Problems & Puzzles: Puzzles

Puzzle 545.- 14 or 17<->25

Jeff Hellen sent the following puzzle.

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?

____
(*) 1->1; 2->2; 3->3; 4->4; 5->5; 6->5; 7->7: 8->5; 9->5; 10->5; 12->5

Torbjörn Alm wrote:

I  run a search up to 800000000 with no better solution.
Q1: No number below 800000000 ends up other than at 14 or 17/25.
Observation #2 is that at start the Jeff-numbers drop very fast until about 10000000.
If an alternative number existed, it has to be > 800000000.
There a loop has to stay for infinity. This in contradicted by the fact
that all Jeff(n) are < n for n> 1000000.

Somewhere at about 1000 you get Jeff(n)>n, but finally it drops down to 14 or 17/25.

The champion has come across a very long loop. I found a number of
n-values giving chain lengths just below 30.

Q2: The fact that Jeff(n) drops so quickly, may be with a factor 100 in many cases,
moves a possible solution with 40 elements far far away.
, probably >> 10^20

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