All this talk of the 2^n sequence and the repeating pattern of 2 4 8 6 got me wondering: Is there a repeating pattern in the tens column? What about the hundreds column and so forth? I would need to calculate the sequence 2 4 8 16 32 64 128 256 512 1024... until I see a pattern.
I couldn't get very far with a direct calculation on the calculator or computer because I would need to make some
really BIG numbers with every digit explicitly stated. (Actually, for this purpose, I only need the last 5 or 6 digits of each number, but I like to be consistent.)
It is easy to see on a calculator that the tens column has a repeating pattern, the first of which begins at 2^2 = 4 and ends at 2^21 = 20947152. In fact you can see below that the two patterns in the tens an units columns are closely linked. The tens pattern begins at n = 2, 22, 42, 62, etc. Each of those numbers (2^2, 2^22, 2^42, 2^62,...) end with 04. So the repeating pattern we have here is 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52.
2^n n
-----------------
00000002 1
000000
04 2 <-- tens pattern begins
00000008 3
00000016 4
00000032 5
00000064 6
00000128 7
00000256 8
00000512 9
00001024 10
00002048 11
00004096 12
00008192 13
00016384 14
00032768 15
00065536 16
00131072 17
00262144 18
00524288 19
01048576 20
02097152 21 <-- tens pattern ends
041943
04 22 <-- tens pattern repeats
08388608 23
16777216 24
...
I used Excel for the really big number crunching. I started with 2, then wrote a formula to find each digit in its own individual cell. I assumed that the hundreds pattern would be 100 numbers long. Consider the units pattern (2 4 8 6) is four numbers long. The tens pattern (04 08 ... 76 52) is 20 numbers long. 4 * 5 = 20. So I did 20 * 5 = 100. I also assumed that it would start with 2^3 = 008 since the tens pattern starts with 2^2 = 04 and the ones pattern starts with 2^1 = 2.
Sure enough, I got first repetition of the hundreds pattern at
2^103 =
10141204801825835211973625643
008
I then derived that each pattern has length
L = 5^C - 5^(C-1)
where C is the column number, beginning with C = 1 for units, C = 2 for tens, C = 3 for hundreds, etc.
Also, each pattern has its first repetition at
n = L + C
For the thousands pattern, we have
C = 4
L = 5^4 - 5^3 = 500
n = 500 + 4 = 504
So the thousands pattern is 500 numbers long, begins at 2^4 = 0016 and repeats at
2^504 = 52374249726338269920211035149241586435466272736689036631732661889538140742474792 87813232147721446651441418694604096113614747610473416628885325644143
0016
By the way, when I say "500 numbers long", I mean all the numbers from 2^4 to 2^503. The above number for 2^504 is 152
digits long.
Before using Excel, I calculated 2^121 by hand, some of it in my head, and some of it in 12 digit chunks with my TI-89 calculator. I got the final answer terribly wrong, though, because I missed carrying a 1, half way through.
Thus I have spent a good chunk of my weekend calculating 2^504
JUST FOR FUN ! 
Take THAT, neurotypicals !!!
