cp's OEIS Frontend

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.

%I A367479 #36 Jun 23 2024 16:08:45
%S A367479 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,
%T A367479 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,
%U A367479 7,11,10,10,11,10,11,11
%N 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.
%C A367479 next_prime(x) is the next prime >= x, and previous_prime(x) is the next prime <= x.
%C A367479 Conjecture: This hailstone operation on prime numbers will always reach 2.
%C A367479 The map does not go into a loop for any starting prime.
%e A367479 For n=1, prime(1)=2, requires a(1)=0 steps to reach 2.
%e A367479 For n=2, prime(2)=3, requires a(2)=1 step: 3 -> 2.
%e A367479 For n=3, prime(3)=5, requires a(3)=8 steps: 5 -> 11 -> 17 -> 23 -> 29 -> 37 -> 7 -> 3 -> 2.
%o A367479 (Python)
%o A367479 from sympy import nextprime, prevprime
%o A367479 from math import isqrt
%o A367479 def hailstone(prime):
%o A367479     if (prime + 1) % 6 == 0:
%o A367479         jump = prime + isqrt(prime-1) + 1
%o A367479         jump = nextprime(jump - 1)
%o A367479     else:
%o A367479         jump = isqrt(prime-1) + 1
%o A367479         jump = prevprime(jump + 1)
%o A367479     return jump
%o A367479 def a(n):
%o A367479     p = nextprime(1,n)
%o A367479     count = 0
%o A367479     while p != 2:
%o A367479         p = hailstone(p)
%o A367479         count += 1
%o A367479     return count
%Y A367479 Cf. A007528.
%Y A367479 Similar sequence: A365048.
%K A367479 nonn
%O A367479 1,3
%A A367479 _Najeem Ziauddin_, Nov 19 2023