Question: Given *day of year*, how can I get the *week of year* by using Bash?

Answer: 
Using the [`date`](http://unixhelp.ed.ac.uk/CGI/man-cgi?date) command, you can show the week of year using the "%V" format parameter:

```
/bin/date +%V

```

You can tell `date` to parse and format a custom date instead of the current one using the "-d" parameter:

```
/bin/date -d "20100215"

```

Then, mixing the two options, you can apply a custom format to a custom date:

```
/bin/date -d "20100215" +%V

```
If you're using GNU `date` you can use relative dates like this:

```
$ doy=193
$ date -d "Jan 1 +$((doy -1)) days" +%U
28

```

This would give you a very simplistic answer, but doesn't rely on `date`:

```
$ echo $((doy / 7))

```

which pays no attention to the day of week.

Here's a demonstration of the week numbering systems:

```
$ printf "\nDate\t\tDOW\tDOY\t%%U %%V %%W\n"; \
for d in "Jan "{1..4}" 2010" \
"Dec "{25..31}" 2010" \
"Jan "{1..4}" 2011"; \
do printf "%s\t" "$d"; \
date -d "$d" +"%a%t%j%t%U %V %W"; \
done

Date            DOW     DOY     %U %V %W
Jan 1 2010      Fri     001     00 53 00
Jan 2 2010      Sat     002     00 53 00
Jan 3 2010      Sun     003     01 53 00
Jan 4 2010      Mon     004     01 01 01
Dec 25 2010     Sat     359     51 51 51
Dec 26 2010     Sun     360     52 51 51
Dec 27 2010     Mon     361     52 52 52
Dec 28 2010     Tue     362     52 52 52
Dec 29 2010     Wed     363     52 52 52
Dec 30 2010     Thu     364     52 52 52
Dec 31 2010     Fri     365     52 52 52
Jan 1 2011      Sat     001     00 52 00
Jan 2 2011      Sun     002     01 52 00
Jan 3 2011      Mon     003     01 01 01
Jan 4 2011      Tue     004     01 01 01

```