## 46 → 280 → 520 →  133 → 55  → 250 →133

### He also found that 12 had a fixed point attractor at 8208, ( 12 --> 17 --> 2402 --> 288 --> 8208 --> 8208 --> 8208 ) which means that 17, 21, 71, 102, 107, 170, 201, 210, 224, 242, 288, 422, 701, 710,828, 882 will all go to the same attractor (and of course many more).

I have now convinced myself that 13, 14, 15, 16, 18, 19,   go to the same cycle as 2 through 9 so now 2 through 9, 11, 13, 14, 15, 16, 18, 19, 20, 22, 23, 24, 25, 26,27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 50, 51, 52, 56, 62, 65, 72, 81, 91 101, 111, 200...
All end in the cycle (13139 --> 6725 --> 4338 --> 4514 --> 1138 --> 4179 --> 9219 --> 13139...

### If my calculating does not contain any big errors, so far I have found end behavior for the following numbers under 50:

Two  have fixed attractors at 1 (1,10)

### forty (listed above) end at the cycle (13139 --> 6725 --> 4338 --> 4514 --> 1138 --> 4179 --> 9219 --> 13139...

Three end in the fixed point 8208 (12, 17, 21 )

5 of the numbers are not yet checked. In time I will reduce these.

### Also, I found that there seems to be a fixed point where some numbers end up at, similar to what Pat found when cubing the digits. This number was 8,208, and satisfies the condition (ie 8^4 + 2^4 + 0^4 + 8^4 = 8208) . What numbers gave me 8,208? I computed a bit more than the first 200 digits (225 to be exact) and found the numbers 12, 17, 21, 46, 64, 71, 102, 107, 120, 137, 145, 154, 170, 173, 201, 210, and 224 (and of course 288) will get to 8,208 and stay there. Now, past these, it’s obvious that 317, 415, 514, 710, 701, 713, 422, etc. also work. When you cubed the digits, Pat found that you reach the number 153 and it repeats forever. However, Pat found that if you have any multiple of three, you will reach 153. There is a pattern there. I am sadly not able to find any pattern with these; they seem to be random, like the happy numbers when you square the digits (1, 7, 10, 13, 19, 23, 28…etc.).

Other four digit

### Another loop number these could go to is 13,139. With 13,139, there is a loop involved (13139 -- 6725 -- 4338 -- 4514 -- 1138 -- 4179 -- 9219 -- 13139…). This has happened with every number I haven’t mentioned (around 90% of the numbers I’ve tried).(lots of numbers take a very long time to drop into their cycle)

Going back to 8208, I keep wondering if there is another fixed point. When cubing the digits, there are five numbers that equal the sum of their digits cubed: 1, 153, 370, 371, and 407. When squaring the digits, there is only one number that equals the sum of its digits squared: 1. But, when taking the fourth power, I have only found two numbers that work: 1 and 8,208.  I do wonder if there are more or not; perhaps a good problem for a computer programmer because doing these manually, though possible, takes up a lot of time. (Derek didn't find the other two fixed point attractors, 1634 and 9474)

### It would seem that the only sum of four fourth powers that sum to 1634 are the actual digits of 1,6,3 and 4. My reasoning is that there can not be any digit greater than 6, since the fourth powers of 7,8, and 9 all exceed 1634.  No four digits picked from 0, 1, 2, 3, and 4 can have a sum large enough since 1634 exceeds 4 times 4^4 .  So there must be at least one five or one six, but not one of each , nor two sixes.  So we need only to check using one six and three less than 5, or one or two fives with the rest less than five. trying each of these combinations fails.

9474 seems to suffer a similar fate of attracting only the numbers made up of the same four digits.