A. Take any number, say ABCDE, and from it
obtain:
1) all single digit numbers A, B, C, D,
E
2) all two-digit numbers from left to right AB, BC, CD, DE
3) all three-digit numbers from left to right ABC, BCD, CDE
4) all four-digit numbers from left to right ABCD, BCDE
5) and last the number itself ABCDE
This gives (for a 5 digit initial number)
15 separate numbers.
B. Break down all 15 numbers into their
prime factors. (For a zero or one use zero or one, respectively, and for
a prime number use just the prime itself.)
C. Take all these factors and add them
together to yield a new number.
D. Repeat the process until reach and
ending point or ending cycle.
With the only exceptions being 0 through
10 and 12 (*), all starting numbers seem to eventually cycle down to an
endpoint of either 14 or the two-number loop 17<-> 25.
I have taken this up to 100,000,011 and
there have been no other exceptions. Twenty-eight percent of
the numbers reach the endpoint 14 with the remainder going to the loop
17-25.
Q1. Besides the 0-10 & 12, exceptions, are
there no others? ( that is to say,
is there any
other ending point [as 14] or ending cycle
[as 17 <-> 25]?)
I (CR) counted the
cycles needed to reach the ending point, and I saved the smallest
champions.
For example: The
smallest number needing 3 cycles is 19:
19 ->
1+3+3+19=26 -> 2+2+3+2+13 = 22 -> 2+2+2+11 = 17
(end)
The largest cycle I
got was 30 for the champion (minimal) number
279799231.
Q2. Can you get the minimal number for 40 & 50
cycles?
