Weird behavior in zsh random generator

Question

I have this script to generate $1 digit(s) of mixed random char from the defined charlists as arrays.

#!/bin/bash

charlist1=(a b c d e f g h i j k l m n o p q r s t u v w x y z)
charlist2=(0 1 2 3 4 5 6 7 8 9)
charlists=(${charlist1[*]} ${charlist2[*]})

i="1"
while [ $i -le $1 ]; do
    out=$out`echo -n ${charlists[$(($RANDOM % ${#charlists[*]}))]}`
    i=$(( i + 1 ))
done

echo $out

It runs fine in bash, but then if I invoke it with zsh by zsh that_script_above.sh 6 it just generates 6 digit of some same character like this:

>>> zsh that_script_above.sh 6
llllll
>>> zsh that_script_above.sh 6
bbbbbb

And if I modified that script to be like this:

#!/bin/bash

charlist1=(a b c d e f g h i j k l m n o p q r s t u v w x y z)
charlist2=(0 1 2 3 4 5 6 7 8 9)
charlists=(${charlist1[*]} ${charlist2[*]})

i="1"
while [ $i -le $1 ]; do

    echo -n ${charlists[$(($RANDOM % ${#charlists[*]}))]}

    i=$(( i + 1 ))
done

echo

it works well as I desired for both bash and zsh.

So, here is my questions:

1. Can someone explain to me about my problem with zsh bash behaviour stated above?
2. And how can I use for loop in bash with customizable variable? because it seems for i in {1..$1} doesn't work in bash.

Answer 1

Q1. Command substitution (backticks) uses a subshell, and in zsh a subshell's RNG state not reseeded. Since you repeatedly create a new subshell without using $RANDOM in the parent, you get the same value in each subshell. See:
https://stackoverflow.com/questions/32577117/references-to-random-in-subshells-all-returning-identical-values
https://superuser.com/questions/1210435/different-behavior-of-in-zsh-and-bash-functions

You don't need the command-substitution and echo, and you also don't need the $((..)) because an array subscript is already evaluated as an arithmetic expression, but you do need +1 because zsh arrays are 1-origin (you were lucky you didn't happen to hit 0):

 out=$out${charlists[ $RANDOM % ${#charlists[*]} + 1 ]}

Aside: even if you did need an echo in command substitution you don't need -n because command substitution itself removes any trailing newline(s) from the data captured and substituted.

Q2. bash does brace-expansion before it does parameter-expansion (and command and arithmetic substitution/expansions), but zsh (and ksh) does it after. You can use for i in $(seq 1 $1) or yucky but builtin for i in $(eval "echo {1..$n}")