Print Matching line and nth line from the matched line

In awk, you'd do it as follows

awk '/pattern/{nr[NR]; nr[NR+4]}; NR in nr' file > new_file`

or

awk '/pattern/{print; nr[NR+4]; next}; NR in nr' file > new_file`

Explanation

The first solution finds all lines that match pattern. When it finds a match it stores the record number (NR) in the array nr. It also stores the 4th record from NR in the same array. This is done by the nr[NR+4]. Every record (NR) is then checked to see if it's present in the nr array, if so the record is printed.

The second solution works essentially the same way, except when it encounters th e pattern it prints that line, and then stores the 4th record ahead of it in the array nr, then goes to the next record. Then when awk encounters this 4th record the NR in nr block will get executed and print this +4 record there after.

Example

Here's an example data file, sample.txt.

$ cat sample.txt 
1
2
3
4 blah
5
6
7
8
9
10 blah
11
12
13
14
15
16

Using the 1st solution:

$ awk '/blah/{nr[NR]; nr[NR+4]}; NR in nr' sample.txt 
4 blah
8
10 blah
14

Using the 2nd solution:

$ awk '/blah/{print; nr[NR+4]; next}; NR in nr' sample.txt 
4 blah
8
10 blah
14

You can try the -A option with grep, which specifies how many lines after the matching line should be printed. Couple this with sed, and you would get the required lines.

grep -A 4 pattern input.txt | sed -e '2,4d'

Using sed, we delete the from the second line until the fourth.

sed -n 's/^[ \t]*/; /img class=\"devil_icon/,+4 { 3,5d ; p }' input.txt

I'm simply adding a deletion of the appropriate lines, before printing { 3,5d ; p }.