print n lines, then skip n lines in large text file

Question

I want to use sed to print n lines, skip n lines, print n lines etc until the end of a text file, starting at a certain line. eg stating at line 4, print 5-9, the skip 10-14, print 15-19 etc
From the file

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

I want

5
6
7
8
9
15
16
17
18
19
25
26
27
28
29
35
36
37
38
39
etc

If I try

sed -n '4~5p' FILE.txt > NEWFILE.txt 

will give me
4
9
14
19

which I don't want.

Answer 1

I don't think there's an easy way to do this in sed , as it can't do arithmetic. awk is better:

awk 'NR%10 >= 5' FILE.txt > NEWFILE.txt

NR%10 is the record number modulo 10 (ie the last digit of the line number). So this prints any line where the last digit of the line number is at least 5: 5-9, 15-19, 25-29, etc.

Answer 2

sed is for simple subsitutions on individual lines, that is all. Just use awk:

$ awk '!(NR%5){f=!f} f' file
5
6
7
8
9
15
16
17
18
19

Answer 3

Here's a sed solution. Try figuring that out ;)

sed -n 'n ; n ; n ; n ; n ; h ; n ; H ; n ; H ; n ; H ; n ; H ; x ; p' file

Answer 4

This might work for you (GNU sed):

sed -n '5~10,+4p' file

Use a range where the first address steps ever 10 lines from line 5 and the second address is 4 lines following the first address. See here for details.

BTW sed -n '4~5p' does not give the answer you thought.