A354747 Start with 2*n-1; repeatedly triple and add 2 until reaching a prime. a(n) = number of steps until reaching a prime > 2*n-1, or 0 if no prime is ever reached.

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 1, 2, 10, 1, 1, 2, 1, 2, 4, 1, 1, 1, 2, 1, 1, 4, 3, 2, 3, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 2, 1, 3, 3, 1, 1, 2, 3, 3, 5, 1, 1, 1, 2, 3, 9, 1, 1, 2, 1, 2, 4, 1, 2, 1, 6, 1, 1, 2, 1, 1, 5, 1, 3, 1, 2, 1, 1, 3, 1

Offset: 1

Author

Felix Fröhlich, Jun 06 2022

Keywords

• nonn

Comments

a(n) is the smallest m >= 1 such that 2*n*3^m - 1 is prime.

The smallest unknown case is n = 100943. Is a(100943) = 0?

If it exists, a(100943) > 30000. - Michael S. Branicky and Jon E. Schoenfield, Jun 07 2022

Examples

			For n = 21: Successively applying the map x -> 3*x+2 to 2*21-1 = 41 yields the sequence 41, 125, 377, 1133, 3401, 10205, 30617, 91853, 275561, 826685, 2480057, reaching the prime 2480057 after 10 steps, so a(21) = 10.
		

Crossrefs

Cf. A016789, A050412, A354748.

Programs

PARI

a(n) = my(x=2*n-1, i=0); while(1, x=3*x+2; i++; if(ispseudoprime(x), return(i)))

Python

from sympy import isprime
def f(x): return 3*x + 2
def a(n):
    fn, c = f(2*n-1), 1
    while not isprime(fn): fn, c = f(fn), c+1
    return c
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Jun 06 2022