How do I print every nth entry of the mth column, starting from a particular line of a file?

Consider the following data in a file file.txt:

$
$
$
FORCE   10   30   40
*        1    5    4
FORCE   11   20   22
*        2    3    0
FORCE   19   25   10
*       16   12    8
.
.
.

I want to print every 2nd element of the third column, starting from line 4, resulting in:

30
20
25

I have tried:

cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'

However, this is not resulting in anything being printed and no errors generated either.

You might use awk checking that the row number > 3 and then check for an even row number with NR%2==0.

Note that you don't have to use cat

awk 'NR > 3 && NR%2==0 {
  print $3
}' file.txt

Output

30
20
25

Using sed

$ sed -En '4~2s/([^ \t]*[ \t]+){2}([^ \t]*).*//p' input_file
30
20
25

> I have tried:
>
> cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'
>
> However, this is not resulting in anything being printed and no errors
> generated either.

You do not need cat whilst using GNU sed as it can read file on its' own, in this case it would be sed 's/\|/ /' file.txt.

You should consider if you need that part at all, your sample input does not have pipe character at all, so it would do nothing to it. You might also drop that part if lines holding values you want to print do not have that character.

Output is empty as NR%2==4 does never hold, remainder of division by x is always smaller than x (in particular case of %2 only 2 values are possible: 0 and 1)

This might work for you (GNU sed):

sed -nE '4~2s/^((\S+)\s*){3}.*//p' file

Turn off implicit printing by setting the -n option and reduce back slashes in regexps by turning on -E.

From the fourth line and then every second line thereafter, capture the third column and print it.

N.B. The \2 represents the last inhabitant of that back reference which in conjunction with the {3} means the above.

Alternative:

sed -n '4,${s/^\(\(\S\+\)\s*\)\{3\}.*//p;n}' file