A246009 Length of Collatz cycles '3*n + 1' of prime numbers.

2, 8, 6, 17, 15, 10, 13, 21, 16, 19, 107, 22, 110, 30, 105, 12, 33, 20, 28, 103, 116, 36, 111, 31, 119, 26, 88, 101, 114, 13, 47, 29, 91, 42, 24, 16, 37, 24, 68, 32, 32, 19, 45, 120, 27, 120, 40, 71, 14, 35, 84, 53, 22, 66, 123, 79, 30, 43, 17, 43, 61, 118, 38, 87, 131, 38, 25, 113, 126, 33, 126, 51, 46, 20, 59

Offset: 1

Author

Freimut Marschner, Aug 12 2014

Keywords

• nonn

Comments

Define a Collatz cycle C(prime(n)) = {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) + 1}, z >= 1, c(1) = prime(n), n >= 1}; then the length of C(prime(n)) depends only on the starting point c(1) if C ends with c(z) = 1. The length of C(prime(n)) is z, so a(n) = z.

The longest C(prime(n)) out of 10^5 prime numbers is C(prime(96648)) = C(1252663) with a(96648) = 510.

Until now C is not proved mathematically. So if the ending point c(z) is not equal to 1 then C(prime(n)) is not a 'true' Collatz cycle or does not exist.

Examples

			a(1) = {c(1) = prime(1) = 2, 2 mod 2 = 0, c(2) = 2/2 = 1, z=2} = 2;
a(3) = {c(1) = prime(3) = 5, 5 mod 2 = 1, c(2) = 3*5 + 1 = 16; 16 mod 2 = 0, c(3) = 16/2 = 8; 8 mod 2 = 0, c(4) = 8/2 = 4; 4 mod 2 = 0, c(5) = 4/2 = 2; 2 mod 2 = 0, c(6) = 2/2 = 1, z=6} = 6.
		

Links

• Freimut Marschner, Table of n, a(n) for n = 1..100000

Crossrefs

A006577 (Number of halving and tripling steps to reach 1 in '3x+1' problem), A070975 (Number of steps to reach 1 in '3x+1' (or Collatz) problem starting with prime(n)).

Programs

PARI

a(n)=n=prime(n);A=List;while(n != 1,listput(A,n);if(n%2==0,n=n/2,n=3*n+1));listput(A,1);return(#Vec(A)) \\ Edward Jiang, Sep 06 2014

Formula

a(n) = z where {c(z+1) = c(z)/2 if c(z) mod 2 = 0, otherwise c(z+1) = 3*c(z) + 1}, z >= 1, c(1) = prime(n), n >= 1}, a(n) = A006577(prime(n)) + 1 = A070975(n) + 1.