The main result of this article is given in the following theorem

**Theorem 2.1**. *For all* $m\ge 1\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}n\ge 4\lceil (2nm{)}^{1/3}\rceil \u2a7d$$pd({W}_{n,m})\le 2\lceil (nm{)}^{\frac{1}{2}}\rceil +1$

We need following lemmas to prove our main results.

**Lemma 2.2**. *Let Π* = *{S*_{1}, *S*_{2}, . . . , *S*_{q}} be a resolving q partition of $V({W}_{n,m}).Ifc\in {S}_{1},then\mid {S}_{1}\mid \u2a7d1+\left(\genfrac{}{}{0em}{}{q-1}{2}\right)+\left(\genfrac{}{}{0em}{}{q-1}{1}\right)+\left(\genfrac{}{}{0em}{}{q-1}{0}\right)$*where c is the central vertex*.

*Proof*. We analyze that *r*(*c|Π*) = (0, 1, 1, . . . , 1) and *r*(*v|Π*) = (0, . . .) for *v ∈ S*_{1}*∖{c}*. We also know that the diameter of *W*_{n}_{,m} is 2, the elements of the q vector representation *r*(*v|Π*) of each rim vertex *v ∈ S*_{1}*∖{c}* except the first element can be 1 or 2. But in vector representation there can be at most two elements which will be equal to 1 apart from the first position of the vector. For the rim vertices in the vector representation there are *q* − 1 positions which can be filled by at most two 1’s and the other can be filled by 2’s. Thus , for all vertices *v ∈ S*_{1}*∖{c}* there are at most $\left(\genfrac{}{}{0em}{}{q-1}{2}\right)+\left(\genfrac{}{}{0em}{}{q-1}{1}\right)+\left(\genfrac{}{}{0em}{}{q-1}{0}\right)$different vector representations. Together with these vector representations of the center, we arrive at most $1+\left(\genfrac{}{}{0em}{}{q-1}{2}\right)+\left(\genfrac{}{}{0em}{}{q-1}{1}\right)+\left(\genfrac{}{}{0em}{}{q-1}{0}\right)$different representations. Therefore $\mid {S}_{1}\mid \u2a7d1+\left(\genfrac{}{}{0em}{}{q-1}{2}\right)+\left(\genfrac{}{}{0em}{}{q-1}{1}\right)+\left(\genfrac{}{}{0em}{}{q-1}{0}\right).$

**Lemma 2.3**. *Let Π* = *{S*_{1}, *S*_{2}, . . . , *S*_{q}} be a resolving q partition of V(*W*_{n}_{,m})*. If c ∈ S*_{1}*, then* $\mid {S}_{i}\mid \u2a7d\left(\genfrac{}{}{0em}{}{q-2}{2}\right)+\left(\genfrac{}{}{0em}{}{q-2}{1}\right)+\left(\genfrac{}{}{0em}{}{q-2}{0}\right)$*for each* 2 6 *i* 6 *q*.

*Proof*. We consider another class not *S*_{1}, say *S*_{2}, where c does not belong to *S*_{2}. Then for *w ∈ S*_{2}, the vector representation becomes *r*(*w|Π*) = (1, 0, . . . ). For rim vertices in the vector representation there are *q* − 2 which can be filled by at most two 1’s and the other can be filled by 2’s. Thus for all vertices *w ∈ S*_{2}, there are at most $\left(\genfrac{}{}{0em}{}{q-2}{2}\right)+\left(\genfrac{}{}{0em}{}{q-2}{1}\right)+\left(\genfrac{}{}{0em}{}{q-2}{0}\right)$different representations. Therefore$\mid {S}_{i}\mid \u2a7d\left(\genfrac{}{}{0em}{}{q-2}{2}\right)+\left(\genfrac{}{}{0em}{}{q-2}{1}\right)+\left(\genfrac{}{}{0em}{}{q-2}{0}\right)$for each 2 6 *i* 6 *q*.

With the help of above two lemmas, we can now find the lower bound for the partition dimension of *W*_{n}_{,m}.

**Proposition 2.4**. *We have ⌈*(2*nm*)^{1/3}*⌉* 6 *pd*(*W*_{n}_{,m})*, for every n* > 4 *and m* ≥ 1.

*Proof*. Let us assume *pd*(*W*_{n}_{,m}) = *q* and *Π* = *{S*_{1}, *S*_{2}, . . . , *S*_{q}} be a resolving q partition of *V*(*W*_{n}_{,m}). Let *c ∈ S*_{1} and by lemma (2.2) we have $\mid {S}_{1}\mid \u2a7d1+\left(\genfrac{}{}{0em}{}{q-1}{2}\right)+\left(\genfrac{}{}{0em}{}{q-1}{1}\right)+$$\left(\genfrac{}{}{0em}{}{q-1}{0}\right)$and by lemma (2.3) we have $\mid {S}_{i}\mid \u2a7d\left(\genfrac{}{}{0em}{}{q-2}{2}\right)+\left(\genfrac{}{}{0em}{}{q-2}{1}\right)+\left(\genfrac{}{}{0em}{}{q-2}{0}\right)$

for 2 6 *i* 6 *q*. Using these two lemmas we get | *V*(*W*_{n}_{,m}) |= $nm+1=\sum _{1}^{q}\mid {S}_{i}\mid \u2a7d1+\sum _{0}^{2}\left(\genfrac{}{}{0em}{}{q-1}{i}\right)+(q-1)\left(\genfrac{}{}{0em}{}{q-2}{i}\right).$$nm<({q}^{3}-3{q}^{2}+6q-2)/2\le \frac{{q}^{3}}{2}$for every *q* > 2. It follows that *q* > *⌈*(2*nm*)^{1/3}*⌉*.

We now move towards the upper bounds. Following two results give upper bounds for the partition dimension of *W*_{n}_{,m} but these bounds are not sharp enough.

**Proposition 2.5**. *For every n* > 2 *we have pd*(*W*_{n}_{,m}) ≤ $\lfloor \frac{2n+2}{5}\rfloor +(m-1)\lfloor \frac{2n+4}{5}\rfloor +1.$

*Proof*. We easily obtain $dim({W}_{n,m})=\lfloor \frac{2n+2}{5}\rfloor +(m-$$1)\lfloor \frac{2n+4}{5}\rfloor ,$by ([29], theorem 2.2). By using elementary inequality relating metric and partition dimension we arrive at the required result.

**Proposition 2.6**. *For every n* > 2 *we have pd*(*W*_{n}_{,m}) ≤ $\frac{2}{5}(mn+2m-1)+1.$

*Proof*. From above, we obtain *dim*$({W}_{n,m})=\lfloor \frac{2n+2}{5}\rfloor +(m-$$1)\lfloor \frac{2n+4}{5}\rfloor $But using definition of floor function we have $\lfloor \frac{2n+2}{5}\rfloor +(m-1)\lfloor \frac{2n+4}{5}\rfloor <\frac{2n+2}{5}+\frac{2}{5}(m-1)(2n+2)=$*mn* + 2*m* − 1. Using relation between *pd* and *dim* we obtain the required result.

We now move forward to find upper bound which is sharper then the above bounds.

**Proposition 2.7**. *We have pd*(*W*_{n}_{,m}) ≤ *p* +1*, For every n* > 2 *and p is the least prime number, such that p*(*p* − 1) ≥ *nm*

*Proof*. Let *p* be a the smallest prime number in the sense that *p*(*p* − 1) > *nm*. Since *p* is the prime number, so the sequence *{*0, *i*, 2*i*, 3*i*, . . . , (*p* − 1)*i}*, where 1 6 *i* 6 *p* − 1 and all the above numbers 0, *i*, 2*i*, 3*i*, . . . , (*p* − 1)*i* are reduced modulo *p*. It is obvious that these numbers are the permutation of the set *{*0, 1, . . . , *p* − 1*}*. Now assume that the sequence

$$({x}_{k}{)}_{k}=1,2,\dots ,p(p-1)={X}_{1},{X}_{2},\dots ,{X}_{(p-1)/2}$$where for every 1 6 *i* 6 (*p* − 1)/2 there is a subsequence

$${X}_{i}=0,0,i,i,2i,2i,\dots (p-1)i,(p-1)i$$which contains 2*p* terms and every pair of equal elements in above sequence different from 0, 0. And we can obtain all these numbers from the previous one by adding *i* module *p* to every component. Thus, for *V*(*W*_{n}_{,m}) the resolving partition *Π* = *{S*_{1}, *. . . S*_{p}_{+1}*}* is defined as:

a) if *nm* = *p*(*p*−1) then *S*_{p}_{+1} = *{c}* and also in this case every element *V*_{i}(0 6 *i* 6 *nm* − 1) is assigned to the class *S*_{xi}_{+1+1};

b) if *nm* < *p*(*p* − 1) then *S*_{p} = *{c*, *v*_{nm}_{−1}*}* and also every element *v*_{i}(0 6 *i* 6 *nm* − 2) is assigned to the class *S*_{xi}_{+1+1}.

so,Π is a resolving connected partition of *V*(*W*_{n}_{,m}) having *p* + 1 classes, which implies *pd*(*W*_{n} , *m*) 6 *p* + 1.

**Proposition 2.8**. *For every n* ≥ 4 *we have* [(2*nm*)^{1/3}] ≤ $pd({W}_{n,m})\le 2\lceil (nm{)}^{\frac{1}{2}}\rceil +1$

*Proof*. Since *p* is prime then it should satisfy that *p*(*p*−1) ≥ *nm* so select $p\ge \lceil (nm{)}^{\frac{1}{2}}\rceil +1.$Bertrand’s postulate can be used here which states that for every *n* ≥ 1, there always exists a prime with

$$\begin{array}{r}n<p\le 2n.\end{array}$$It was proved by Chebyshev for the first time in the history, see [30]. So we can easily infer that

$$\lceil (nm{)}^{\frac{1}{2}}\rceil <p\le 2\lceil (nm{)}^{\frac{1}{2}}\rceil .$$So it can be derived that

$$pd({W}_{n,m})\le p+1\le 2\lceil (nm{)}^{\frac{1}{2}}\rceil +1.$$which is the required result.

It is clear that by putting *m* = 1 we obtain simple wheel *W*_{n} with the result, for 4 6 *n* 0 7 the *pd*(*W*_{n}) = 3 , and for 8 6 *n* 6 19 the *pd*(*W*_{n}) = 4 and *pd*(*W*_{3}) = 4 see [25]. So our results is an extension of this result.

*Example*. Let *n* = 3 and *m* = 10, we have *⌈*(2(3)(10))^{1/3}*⌉* 6 *pd*(*W*_{3},10) 6 2*⌈*((3)(10))^{1/2}*⌉* + 1 *⌈*(60)_{1/3} 6*⌉pd*(*W*_{3},10) 6 2*⌈*(30)^{1/2}*⌉* + 1 3 6 *pd*(*W*_{3},10) 6 11

*Example*. Take *W*_{2,10} with resolving partition Π = *{S*_{1}, *S*_{2}, *S*_{3}, *S*_{4}, *S*_{5}, *S*_{6}*}* and *S*_{1} = *{v*_{2}, *v*_{7}, *v*_{8}, *v*_{13}*}* , *S*_{2} = *{v*_{3}, *v*_{11}, *v*_{12}, *v*_{19}*}*,*S*_{3} = *{v*_{5}, *v*_{9}, *v*_{10}, *v*_{16}*}*, *S*_{4} = *{v*_{6}, *v*_{14}, *v*_{15}, *v*_{17}*}*,*S*_{5} = *{v*_{0}, *v*_{1}, *v*_{4}, *v*_{18}*}*

and *S*_{6} = *{c}*

$$\begin{array}{}c|\prod )=(1,1,1,1,1,0),({v}_{0}|\prod )=(2,2,1,2,0,1),& \\ ({v}_{1}|\prod )=(1,2,2,2,0,1),& \\ ({v}_{2}|\prod )=(0,1,2,2,1,1),({v}_{3}|\prod )=(1,0,2,2,1,1),& \\ ({v}_{4}|\prod )=(2,1,1,2,0,1),& \\ ({v}_{5}|\prod )=(2,2,0,2,1,1),({v}_{6}|\prod )=(1,2,1,0,2,1),& \\ ({v}_{7}|\prod )=(0,2,2,1,2,1),& \\ ({v}_{8}|\prod )=(0,2,1,2,2,1),({v}_{9}|\prod )=(1,2,0,2,1,1),& \\ ({v}_{10}|\prod )=(2,1,0,2,2,1),& \\ ({v}_{11}|\prod )=(2,0,1,2,2,1),({v}_{12}|\prod )=(1,0,2,2,2,1),& \\ ({v}_{13}|\prod )=(0,1,2,1,2,1),& \\ ({v}_{14}|\prod )=(1,2,2,0,2,1),({v}_{15}|\prod )=(2,2,1,0,2,1),& \\ ({v}_{16}|\prod )=(2,2,0,1,2,1),\end{array}$$$$\begin{array}{}{v}_{17}|\prod )=(2,2,1,0,1,1),({v}_{18}|\prod )=(2,1,2,1,0,1),& \\ ({v}_{19}|\prod )=(2,0,1,2,1,1).\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\end{array}$$Also in the case of *W*_{3,10} the partition dimension is 8 which is also ≤ *p* + 1, where *p* is 7.

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