Calculate days on which Friday the 13th occurs (inspired from the work of the user justsomeguy)

for i in {2018..2025}-{01..12}-13; do [[ $(date --date $i +"%u" | grep 5) != 5 ]] || echo "$i Friday the 13th"; done
Friday is the 5th day of the week, monday is the 1st. Output may be affected by locale.
Sample Output
2018-04-13 Friday the 13th
2018-07-13 Friday the 13th
2019-09-13 Friday the 13th
2019-12-13 Friday the 13th
2020-03-13 Friday the 13th
2020-11-13 Friday the 13th
2021-08-13 Friday the 13th
2022-05-13 Friday the 13th
2023-01-13 Friday the 13th
2023-10-13 Friday the 13th
2024-09-13 Friday the 13th
2024-12-13 Friday the 13th
2025-06-13 Friday the 13th

By: test666v2
2018-07-10 21:31:02

3 Alternatives + Submit Alt

  • I removed the dependency of the English language Show Sample Output

    for y in $(seq 1996 2018); do echo -n "$y -> "; for m in $(seq 1 12); do NDATE=$(date --date "$y-$m-13" +%w); if [ $NDATE -eq 5 ]; then PRINTME=$(date --date "$y-$m-13" +%B);echo -n "$PRINTME "; fi; done; echo; done
    ginochen · 2018-06-25 09:20:57 1
  • Simply change the years listed in the first seq, and it will print out all the months in that span of years that have Friday the 13ths in them. Show Sample Output

    for y in $(seq 1996 2018); do echo -n "$y -> "; for m in $(seq 1 12); do NDATE=$(date --date "$y-$m-13" +%A); if [ $NDATE == 'Friday' ]; then PRINTME=$(date --date "$y-$m-13" +%B);echo -n "$PRINTME "; fi; done; echo; done
    suspenderguy · 2018-06-13 20:11:46 0
  • Alter the years in the first brace expansion to select your year range. Modify date format to your liking but leave " %w" at the end. Show Sample Output

    for i in {2018..2022}-{01..12}-13; do date --date $i +"%Y %B %w" | sed '/[^5]$/d; s/ 5*$//'; done
    justsomeguy · 2018-07-09 15:47:39 0

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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