Printing nth rows after a row number in awk

    1,A
    2,B
    3,C
    4,D
    5,E
    6,F
    7,G
    8,H
    9,I
   10,J
   11,K
   12,L
   13,M
   14,N

How do I print row 4 first and then the 9th row and then the 14th row using awk? I want to first print the 4th row and then continue printing every 5th row after the 4th row, until the end of the file.
I tried this:

awk '{if(NR==4 || (NR>4 && NR==NR+7)) print $0}' file

But this doesn’t work. Any help is appreciated.

The output should be:

4,D
9,I
14,N

>Solution :

Assuming you want to print every 5th line starting from a specific line number:

$ seq 20 | awk 'NR==4{c=4} c && !((++c) % 5)'
4
9
14
19

$ seq 20 | awk 'NR==2{c=4} c && !((++c) % 5)'
2
7
12
17

$ seq 20 | awk 'NR==6{c=4} c && !((++c) % 5)'
6
11
16

c && !((++c) % 5) says:

If c is set then increment c and test if that new value modulo
5 is zero.

So no line before NR==4 can be printed as c is never populated before that happens, and then when c is set to 4, it’s then increment to 5 and 5 % 5 is 0 so the line is printed. c gets incremented for every line after that and so c % 5 continually rotates through 1 2 3 4 0 thus printing every 5th line when 0 occurs and so !0 is true.

To do the above using values set on the command line rather than hard-coded in the script would be:

$ seq 20 | awk -v b=4 -v n=5 'NR==b{c=n-1} c && !((++c) % n)'
4
9
14
19