. Since *d*_{G}(*x*, *x*_{0}) ≤ 3/2 + ⌊ *a*_{k}/2⌋, we have 1/2+ ⌊ *a*_{k}/2⌋=*d*_{G}(*p*, [*xz*]∪[*yz*]) ≤ *d*_{G}(*x*, *p*) ≤ 1/2 + ⌊ *a*_{k}/2⌋. Therefore, *d*_{G}(*x*, *p*) = 1/2 + ⌊ *a*_{k}/2⌋, *d*_{G}(*p*, *x*_{0}) = 1, *p* ∈ *V*(*G*) and *x* is the midpoint of [*x*_{1}, *x*_{2}]∈ *E*(*G*). Thus *d*_{G}(*p*, *x*_{1})≥⌊ *a*_{k}/2⌋, *d*_{G}(*p*, *x*_{2})≥⌊ *a*_{k}/2⌋ and *p* ≤ *x*_{1} ≤ *x*_{2} ≤ *x*_{0}. By Lemma 3.3, we have *d*_{G}(0, ⌊ *a*_{k}/2⌋) ≤ ⌊ *a*_{k}/2⌋−1 or *d*_{G}(0, ⌊ *a*_{k}/2⌋+1) ≤ ⌊ *a*_{k}/2⌋−1.

If *d*_{G}(0, ⌊ *a*_{k}/2⌋) ≤ ⌊ *a*_{k}/2⌋−1, then *x*_{1}≥⌊ *a*_{k}/2⌋+1, *x*_{2} ≤ *a*_{k}−(⌊ *a*_{k}/2⌋−1) and
$$\begin{array}{}1\le {x}_{2}-{x}_{1}\le {a}_{k}-(\lfloor {a}_{k}/2\rfloor -1)-\lfloor {a}_{k}/2\rfloor -1={a}_{k}-2\lfloor {a}_{k}/2\rfloor \le 0,\end{array}$$
which is a contradiction.

If *d*_{G}(0, ⌊ *a*_{k}/2 ⌋ + 1) ≤ ⌊ *a*_{k}/2⌋−1, then *x*_{2} ≤ ⌊ *a*_{k}/2⌋, *x*_{1}≥⌊ *a*_{k}/2⌋ and 1 ≤ *x*_{2} − *x*_{1} ≤ ⌊ *a*_{k}/2⌋ − ⌊ *a*_{k}/2⌋ ≤ 0, which is a contradiction.

Therefore, we conclude in every case that *a*_{2}≥*a*_{k}−1 or *G* ∈ 𝓔. Hence, Lemma 3.3 gives *k* = 2, or *k* = 3 and *a*_{2} = *a*_{3}−1, or *k* = 4, *a*_{2} = *a*_{4} − 2, *a*_{3} = *a*_{4} − 1 and *a*_{4} is odd. □

We also have a sharp lower bound for the hyperbolicity constant.

#### Theorem 3.8

*For any integers* *k* > 1 *and* 1 < *a*_{2} < ⋯ < *a*_{k} *we have*
$$\delta ({C}_{\mathrm{\infty}}(1,\phantom{\rule{thinmathspace}{0ex}}{a}_{2},\dots ,{a}_{k}))\ge \frac{3}{2},$$

*and the equality is attained if* *a*_{k} = *k*.

#### Proof

Define *G* = *C*_{∞} (1, *a*_{2}, …, *a*_{k}), and consider the geodesics *γ*_{1}, *γ*_{2} in *G* given by
$$\begin{array}{}{\gamma}_{1}:=[0,{a}_{k}]\cup [{a}_{k},2{a}_{k}]\cup [2{a}_{k},2{a}_{k}+1],\\ {\gamma}_{2}:=[0,1]\cup [1,1+{a}_{k}]\cup [1+{a}_{k},1+2{a}_{k}].\end{array}$$

Let *T* be the geodesic bigon *T* = {*γ*_{1}, *γ*_{2}} in *G*. If *p* is the midpoint of [*a*_{k}, 2*a*_{k}], then
$$\delta (\mathit{G})\ge {d}_{G}(p,{\gamma}_{2})=min\{{d}_{G}(p,{a}_{k})+{d}_{G}({a}_{k},1+{a}_{k}),{d}_{G}(p,2{a}_{k})+{d}_{G}(2{a}_{k},2{a}_{k}+1),\frac{1}{2}L({\gamma}_{1})\}=\frac{3}{2}.$$

Assume now that *a*_{k} = *k*, i.e., *G* = *C*_{∞} (1, *a*_{2}, …, *a*_{k}) = *C*_{∞}(1, 2, ⋯, *k*). Therefore, *d*_{G}(*m*, *m* + *w*) = 1 for every *m*, *w* ∈ ℤ with |*w*| ≤ *k*. Note that if *x*, *y* ∈ *J*(*G*) and *x* and *y* are not related, then |*x*_{1} − *y*_{1}| < *k*, *d*_{G}(*x*_{1}, *y*_{1}) = 1 and
$${d}_{G}(x,y)\le {d}_{G}(x,{x}_{1})+{d}_{G}({x}_{1},{y}_{1})+{d}_{G}({y}_{1},y)=2.$$(7)

Let us consider a geodesic triangle *T* = {*x*, *y*, *z*} in *G* and *p* ∈ [*xy*]; by Theorem 3.2 we can assume that *T* is a cycle with *x*, *y*, *z* ∈ *J*(*G*). We consider several cases.

(A)

If *x* and *y* are not related, then (7) gives
$${d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,\{x,y\})\le \frac{1}{2}{d}_{G}(x,y)\le 1.$$

(B)

Assume that *xR y*. Without loss of generality we can assume that *xL y*. Define *w*_{j}, *i*_{0}, *x*_{0} and *y*_{0} as in the proof of Theorem 3.7. Then *w*_{i0−1} < *x*_{2} ≤ *w*_{i0} = *x*_{0}.

If *x*_{0} = *x*_{2}, then *L*([*x**x*_{0}]) ≤ 1/2. If *x*_{0} > *x*_{2}, then *w*_{i0−1} < *x*_{2} < *w*_{i0}. Since *x*_{2} − *w*_{i0}≤*w*_{i0}−*w*_{i0 −1} ≤ *k*, *d*_{G}(*x*_{2}, *x*_{0}) = 1 and
$$L([x{x}_{0}])={d}_{G}(x,{x}_{0})\le {d}_{G}(x,{x}_{2})+{d}_{G}({x}_{2},{x}_{0})\le \frac{1}{2}+1=\frac{3}{2}.$$

Therefore, in both cases, if *p* ∈ [*x**x*_{0}], then
$${d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,x)\le L([x{x}_{0}])\le \frac{3}{2}.$$

A similar argument to the previous one shows that if *p* ∈ [*y*_{0}*y*], then *d*_{G}(*p*, [*xz*] ∪ [*yz*]) ≤ 3/2. If *y*_{0} ∈ [*x**x*_{0}], then *d*_{G}(*p*, [*xz*] ∪ [*yz*]) ≤ 3/2 holds for every *p* ∈ [*xy*]. Consider now the case *y*_{0} ∉ [*x**x*_{0}].

Since *x*_{2} ≤ *x*_{0} and *y*_{0} ≤ *y*_{1}, every *p* ∈ *V*(*G*) ∩ [*x*_{0}*y*_{0}] ⊂ [*xy*] verifies *x*_{2} ≤ *p* ≤ *y*_{1}. Since *T* is a continuous curve, we obtain
$$([xz]\cup [yz])\cap \{{x}_{1},{x}_{2}\}\ne \mathrm{\varnothing},\phantom{\rule{1em}{0ex}}([xz]\cup [yz])\cap \{{y}_{1},{y}_{2}\}\ne \mathrm{\varnothing}.$$

Since [*xz*] ∪ [*yz*] is a continuous curve joining *x* and *y*, if *p* ∈ *V*(*G*) ∩ [*x*_{0}*y*_{0}], then there exist *u*, *v* ∈ *V*(*G*) ∩ ([*xz*] ∪ [*yz*]) with [*u*, *v*] ∈ *E*(*G*) and *u* ≤ *p* ≤ *v*. Hence, *p* − *u* ≤ *v* − *u* ≤ *k*, *v* − *p* ≤ *v* − *u* ≤ *k* and
$${d}_{G}(p,[xz]\cup [yz])\le {d}_{G}(p,\{u,v\})=1.$$

Therefore, if *p* ∈ [*x*_{0}*y*_{0}], then
$$\begin{array}{}{\displaystyle {d}_{G}(p,[xz]\cup [yz])\le \frac{3}{2}.}\end{array}$$

These inequalities give *δ*(*T*) ≤ 3/2 and, hence,
$$\begin{array}{}{\displaystyle \delta ({C}_{\mathrm{\infty}}(1,2,\dots ,k))\le \frac{3}{2}.}\end{array}$$

Since we have proved the converse inequality, we conclude that the equality holds. □

As usual, the *complement* *G* of the graph *G* is defined as the graph with *V*(*G*) = *V*(*G*) and such that *e* ∈ *E*(*G*) if and only if *e* ∉ *E*(*G*). We are going to bound the hyperbolicity constant of the complement of every infinite circulant graph. In order to do it, we need some preliminaries.

For any graph *G*, we define,
$$\begin{array}{}\text{diam\hspace{0.17em}}V(G):={\displaystyle sup\{{d}_{G}(v,w)|v,w\in V(G)\},}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{diam\hspace{0.17em}}G:={\displaystyle sup\{{d}_{G}(x,y)|x,y\in G\}.}\end{array}$$

We need the following well-known result (see a proof, *e*.*g*., in [36, Theorem 8]).

#### Theorem 3.9

*In any graph G the inequality δ*(*G*) ≤ (diam *G*) /2 *holds*.

We have the following direct consequence.

#### Corollary 3.10

*In any graph G the inequality δ*(*G*) ≤ (diam *V*(*G*) + 1)/2 *holds*.

From [34, Proposition 5 and Theorem 7] we deduce the following result.

#### Lemma 3.11

*Let G be any graph with a cycle g*. *If L*(*g*) ≥ 3, *then δ*(*G*) ≥ 3/4. *If L*(*g*) ≥ 4, *then δ*(*G*) ≥ 1.

We say that a vertex *v* of a graph *G* is a *cut-vertex* if *G* ∖ {*v*} is not connected. A graph is *two-connected* if it does not contain cut-vertices.

We need the following result in [26, Proposition 4.5 and Theorem 4.14].

#### Theorem 3.12

*Assume that G is a two-connected graph*. *Then G verifies δ*(*G*) = 1 *if and only if* diam *G* = 2. *Furthermore*, *δ*(*G*) ≤ 1 *if and only if* diam *G* ≤ 2.

#### Definition 3.13

*Let us consider integers k* ≥ 1 *and* 1 ≤ *a*_{1} < *a*_{2} < ⋯ <*a*_{k}. *We say that* {*a*_{1}, *a*_{2}, …, *a*_{k}} *is a* 1-modulated *sequence if for every x*, *y* ∈ ℤ *with* |*y*| ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}} *we have* |*x*| ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}} *or* |*x* − *y*| ∉ {*a*_{1}, *a*_{2}; ⋯,*a*_{k}}.

We have the following sharp bounds for the hyperbolicity constant of the complement of every infinite circulant graph. In particular, they show that the complement of infinite circulant graphs are hyperbolic.

#### Theorem 3.14

*For any integers k* ≥ 1 *and* 1 ≤ *a*_{1} < *a*_{2} < ⋯ <*a*_{k} we have
$$\begin{array}{}{\displaystyle 1\le \delta (\overline{{C}_{\mathrm{\infty}}({a}_{1},{a}_{2},\dots ,{a}_{k})})\le \frac{3}{2}.}\end{array}$$

*Furthermore*,
$\begin{array}{}\delta (\overline{{C}_{\mathrm{\infty}}({a}_{1},{a}_{2},\dots ,{a}_{k})})=1\end{array}$ *if and only if* {*a*_{1}, *a*_{2}, …, *a*_{k}} *is* 1-*modulated*. *If there is* 1 ≤ *j* < *a*_{k}/5 *with j*,5*j* ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}} *and* 2*j*,3*j*,4*j* ∈ {*a*_{1}, *a*_{2}, …, *a*_{k}}, *then*
$\begin{array}{}\delta (\overline{{C}_{\mathrm{\infty}}({a}_{1},{a}_{2},\dots ,{a}_{k})})=3/2.\end{array}$

#### Proof

Define *G* : = *C*_{∞} (*a*_{1}, *a*_{2}, …, *a*_{k}). Given *u*, *v* ∈ ℤ, consider *w* ∈ ℤ with *w* > *u* + *a*_{k} and *w* > *v* + *a*_{k}. Since [*u*, *w*],[*v*, *w*] ∉ *E*(*G*), we have [*u*, *w*],[*v*, *w*] ∈ *E*(*G*) and *d*_{G}(*u*, *v*) ≤ *d*_{G}(*u*, *w*) + *d*_{G}(*w*, *v*) = 2. Hence, diam *V*(*G*) ≤ 2 and Corollary 3.10 gives *δ*(*G*) ≤ 3/2.

Since [0, *a*_{k} + 1], [*a*_{k} + 1, 2*a*_{k} + 2], [2*a*_{k} + 2, 3*a*_{k} + 3], [3*a*_{k} + 3, 0] ∉ *E*(*G*), we have [0, *a*_{k} + 1], [*a*_{k} + 1, 2*a*_{k} + 2], [2*a*_{k} + 2, 3*a*_{k} + 3], [3*a*_{k} + 3, 0]∈ *E*(G). Since the cycle *C* := {0, *a*_{k} + 1, 2*a*_{k} + 2, 3*a*_{k} + 3, 0} in G has length 4, Lemma 3.11 gives *δ* (G) ≥ 1.

Since *G* is a circulant graph, the sequence {*a*_{1}, *a*_{2}, …, *a*_{k}} is 1-modulated if and only if for every *x*, *y*_{1}, *y*_{2} ∈ ℤ with |*y*_{2} − *y*_{1}|∉{*a*_{1}, *a*_{2}, …, *a*_{k}}, we have |*x* − *y*_{1}| ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}} or |*x* − *y*_{2}| ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}}. This happens if and only if *d*_{G}(*x*, [*y*_{1}, *y*_{2}]) ≤ 1 for every *x* ∈ *V*(G) and [*y*_{1}, *y*_{2}]∈ *E*(G), and this condition is equivalent to diam *V*(G) ≤ 2. Since G is a two-connected graph, Theorem 3.12 that diam G ≤ 2 if and only if *δ* (G) ≤ 1. Since *δ* (G)≥1, we conclude that *δ* (G) = 1 if and only if {*a*_{1}, *a*_{2}, …, *a*_{k}} is 1-modulated.

Assume that there is 1 ≤ *j* < *a*_{k}/5 with *j*, 5*j* ∉ {*a*_{1}, *a*_{2}, …, *a*_{k}} and 2*j*, 3*j*, 4*j* ∈ {*a*_{1}, *a*_{2}, …, *a*_{k}}, and consider the cycle *C* := {0, *j*, 2*j*, 3*j*, 4*j*, 5*j*, 0} in G with length 6. Let *x* and *y* be the midpoints of the edges [2*j*, 3*j*] and [5*j*, 0], respectively. Since *d*_{G}([2*j*, 3*j*], [5*j*, 0]) = 2, *d*_{G}(*x*, *y*) = 3 and *C* contains two geodesics *g*_{1}, *g*_{2} joining *x* and *y*, with *g*_{1} ∩ *V*(G) = {0, *j*, 2*j*} and *g*_{2} ∩ *V*(G) = {3*j*, 4*j*, 5*j*}. Since *d*_{G}(*j*, {3*j*, 4*j*, 5*j*})≥ 2, we have *δ* (G)≥*d*_{G}(*j*, *g*_{2}) = *d*_{G}(*j*, {*x*, *y*}) = 3/2, and we conclude *δ* (G) = 3/2. □

Since Paley graphs is an important class of circulant graph, which is attracting great interest in recent years (see, *e*.*g*., [45]), we finish this paper with a result on the hyperbolicity constant of Paley graphs.

Recall that the Paley graph of order *q* with *q* a prime power is a graph on *q* nodes, where two nodes are adjacent if their difference is a square in the finite field *GF*(*q*). This graph is circulant when *q* ≡ 1(mod 4). Paley graphs are self-complementary, strongly regular, conference graphs, and Hamiltonian.

#### Proposition 3.15

*For any Paley graph G we have*
$$\begin{array}{}1\le \delta (G){\displaystyle \le \frac{3}{2}.}\end{array}$$

#### Proof

Let us denote by *n* the cardinality of *V*(*G*).

Since Paley graphs are self-complementary (the complement of any Paley graph is isomorphic to it), the degree of any vertex is (*n* − 1)/2. Hence, given *u*, *v* ∈ *V*(*G*) with [*u*, *v*] ∉ *E*(*G*), there exists a vertex *w* ∈ *V*(*G*) with *d*_{G}(*u*, *w*) = *d*_{G}(*u*, *v*) = 1, and we conclude that diam *V*(*G*) ≤ 2. Therefore, Corollary 3.10 gives *δ*(*G*) ≤ 3/2.

Since *G* is a Hamiltonian graph, there exists a Hamiltonian cycle *g*. Since *L*(*g*) = *n* ≥ 5, Lemma 3.11 gives *δ*(*G*) ≥ 1. □

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