delete from beginning of file to nth occurrence of pattern

Question

Looking for command line tool (awk, sed, etc.) command to omit lines from the beginning of a file to the nth occurrence of a pattern. Stated another way, print only from the nth occurrence of pattern to end of file. Assume one match per line; line does not necessarily begin with pattern.

e.g. omit from beginning of file to second foo in the following:

something
abc foo1
maybe something else
foo2 -this line and anything before is gone- 
maybe not
foo3
something

Desired result:

maybe not
foo3
something

Bonus points for including/excluding the line with the nth occurrence.

Answer 1

To exclude the n^th occurrence of the pattern

awk -v 'n=3' 'NR == 1, /pattern/ && !--n {next}; 1'

(replace 3 with the number of occurrences you want (2 in your case), and pattern with your actual pattern (foo in your case)).

Example:

$ seq 30 | awk -v 'n=3' 'NR == 1, /6/ && !--n {next}; 1'
27
28
29
30

To include the n^th occurrence of the pattern

awk -v 'n=3' '/pattern/ && !--n, 0'

Example:

$ seq 30 | awk -v 'n=3' '/6/ && !--n, 0'
26
27
28
29
30

Answer 2

% perl -ne 'BEGIN{$NTH=2;$p=0} print if $p; /foo/ && $NTH--; $p=1 if !$NTH' input
maybe not
foo3
something

Where NTH is how many times, and then you need a to-print flag; decrement NTH on match and enable printing if NTH decremented enough. Move print to end to catch the ultimate desired foo.