awk to print all columns from the nth to the last with spaces

Question

I have the following input file:

a 1  o p
b  2 o p p
c     3 o p p  p

in the last line there is a double space between the last p\'s

Answer 1

GNU sed

remove first n fields

sed -r 's/([^ ]+ +){2}//' file

GNU awk 4.0+

awk '{sub("([^"FS"]"FS"){2}","")}1' file

GNU awk <4.0

awk --re-interval '{sub("([^"FS"]"FS"){2}","")}1' file

Incase FS one doesn't work(Eds suggestion)

awk '{sub(/([^ ] ){2}/,"")}1' file

Replace 2 with number of fields you wish to remove

EDIT

Another way(doesn't require re-interval)

awk '{for(i=0;i<2;i++)sub($1"[[:space:]]*","")}1' file

Further edit

As advised by EdMorton it is bad to use fields in sub as they may contain metacharacters so here is an alternative(again!)

awk '{for(i=0;i<2;i++)sub(/[^[:space:]]+[[:space:]]*/,"")}1' file

Output

o p
o p p
o p p  p