I can beat that by one.

$$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$

It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it could be interesting to see what constant you could get. The approach of using squares of successive generators together with the binary expansion of $n$ gives a bound of $4 \log_2 n$, which could be reduced to about $3.5 \log_2 n$ by using positives and negatives in the binary expansion of $n$.

But I think using cubes and the ternary expansion of $n$ is better in general. That seems to give a bound of $5 \log_3 n \sim 3.155 \log_2 n$.

I can beat that by one.

$$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$

It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it could be interesting to see what constant you could get. The approach of using squares of successive generators together with the binary expansion of $n$ gives a bound of $4 \log_2 n$, which could be reduced to about $3.5 (\log_2 n+1)$ by using positives and negatives in the binary expansion of $n$.

But I think using cubes and the ternary expansion of $n$ is better in general. That seems to give a bound of $5 (\log_3 n +1)\sim 3.155 \log_2 n+5$.

I can beat that by one.

$$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$

It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it could be interesting to see what constant you could get. The approach of using squares of successive generators together with the binary expansion of $n$ gives a bound of $4 \log_2 n$, which could be reduced to about $3.5 \log_2 n$ by using positives and negatives in the binary expansion of $n$. In some cases it might be better to use cubes.

I can beat that by one.

$$C_{173} \cong \langle a,b,c,d,e \mid a^4=b, b^2=c,c^4=d,d^4=e, abcde=1 \rangle.$$

It might be hard to prove minimality. I am sure the best you can do is $O(\log n)$, but it could be interesting to see what constant you could get. The approach of using squares of successive generators together with the binary expansion of $n$ gives a bound of $4 \log_2 n$, which could be reduced to about $3.5 \log_2 n$ by using positives and negatives in the binary expansion of $n$.

But I think using cubes and the ternary expansion of $n$ is better in general. That seems to give a bound of $5 \log_3 n \sim 3.155 \log_2 n$.