I generally don't help people with homework assignments (I get quite a few requests every week) but this one is pretty interesting, all in all, so let's have a shot at trying to figure out how to solve this problem with a Unix shell script.

The first step is to decide how complicated to make it: we can either use cal and parse the results or, if we're lucky, we'll have a version of the command-line date command that can take arbitrary "epoch values" and turn them into date strings. An epoch value, by the way, is the core date unit in Unix, the number of seconds since the beginning of "Unix time". That value? Well, we can let the command tell us:

$ date -r0
Wed Dec 31 18:00:00 CST 1969

So one way to solve this problem is to simply feed the "date" command epoch values for each day in the date range we need, testing each to see if it's a "Friday" and a "13".

The only hard part is figuring out the epoch date for, say, high noon on January 1st, 2007.

Here's how I'd solve that:

$ date 
Sat Apr 28 14:35:55 CDT 2007
$ date +%s
1177788958

Now we know that epoch date 1177788958 = 28 April, 2007 at 2:23pm

$ expr 60 \* 60 \* 24
86400

That's seconds in a day, the basic value to step from one day to another

$ date -r `expr 1177788958 - 86400`
Fri Apr 27 14:35:58 CDT 2007

Proof: you can see we now see the exact same time of day, one day earlier.
Now, we can also use "date" to figure out how many days into the year we are:

$ date +%j
118

Now, finally, we can just subtract 118 * 88400 to get to January 1st:

$ date -r `expr 1177702558 - 10195200`
Sat Dec 30 13:35:58 CST 2006

Nope. I'm guessing it's a timezone issue, so we can just recalculate...

$ expr 86400 \* 116
10022400
$ expr 1177702558 - 10022400
1167680158
$ date -r 1167680158
Mon Jan  1 13:35:58 CST 2007

Okay, so that's the tough part figured out. We now know the epoch date for 1 January, 2007, mid-day. It's not at noon, but since we'll be at exactly the same time of day for every day it'll work just fine...

Now the script is actually pretty straightforward, believe it or not. As a first stab, let's just see if we can produce a week's worth of dates in a loop:

starton=1167680158      # 1 jan, 2007, midday
oneday=86400
daystotest=7                    # test one week to start

span=$(( $oneday * $daystotest ))
endon=$(( $starton + $span ))

theday=$starton

while [ $theday -le $endon ]; do
  date -r $theday
  theday=$(( $theday + $oneday ))
done

Let's run those settings:

$ sh fridays.sh
Mon Jan  1 13:35:58 CST 2007
Tue Jan  2 13:35:58 CST 2007
Wed Jan  3 13:35:58 CST 2007
Thu Jan  4 13:35:58 CST 2007
Fri Jan  5 13:35:58 CST 2007
Sat Jan  6 13:35:58 CST 2007
Sun Jan  7 13:35:58 CST 2007
Mon Jan  8 13:35:58 CST 2007

Good. Now, let's do another interim solution by counting how many Fridays there are. This we can do by just modifying the while loop:

while [ $theday -le $endon ] 
do
  if [ ! -z "$(date -r $theday  | grep "^Fri ")" ] ; then
    echo "Match: $(date -r $theday)"
    matches=$(( $matches + 1 ))
  fi
  theday=$(( $theday + $oneday ))
done

Now this time I'll run for 60 days by changing "daystotest" to 60, and run it:

$ sh fridays.sh
Match: Fri Jan  5 13:35:58 CST 2007
Match: Fri Jan 12 13:35:58 CST 2007
Match: Fri Jan 19 13:35:58 CST 2007
Match: Fri Jan 26 13:35:58 CST 2007
Match: Fri Feb  2 13:35:58 CST 2007
Match: Fri Feb  9 13:35:58 CST 2007
Match: Fri Feb 16 13:35:58 CST 2007
Match: Fri Feb 23 13:35:58 CST 2007
Match: Fri Mar  2 13:35:58 CST 2007
Matches: 9

Good!! So now let's add one more step: a test for "13" in the grep:

while [ $theday -le $endon ]
do
  if [ ! -z "$(date -r $theday  | grep -E "^Fri.* 13 ")" ] ; then
    echo "Match: $(date -r $theday)"
    matches=$(( $matches + 1 ))
  fi
  theday=$(( $theday + $oneday ))
done

As you can see, I'm using a bit of a fancy regular expression (hence the addition of the "-E" flag) to test for lines that start with "Fri" and contain space-13-space. Now we'll run this for the entire year of 2007 by changing "daystotest" to 365, and...

$ sh fridays.sh
Match: Fri Apr 13 14:35:58 CDT 2007
Match: Fri Jul 13 14:35:58 CDT 2007
Matches: 2

Just about done. Now we just need to figure out how many days are in 10 years. Hopefully you can do that in your head! 3650 days:

sh fridays.sh
Match: Fri Apr 13 14:35:58 CDT 2007
Match: Fri Jul 13 14:35:58 CDT 2007
Match: Fri Jun 13 14:35:58 CDT 2008
Match: Fri Feb 13 13:35:58 CST 2009
Match: Fri Mar 13 14:35:58 CDT 2009
Match: Fri Nov 13 13:35:58 CST 2009
Match: Fri Aug 13 14:35:58 CDT 2010
Match: Fri May 13 14:35:58 CDT 2011
Match: Fri Jan 13 13:35:58 CST 2012
Match: Fri Apr 13 14:35:58 CDT 2012
Match: Fri Jul 13 14:35:58 CDT 2012
Match: Fri Sep 13 14:35:58 CDT 2013
Match: Fri Dec 13 13:35:58 CST 2013
Match: Fri Jun 13 14:35:58 CDT 2014
Match: Fri Feb 13 13:35:58 CST 2015
Match: Fri Mar 13 14:35:58 CDT 2015
Match: Fri Nov 13 13:35:58 CST 2015
Match: Fri May 13 14:35:58 CDT 2016
last day tested was Fri Dec 30 13:35:58 CST 2016
Matches: 18

As you can see, I added a test to see the last day we tested, because we need to take into account leapyears. Not bad, though, we're only off one day in ten years, so I'll readjust it to 3651, but since that penultimate day is Friday, the 31st of December, 2016 can't be a Friday the 13th, so I'll stick with 18 as the answer.

Just for completeness, here's the full script "fridays.sh" for you:

#!/bin/sh

# count occurences of Friday the 13th in the next four years.

starton=1167680158              # 1 jan, 2007, midday
oneday=86400
daystotest=3651                     # a decade, compensating for leap years
matches=0

span=$(( $oneday * $daystotest ))
endon=$(( $starton + $span ))

theday=$starton

while [ $theday -le $endon ]
do
  if [ ! -z "$(date -r $theday  | grep -E "^Fri.* 13 ")" ] ; then
    echo "Match: $(date -r $theday)"
    matches=$(( $matches + 1 ))
  fi
  theday=$(( $theday + $oneday ))
done

echo "last day tested was $(date -r $theday)"
echo "Matches: $matches"

exit 0

There ya go. 18 Friday the thirteenths in ten years means, logically, that they occur, on average, 1.8 times per year. Next, we can try to figure out how often a blue moon occurs! (actually, blue moons are the second full moon in a month, so we could actually solve that one as a shell script. But I'll skip that for now)