That's because Mac OS X doesn't actually have a command called "seq", but fortunately the basic task you want to accomplish is easily done as a shell script, and if you're so inclined, drop it in $HOME/bin, add that to your PATH and name it "seq" and you're good to go:
#!/bin/sh
# count - step through numeric values until you get to the max value
if [ "$#" -lt 2 ] ; then
echo "Usage: count {increment}"
exit 1
fi
counter="$1"
max="$2"
if [ "$#" -eq 2 ] ; then
step=1
else
step=$3
fi
while [ $counter -le $max ] ; do
echo $counter
counter=$( expr $counter + $step )
done
exit 0
This is actually kind of fancy as a shell replacement script because not only does it let you step from any value to any secondary value, but you can also specify the increment, so if you need to count only even numbers for some obscure reason, you could do that.
If you really want to experiment with scripting, try writing it so that if the second value is less than the former value that it automatically decrements instead of incrementing. That is, count 1 5 would produce 1 2 3 4 5, but count 10 5 would correctly produce 10 9 8 7 6 5 rather than just produce the first value and then find that the test to see if the current value is greater than the max value succeeds.
If you really want to explore, also tweak it so you can do Fibonacci sequences and other complicated increments (a Fibonacci sequence is where each value is the sum of the previous two, e.g., 1 1 2 3 5 8 13 21 34 55...). Of course, if you do explore in that direction it might well indicate that you have too much time on your hands, in which case you would probably enjoy my book Wicked Cool Shell Scripts too. :-)
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Another solution (on some OS, like Solaris) is to use a special feature of the shell :
echo {1..4}
1 2 3 4
echo {8..4}
8 7 6 5 4