A367479 a(n) is the number of steps required for prime(n) to reach 2 when iterating the following hailstone map: If P == 5 (mod 6), then P -> next_prime(P + ceiling(sqrt(P))), otherwise P -> previous_prime(ceiling(sqrt(P))); or a(n) = -1 if prime(n) never reaches 2.

0, 1, 8, 2, 7, 2, 6, 9, 5, 4, 9, 3, 5, 3, 5, 4, 4, 3, 3, 5, 3, 3, 4, 11, 3, 10, 8, 9, 8, 9, 8, 5, 5, 8, 4, 3, 3, 3, 4, 5, 4, 3, 4, 3, 4, 3, 3, 3, 15, 3, 15, 9, 3, 14, 8, 9, 13, 7, 7, 8, 7, 12, 7, 11, 7, 11, 10, 10, 11, 10, 11, 11

Offset

1,3

Comments

next_prime(x) is the next prime >= x, and previous_prime(x) is the next prime <= x.

Conjecture: This hailstone operation on prime numbers will always reach 2.

The map does not go into a loop for any starting prime.

Links

Table of n, a(n) for n=1..72.

Example

For n=1, prime(1)=2, requires a(1)=0 steps to reach 2.
For n=2, prime(2)=3, requires a(2)=1 step: 3 -> 2.
For n=3, prime(3)=5, requires a(3)=8 steps: 5 -> 11 -> 17 -> 23 -> 29 -> 37 -> 7 -> 3 -> 2.

Prog

(Python)
from sympy import nextprime, prevprime
from math import isqrt
def hailstone(prime):
    if (prime + 1) % 6 == 0:
        jump = prime + isqrt(prime-1) + 1
        jump = nextprime(jump - 1)
    else:
        jump = isqrt(prime-1) + 1
        jump = prevprime(jump + 1)
    return jump
def a(n):
    p = nextprime(1, n)
    count = 0
    while p != 2:
        p = hailstone(p)
        count += 1
    return count

Crossrefs

Cf. A007528.

Similar sequence: A365048.

Sequence in context: A153203 A261829 A244686 * A317386 A306994 A152179

Adjacent sequences: A367476 A367477 A367478 * A367480 A367481 A367482

Keyword

nonn

Author

Najeem Ziauddin, Nov 19 2023

Status

approved