Problems & Puzzles: Puzzles

Puzzle 354. Another prime game. Werner D. Sand sent the following puzzle: If you sum 3 consecutive odd prime numbers p,q,r, you get a number s which is either prime or not: p+q+r=s. If s is prime, you call it p and repeat the game. If s is not prime, you call the greatest prime factor p and repeat the game. Finally, you get into an infinite cycle, which is one of the following three: 1) 7,31,109,349,1061,103,29,97,43,13 2) 11,41,131,37 3) 17,59 no matter what starting prime numbers you choose Questions: 1) Can you prove or argument that all the 'games' will end in a cycle? 2) Can you discover another cycle or prove that these three are the only one existing?

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Contributions came from Farideh Firoozbakht and Enoch Haga.

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Farideh wrote:

Answer to Q1:

I think all the games will end in a cycle. May be it can be proved

by induction but I don't know the exact proof.

 

Enoch and I tried to find a prime that if we start the game with it before entering the cycle, we get more than 27 primes. So we have found the following sequence with

maximal number of iterations to entering the cycle .

 

43, 61, 79, 101, 271, 337, 1009, 4079, 22571, 136519, 432073, 6346297,

6699743, 18480589, 292709303, 433640083

 

It's interesting that for these primes except for the first, all the games will end

to the cycle (41, 131, 37, 11).

And the corresponding sequence of number of iterations is :

 

5, 6, 8, 10, 11, 13, 17, 18, 21, 25, 26, 27, 28, 30, 31, 34

 

 Example : p=433640083;  n=34

 

433640083, 1300920277, 3902760919, 1300920311, 3902760991, 285567881,

19923341, 59770063, 432073, 432097, 259271, 777857, 2333579, 72173, 43321,

130043, 390151, 40361, 121171, 363541, 4211, 12647, 12653, 1151, 3467, 10427,

467, 1433, 617, 1867, 181, 113, 53, 173, (41, 131, 37, 11)

 

Note that before entering the cycle we need 34 iterations so the statement "At the beginning, before entering the cycle, you may have a leader of numbers, but no more than 28." isn't correct (please see A109756).

 

Answer to Q2:

I checked up to prime(45470000)=888875167 and it seems that there is no other cycle.

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Enoch wrote:

Q1) By running a program to generate random primes, I found all three cycles and no more. I also found that specific "trigger" sums were followed by the cycles. Perhaps a study of these triggers might lead to an explanation of how and why these cycles occur. I hypothesize that since the largest prime factor of any nonprime sum is smaller than the sum, this begins the downward trend to the cycle.

Q2) I found no other cycles. I feel that perhaps there could be other cycles as the limit of the primes studied increases. I have no evidence, just a guess.
 

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Werner D. Sand wrote:

My own ideas about my puzzle:  It is not self-evident that the sequence does not diverge.  The sequence s=p²+q²+r² seems to diverge.  The calculation specification s must only be “strong” enough.  I assume "convergence" (cycle) for s=linear and divergence for s=quadratic or more.  More exactly:  the "convergence radius" in s=p^c+q^c+r^c is c=1.5.   If the sequence s does not diverge, thus is upward limited, then it necessarily flows in a cycle (which may extremely be a constant, in our case containing at least 2 elements),  because sometime it must meet an s, which already was met, at the latest after b steps, if the upper barrier is b.  There are many possible variants of the play.  The simplest is:  s=2p+1.  Here we get only 1 cycle:  3,7,5,11,23,47,19,13 (that would be a further puzzle).  One can also choose 5 or 7 or more sequential prime numbers.

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