in reply to Re^2: Powers of three
in thread Powers of three

A simple explanation of why it works, since you wondered:
If you continue your triangle, you end up with this:
               1
              1 1
             1 0 1
            1 1 1 1
           1 0 0 0 1
          1 1 0 0 1 1
         1 0 1 0 1 0 1
        1 1 1 1 1 1 1 1
       1 0 0 0 0 0 0 0 1
      1 1 0 0 0 0 0 0 1 1
     1 0 1 0 0 0 0 0 1 0 1
    1 1 1 1 0 0 0 0 1 1 1 1
   1 0 0 0 1 0 0 0 1 0 0 0 1
  1 1 0 0 1 1 0 0 1 1 0 0 1 1
 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
You probably see a pattern by now, but if not, change all the 0s to spaces, and all the 1s to #s:
               #
              # #
             #   #
            # # # #
           #       #
          # #     # #
         #   #   #   #
        # # # # # # # #
       #               #
      # #             # #
     #   #           #   #
    # # # #         # # # #
   #       #       #       #
  # #     # #     # #     # #
 #   #   #   #   #   #   #   #
# # # # # # # # # # # # # # # #
No matter how far down you go, you always end up with a triangle, and you can continue by copying that triangle to its bottom left and right sides. This is easy to see, but I can't find the words that actually explain it, so please take a look for yourself if you have doubts about that. Now take a look at this: the first row has 1 as its sum. This triangle-point is copied twice, so the first 2 * 1 rows have 3 * 1 as their sum. This triangle is copied twice, so the first 2 * 2 * 1 rows have 3 * 3 * 1 as their sum. The first 2^3 * 1 rows have 3^3 * 1 as their sum. I'm sure you get the idea by now :)
It's not proof, but hopefully it's good enough.