2 A preliminary lemma

Lemma 2.1. Given positive integers $h\geq 2$ , $g\geq 5$ and $u\geq 2$ , let the g-adic representation of n be

$$ \begin{align*} n=e_{1}g^{i_1}+\cdots+e_{u-1}g^{i_{u-1}}+g^{i_{u}}, \end{align*} $$

where $0 \leq i_1<\cdots <i_{u}$ and $1\leq e_j\leq g-1$ for $j=1,\ldots ,u-1$ . Then $n\in (u+1)^{\wedge }\mathcal {A}_{g,h}$ .

Proof. If $i_{u-1}<i_{u}-1$ , or if $i_{u-1}= i_{u}-1$ , $e_{u-1}\notin \{1,g-1\}$ , then because $g\geq 5$ and

$$ \begin{align*} n=e_{1}g^{i_1}+\cdots+e_{u-1}g^{i_{u-1}}+g^{i_u-1}+(g-1)g^{i_u-1}, \end{align*} $$

we see that $n\in (u+1)^{\wedge }\mathcal {A}_{g,h}$ .

If $i_{u-1}= i_{u}-1$ and $e_{u-1}=1 $ , then because $g\geq 5$ and

$$ \begin{align*} n=e_{1}g^{i_1}+\cdots+e_{u-2}g^{i_{u-2}}+g^{i_{u}-1}+2g^{i_{u}-1}+(g-2)g^{i_{u}-1}, \end{align*} $$

we see that $n\in (u+1)^{\wedge }\mathcal {A}_{g,h}$ .

If $i_{u-1}= i_{u}-1$ and $e_{u-1}=g-1 $ , then because $g\geq 5$ and

$$ \begin{align*} n=e_{1}g^{i_1} + \cdots +e_{u-2}g^{i_{u-2}} + g^{i_{u}-1}+(g-2)g^{i_{u}-1}+g^{i_u}, \end{align*} $$

again $n\in (u+1)^{\wedge }\mathcal {A}_{g,h}$ .

This completes the proof of Lemma 2.1.

3 Proof of Proposition 1.2

(1) Assume that $2g^{2k+1}=a_0+a_1\in 2^{\wedge }\mathcal {A}_{g,2}$ with $0<a_0<a_1$ . Then $a_0<g^{2k+1}$ and $a_1<2g^{2k+1}$ . Since $a_0,a_1\in \mathcal {A}_{g,2}$ ,

$$ \begin{align*} a_0+a_1 &\leq (g-1)(g^0+\cdots+g^{2k-2}+g^{2k})+(g-1)(g^1+\cdots+g^{2k-3}+g^{2k-1})+g^{2k+1}\\ &<2g^{2k+1}, \end{align*} $$

which is a contradiction. Hence, $2g^{2k+1}\notin 2^{\wedge }\mathcal {A}_{g,2}\ \text{for all } g\geq 2$ .

(2) Assume that $2g^{3k+2}+1\in 3^{\wedge }\mathcal {A}_{g,3}$ . Write

(3.1) $$ \begin{align} 2g^{3k+2}+1=a_0+a_1+a_2,\quad a_0<a_1<a_2. \end{align} $$

Then $a_1<g^{3k+2}$ and $a_2<2g^{3k+2}$ . By (3.1), there exists at least one $a_i\in A_g(W_0^{(3)})$ .

If $a_2\notin A_g(W_{2}^{(3)})$ , then $a_2\leq (g-1)(g^1+g^4+\cdots +g^{3k+1}).$ Thus,

$$ \begin{align*} a_0+a_1+a_2&<(g-1)(g^0+g^3+\cdots+g^{3k})+ 2(g-1)(g^1+g^4+\cdots+g^{3k+1})\\ &< 2g^{3k+2}+1, \end{align*} $$

which is a contradiction.

If $a_2\in A_g(W_{2}^{(3)})$ , then $a_2\leq (g-1)(g^2+g^5+\cdots +g^{3k-1})+g^{3k+2}.$ Thus,

$$ \begin{align*} a_0+a_1+a_2 &\leq(g-1)(g^0+g^3+\cdots+g^{3k})+(g-1)(g^1+g^4+\cdots+g^{3k+1})\\ &\quad +(g-1)(g^2+g^5+\cdots+g^{3k-1})+g^{3k+2}\\ &<2g^{3k+2}+1, \end{align*} $$

which is a contradiction.

Hence, $2g^{3k+2}+1\notin 3^{\wedge }\mathcal {A}_{g,3}$ for all $g\geq 2$ .

(3) Assume that $2\cdot 4^{4k+3}+5\in 4^{\wedge }\mathcal {A}_{4,4}$ . Write

(3.2) $$ \begin{align} 2\cdot 4^{4k+3}+5=a_0+a_1+a_2+a_3, \quad a_0<a_1<a_2<a_3. \end{align} $$

Then $a_{3}<2\cdot 4^{4k+3}$ .

If $a_3\notin A_4(W_{3}^{(4)})$ , then $a_3\leq 3(4^2+4^6+\cdots +4^{4k+2})$ . Since $2\cdot 4^{4k+3}+5\equiv 5 \pmod {16},$ it follows from (3.2) that

$$ \begin{align*}a_0+a_1+a_2+a_3 &< 4^0+3\times(4^4+\cdots+4^{4k})+4^1+3\times(4^5+\cdots+4^{4k+1})\\ &\quad + 2\times3\times(4^2+4^6+\cdots+4^{4k+2})\\ &<2\cdot 4^{4k+3}+5, \end{align*} $$

which is a contradiction.

If $a_3\in A_4(W_{3}^{(4)})$ , then $a_3\leq 3(4^3+4^7+\cdots +4^{4k-1})+4^{4k+3}$ . If $a_2\geq 4^{4k+3}$ , then we have $a_2+a_3>2\cdot 4^{4k+3}+5$ , which is a contradiction. It follows that $a_2<4^{4k+3}$ . Again by (3.2) and $2\cdot 4^{4k+3}+5\equiv 5 \pmod {16},$

$$ \begin{align*}a_0+a_1+a_2+a_3 &\leq 4^0+3\times(4^4+\cdots+4^{4k})+4^1+3\times(4^5+\cdots+4^{4k+1})\\ &\quad + 3(4^2+4^6+\cdots+4^{4k+2})+3(4^3+4^7+\cdots+4^{4k-1})+4^{4k+3} \\ &< 2\cdot 4^{4k+3}+5, \end{align*} $$

which is a contradiction.

Hence, $2\cdot 4^{4k+3}+5\notin 4^{\wedge }\mathcal {A}_{4,4}$ .

This completes the proof of Proposition 1.2.

4 Proof of Theorem 1.3

By Theorem 1.1, $\mathcal {A}_{g,h}$ is a minimal asymptotic basis of order h. Thus, we only need to prove that $\mathcal {A}_{g,h}$ is a restricted asymptotic basis of order h.

Let $n\geq g^{(h-2)^2+1}$ and let the g-adic representation of n be

$$ \begin{align*} n=e_{1}g^{i_1}+\cdots+e_{t-1}g^{i_{t-1}}+e_tg^{i_{t}}, \end{align*} $$

where $0 \leq i_1<\cdots <i_{t}$ and $1\leq e_j\leq g-1$ for $j=1,\ldots ,t$ .

Case 1: $t=1$ . Then $i_1\geq h$ . Note that

$$ \begin{align*} n=(e_1-1)g^{i_1} + (g-1)g^{i_{1}-1} + \cdots + (g-1)g^{i_{1}-(h-1-\delta)} + g^{i_{1}-(h-1-\delta)}, \end{align*} $$

where $\delta =0$ if $e_1=1$ and otherwise $\delta =1$ . Hence, $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

Case 2: $2 \leq t \leq h-2$ . If there exists a $k\in \{2,\ldots ,t\}$ such that $i_{k}-i_{k-1}>h-t$ , then

$$ \begin{align*} n=(e_k-1)g^{i_k}+(g-1)g^{i_{k}-1} + \cdots + (g-1)g^{i_{k}-(h-t-\delta)}+g^{i_{k}-(h-t-\delta)} + \sum_{j\in\{1,\ldots,t\} \setminus\{k\}}e_jg^{i_{j}}, \end{align*} $$

where $\delta =0$ if $e_k=1$ and otherwise $\delta =1$ . Hence, $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

If $i_{k}-i_{k-1}\leq h-t$ for all $k\in \{2,\ldots ,t\}$ , then by $n\geq g^{(h-2)^2+1}$ , we have $i_1\geq h-t$ . Otherwise, if $i_1<h-t$ , then $i_t<t(h-t)$ , so that

$$ \begin{align*} n=e_{1}g^{i_1}+ \cdots + e_{t-1}g^{i_{t-1}}+e_tg^{i_{t}} < g^{i_t+1}<g^{(h-2)^2+1}, \end{align*} $$

which is a contradiction. Thus,

$$ \begin{align*} n=(e_1-1)g^{i_1} + ( g-1)g^{i_{1}-1} + \cdots + (g-1)g^{i_{1} - (h-t-\delta)} + g^{i_{1}-(h-t-\delta)} +\sum_{j\in\{2,\ldots,t\}}e_jg^{i_{j}}, \end{align*} $$

where $\delta =0$ if $e_1=1$ and otherwise $\delta =1$ . Hence, $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

Case 3: $t = h-1$ . Then

$$ \begin{align*} n = e_1g^{i_1} + e_2g^{i_2} + \cdots + e_{h-1}g^{i_{h-1}}, \end{align*} $$

where $0 \leq i_1<\cdots <i_{h-1}$ , $1\leq e_j\leq g-1$ for $j=1,\ldots ,h-1$ .

If there exists a $k\in \{1,\ldots ,h-1\}$ such that $3 \leq e_k \leq g-1$ , then

$$ \begin{align*} n =\sum_{j\in I\backslash\{k\}}e_jg^{i_j}+g^{i_{k}} +(e_{k}-1) g^{i_{k}}, \end{align*} $$

where $I=\{1,\ldots ,h-1\}$ , and thus $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

Now we consider what happens if $1\leq e_j\leq 2$ for $j=1,\ldots ,h-1$ .

(a) Suppose $e_{h-1}=1$ . Then by Lemma 2.1, $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

(b) Suppose $e_1=e_2=\cdots =e_{h-2}=e_{h-1}=2$ .

If there exist $i_u<i_v $ with $1\leq u,v\leq h-1$ such that $i_u \equiv i_v \pmod h$ , then because

$$ \begin{align*} 2g^{i_{u}}+2g^{i_{v}}=(2g^{i_{u}}+g^{i_{v}})+g^{i_{v}-1}+(g-1)g^{i_{v}-1} \end{align*} $$

and $i_v \geq h$ , we have $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

If $i_s\not \equiv i_t\pmod h$ for $1\leq s\neq t\leq h-1$ , then because $h\geq 4$ , there exist $i_v\ (\geq 1)$ , $i_u$ with $u,v \in \{1,2,\ldots ,h-1\}$ such that $i_v \equiv i_u+1 \pmod h$ . Since $g\geq 5$ and

$$ \begin{align*} 2g^{i_{u}}+2g^{i_{v}}=(2g^{i_{u}}+g^{i_{v}-1})+(g-1)g^{i_{v}-1}+g^{i_{v}}, \end{align*} $$

$n\in h^{\wedge }\mathcal {A}_{g,h}$ .

(c) Suppose $e_{h-1}=2$ , $e_{k}=1$ for some $k\in \{2,\ldots ,h-2\}$ . Then

$$ \begin{align*} n=\sum_{j\in K}e_jg^{i_{j}}+g^{i_k}+\sum_{j\in I}e_jg^{i_{j}}, \end{align*} $$

where $K=\{1,\ldots ,k-1\}$ and $I=\{k+1,\ldots ,h-1\}$ . By Lemma 2.1,

$$ \begin{align*} \sum\limits_{j\in K}e_jg^{i_{j}}+g^{i_k}\in (k+1)^{\wedge}\mathcal{A}_{g,h} \end{align*} $$

and so $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

(d) Suppose $e_{h-1}=e_{h-2}=\cdots =e_2=2$ , $e_1=1$ . If $i_1>0$ , then

$$ \begin{align*} n=g^{i_1-1}+(g-1)g^{i_1-1}+\sum\limits_{j\in\{2,\ldots,h-1\}}2g^{i_{j}} \end{align*} $$

and so $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

If $i_1=0$ , then

$$ \begin{align*} n=1+\sum\limits_{j\in\{2,\ldots,h-1\}}2g^{i_{j}}. \end{align*} $$

(d1) There exists a $k\in \{2,\ldots ,h-1\}$ such that $i_{k}\equiv 0\pmod h$ . Then

$$ \begin{align*} n=(1+g^{i_{k}})+g^{i_{k}-1}+(g-1)g^{i_{k}-1}+\sum_{j\in\{2,\ldots,h-1\}\setminus\{k\}}2g^{i_{j}}. \end{align*} $$

Thus, $n\in h^{\wedge }\mathcal {A}_{g,h}$ .

(d2) Suppose $i_{k}\not \equiv 0\pmod h$ for all $k\in \{2,\ldots ,h-1\}$ .

If $h\geq 5$ , then one of the following two cases must occur.

Case (i): There exist $i_u<i_v $ with $2\leq u,v\leq h-1$ such that $i_u \equiv i_v \pmod h$ . Since

$$ \begin{align*} 2g^{i_{u}}+2g^{i_{v}}=(2g^{i_{u}}+g^{i_{v}})+g^{i_{v}-1}+(g-1)g^{i_{v}-1} \end{align*} $$

and $i_v>h$ , we have

$$ \begin{align*} \sum\limits_{j\in\{2,\ldots,h-1\}}2g^{i_{j}}\in (h-1)^{\wedge}(\mathcal{A}_{g,h}\backslash \{1\}). \end{align*} $$

Thus, $n \in h^{\wedge }\mathcal {A}_{g,h}$ .

Case (ii): $i_s\not \equiv i_t\pmod h$ for $2\leq s\neq t\leq h-1$ . Then there exist $i_v\ (\geq 1)$ , $i_u$ for some $u,v \in \{2,\ldots ,h-1\}$ such that $i_v \equiv i_u+1 \pmod h$ . Because $g\geq 5$ and

$$ \begin{align*} 2g^{i_{u}}+2g^{i_{v}}=(2g^{i_{u}}+g^{i_{v}-1})+(g-1)g^{i_{v}-1}+g^{i_{v}}, \end{align*} $$

we have

$$ \begin{align*} \sum\limits_{j\in\{2,\ldots,h-1\}}2g^{i_{j}}\in (h-1)^{\wedge}(\mathcal{A}_{g,h}\backslash \{1\}). \end{align*} $$

Thus, $n \in h^{\wedge }\mathcal {A}_{g,h}$ .

Now suppose $h=4$ . Since $i_2,i_3\not \equiv 0\pmod 4$ , there is one further case in addition to (i) and (ii), namely, $\{i_2\pmod 4, i_3\pmod 4\}=\{1\pmod 4, 3\pmod 4\}$ . We may assume that $i_{2}\equiv 1\pmod 4$ . Because $g\geq 5$ and

$$ \begin{align*} n=(1+g^{i_{2}-1})+(g-1)g^{i_{2}-1}+g^{i_{2}}+2g^{i_{3}},\\[-18pt] \end{align*} $$

we have $n\in 4^{\wedge }\mathcal {A}_{g,4}$ .

Case 4: $t \geq h$ .

Write $I=\{i_{1},\ldots ,i_{t}\}$ . Since $\lvert I\rvert \geq h$ , it is possible to write I as a union of h nonempty sets $I_{1},\ldots ,I_{h}$ , where each $I_{j}$ is a subset of some $W^{(h)}_{k}$ . It follows that $n \in h^{\wedge }\mathcal {A}_{g,h}$ .

This completes the proof of Theorem 1.3.