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Problems & Puzzles:
Puzzles
Puzzle 1249 Iteration
GPD(2*P+3)
On Nov 10, 2025 Alain Rochelli sent the following
nice puzzle.
The 2*P+3 problem is as follows:
start with any prime P (>3) and consider the
greatest prime divisor of 2*P+3, denoted A(P) =
GPD(2*P+3).
Then, repeat the operation: A(A(P)),
A(A(A(P))) … enough to fall into a cycle.
For
example :
P=5 --> A(5)=GPD(2*5+3)=13 ;
A(13)=GPD(2*13+3)=29 ; A(29)=GPD(61)=61 ;
A(61)=GPD(125)=5.
P=7 -->
A(7)=GPD(2*7+3)=17 ; A(17)=37 ; A(37)=GPD(77)=11 ;
A(11)=GPD(2*11+3)=5 ; start a cycle beginning and
ending with 5.
P=11 --> A(11)=GPD(25)=5 ;
idem
P=13 --> A(13)=GPD(2*13+3)=29 ;
A(29)=61 ; A(61)=5 ; idem.
For prime 17, we
obtain : 37 ; 11 ; 5.
For primes 19, 23, 29,
31, 37 and 41, we obtain also a cycle with 5.
But for prime 43, we obtain 89, 181, 73, 149, 43
and so on.
For prime 47, we obtain 97, 197,
397, 797, 1597, 139, 281, 113, 229, 461, 37, 11, 5.
After computations carried out with the first
primes up to 10^6, it is conjectured that all primes
(>= 5) fall into a cycle included 5 or a cycle
included 43:
a) Cicle "5": {5 -> 13-> 29 ->
61 -> 5}
b) Cycle "43": {43 -> 89 -> 181 -> 73 ->
149 -> 43}
Q. Are there more cycles other than
the two mentioned above (5 & 43)?
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From Dec 6 to 12, contributions came from Jeff Heleen, Vicente
Felipe Izquierdo,
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Jeff wrote:
There doesn't seem to be any other cycles < 10^10.
The longest chain in that range starts with prime 2701684771, that
finally drops to the "5" cycle.
2701684771, 154381987, 308763977, 617527957, 65002943, 10000453,
20000909, 2105359, 4210721, 1684289, 3368581, 1347433, 2694869, 769963,
53101, 1931, 773, 1549, 443, 127, 257, 47, 97, 197, 397, 797, 1597, 139,
281, 113, 229, 461, 37, 11, 5 (length 35).
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Vicente wrote:
There are no other cycles until Prime 6.28x10^8 (14.013.085.423). But
these cycles hold true for odd numbers, not just primes, except for the
composite numbers as
39; 363; 3279, 29523; 265719; 2391483; 21523359; ...that drops in the cycle
3 -> 3... These composite integers are produced by:
(3^k-3)/2, k = even => 4
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