You can save the date of every calculation in a variable and use it to calculate the next date. I use the default format in standard locale for storing the date because date cannot parse your specified format.
# start date d=$(LC_ALL=C date -d "1998-01-01 00:00") echo "# d=$d" # convert format date -d "$d" +"%d-%m-%Y %H:%M" for i in {1..1825}; do # add 6 hours d=$(LC_ALL=C date -d "$d +6 hours") # convert format date -d "$d" +"%d-%m-%Y %H:%M" done This prints
# d=Thu Jan 1 00:00:00 CET 1998 01-01-1998 00:00 01-01-1998 06:00 01-01-1998 12:00 01-01-1998 18:00 02-01-1998 00:00 02-01-1998 06:00 02-01-1998 12:00 02-01-1998 18:00 03-01-1998 00:00 03-01-1998 06:00 03-01-1998 12:00 03-01-1998 18:00 04-01-1998 00:00 04-01-1998 06:00 ... The results depend on your time zone and daylight saving time rules.
Your starting time specification may be ambiguous because it doesn't specify the time zone.
Edit based on Paul_Pedant's comment:
Make an outer loop for + $i days, and an inner loop hard-coded 00, 06, 12, 18. It runs a quarter as many date processes, and it does not care about DST variations.
# start date d=$(LC_ALL=C date -d "1998-01-01 00:00") echo "# d=$d" for i in {1..456}; do # convert (partial) format out=$(date -d "$d" +"%d-%m-%Y") for h in 00 06 12 18; do echo "$out $h:00" done # add 1 day d=$(LC_ALL=C date -d "$d +1 day") done Please specify in the question what result you expect when switching between normal time and daylight saving time occurs. (strictly printing 00:00, 06:00, 12:00, 18:00 or exactly 6 hours time difference when switching DST)
Edit based on glenn jackman's answer
Adding option -u to all date calls will make the script nearly the same as written by glenn jackman and result in strictly printing 00:00, 06:00, 12:00, 18:00 regardless of DST.