Subtract shellscript input argument value

Question

I have the script below that prints on the screen the even numbers of 1 until the input argument.

#!/bin/bash
for i in $(seq 1 ($1)); do
    if [ $(($i % 2)) -eq 0 ]
    then
    echo "$i \c"
    fi
done

For example, for the execution of the script with argument 10 the script prints the following

2 4 6 8 10 

I would like the script to print only up to the input value minus 1. That is, $1-1 which for this case is 9.

---

I have tried the following modification without success.

for i in $(seq 1 (( $1 - 1 )) ); do
    if [ $(($i % 2)) -eq 0 ]
    then
    echo "$i \c"
    fi
done

Answer 1

I also got it this way

for i in $(seq 1  `expr $1 - 0`); do
    if [ $(($i % 2)) -eq 0 ]
    then
    echo "$i,\c"
    fi
done

Answer 2

You only need one $ in the subtraction expression (and remove spaces):

#!/bin/bash
for i in $(seq 1 $(($1 - 1))); do
  if [ $(($i % 2)) -eq 0 ]
  then
    echo "$i"
  fi
done

Extracting a variable becomes more readable:

#!/bin/bash
last=$(($1 - 1))
for i in $(seq 1 $last); do
  if [ $(($i % 2)) -eq 0 ]
  then
    echo "$i"
  fi
done

Like 1 will never be printed, not even the if it is necessary:

#!/bin/bash
last=$(($1 - 1))
for i in $(seq 2 2 $last); do echo $i; done

Several options! :)