Batch copy and rename multiple files in the same directory

I have 20 files like:

01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
...

Files have a similar format in their names. They begin with 01, and they have 01*AAA*.sh format.

I wish to copy and rename files in the same directory, changing the number 01 to 02, 03, 04, and 05:

02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh
...

03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh
...

04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh
...

05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh
...

I wish to copy 20 of 01*.sh files to 02*.sh, 03*.sh, and 04*.sh. This will make the total number of files to 100 in the folder.

I'm really not sure how can I achieve this. I was trying to use for loop in the bash script. But not even sure what should I need to select as a for loop index.

for i in {1..4}; do
   cp 0${i}*.sh 0${i+1}*.sh
done

does not work.

Solution

There are going to be a lot of ways to slice-n-dice this one ...

One idea using a for loop, printf + brace expansion, and xargs:

for f in 01*.sh
do
    printf "%s\n" {02..05} | xargs -r -I PFX cp ${f} PFX${f:2}
done

The same thing but saving the printf in a variable up front:

printf -v prefixes "%s\n" {02..05}

for f in 01*.sh
do
    <<< "${prefixes}" xargs -r -I PFX cp ${f} PFX${f:2}
done

Another idea using a pair of for loops:

for f in 01*.sh
do
    for i in {02..05}
    do
        cp "${f}" "${i}${f:2}"
    done
done

Starting with:

$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh

All of the proposed code snippets leave us with:

$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh

02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh

03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh

04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh

05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh

NOTE: blank lines added for readability