A376801 a(0) = 397; a(n+1) = a(n)^2 if a(n) is prime, floor(a(n)/2) otherwise.

397, 157609, 78804, 39402, 19701, 9850, 4925, 2462, 1231, 1515361, 757680, 378840, 189420, 94710, 47355, 23677, 560600329, 280300164, 140150082, 70075041, 35037520, 17518760, 8759380, 4379690, 2189845, 1094922, 547461, 273730, 136865, 68432, 34216, 17108, 8554

Offset: 0

Author

Matthew House, Oct 04 2024

Keywords

• nonn

Comments

From most starting points, this sequence trivially reaches the cycle (4,2,4,2,...). No other cycles are known to exist. Heuristic results suggest that every starting point almost surely reaches a cycle, despite the colossal size of the intermediate values. a(0) = 397 is the first starting point which is not known to reach the (4,2) cycle: heuristically, the sequence likely runs for 10^12 terms or more before hitting it.

Examples

			The first few primes in the sequence are 397, 1231, 23677, 1069, 4463, and the value is squared after each of these. Otherwise, the value is halved.
		

Links

• Matthew House, Table of n, a(n) for n = 0..2707
• Matthew House, Table of n, a(n) for a(n) prime, n <= 2341068 (primes up to 10^10000 proven via ECPP)
• Joseph O'Rourke et al., A Collatz-like function that bifurcates on primes, MathOverflow, 2015-2024.
• Strashimir G. Popvassilev et al., Are there always at least *five* divisions?, MathOverflow, 2015.

Programs

Mathematica

NestList[If[PrimeQ[#], #^2, Quotient[#, 2]] &, 397, 50]

Python

from functools import lru_cache
from sympy import isprime
@lru_cache(maxsize=None)
def A376801(n): return (a**2 if isprime(a:=A376801(n-1)) else a>>1) if n else 397 # Chai Wah Wu, Oct 25 2024

Formula

If a(n) is prime and a(n) >= 13, then a(n+6) = floor(a(n)^2/32).