Using "seq", If one column is equal to 5, continue the other column

Question

For example, I want to start from the far right column, and once that reaches 5, continue the count from the second farthest column.

0.0.0.0
0.0.0.1
0.0.0.2
0.0.0.3
0.0.0.4
0.0.0.5
0.0.1.5
0.0.2.5
0.0.3.5
0.0.4.5
0.0.5.5
0.1.5.5
0.2.5.5
0.3.5.5
0.4.5.5
0.5.5.5
1.5.5.5
2.5.5.5
3.5.5.5
4.5.5.5
5.5.5.5

So far, I was thinking of using:

for i in $(seq 0 5); do echo "0.0.0.$i"; done

and once i == 5, then well set i=0, and move the echo to the third position.

Answer 1

Using loop:

{
  for i in $(seq 0 5); do echo "0.0.0.$i"   ; done;
  for j in $(seq 1 5); do echo "0.0.$j.$i"  ; done;
  for k in $(seq 1 5); do echo "0.$k.$j.$i" ; done;
  for l in $(seq 1 5); do echo "$l.$k.$j.$i"; done;
}

Using awk and more flexible to increase the repeat time as well as adjustable number of fields from the single line input:

awk -F"." -v OFS="." -v repeat=5 '
{ print $0;
  for(c=NF; c>=1; c--){
      for (i=1; i<=repeat; i++) { $c=i ; print $0; };
  };
}' <<<'0.0.0.0'