A365048 a(n) is the number of steps required for the n-th odd prime number to reach 3 when iterating the following hailstone map: If P+1 == 0 (mod 6), then the next number = smallest prime >= P + (P-1)/2; otherwise the next number = largest prime <= (P+1)/2.

0, 2, 1, 6, 2, 5, 2, 4, 4, 3, 3, 5, 3, 8, 5, 13, 4, 4, 7, 4, 4, 6, 12, 9, 6, 9, 6, 6, 14, 5, 8, 11, 5, 8, 5, 5, 5, 16, 13, 13, 13, 13, 10, 7, 10, 10, 7, 15, 15, 15, 12, 15, 15, 12, 12, 12, 9, 6, 12, 6, 12, 6, 17, 6, 14, 6, 17, 14, 14, 11, 11, 14, 14, 14, 8, 11, 11, 14, 11, 8, 11, 16

Offset: 1

Author

Najeem Ziauddin, Oct 21 2023

Keywords

• nonn

Comments

Conjecture: This hailstone operation on odd prime numbers will always reach 3.

If the condition "(P + (P-1)/2)" is changed to "(P + (P+1)/2)" then some prime numbers will go into a loop. For example, 449 will loop through 2609.

If the condition "(P+1)/2" is changed to "(P+3)/2" then some prime numbers will go into a loop. For example, 5 will go into the loop 5,7,5,7,....

Examples

			Case 3: 0 steps required.
Case 5: 2 steps required: 5,7,3.
Case 7: 1 step required: 7,3.
Case 11: 6 steps required: 11,17,29,43,19,7,3.
case 17: 5 steps required: 17,29,43,19,7,3.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007528, A065091 (odd primes).

Programs

Mathematica

A365048[n_]:=Length[NestWhileList[If[Divisible[#+1,6],NextPrime[#+(#-1)/2-1],NextPrime[(#+1)/2+1,-1]]&,Prime[n+1],#>3&]]-1;Array[A365048,100] (* Paolo Xausa, Nov 13 2023 *)

Python

from sympy import nextprime, prevprime
def hailstone(prime):
  if (prime + 1) % 6 == 0:
     jump = prime + ((prime - 1) / 2)
     jump = nextprime(jump - 1)
  else:
     jump = ((prime + 1) / 2)
     jump = prevprime(jump + 1)
  return jump
q = 2
lst = []
while q < 3000:
   count = 0
   p = nextprime(q)
   q = p
   while p != 3:
      p = hailstone(p)
      count = count + 1
   lst.append(count)