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

1,1

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.

Links

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

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.

Example

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.

Prog

(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

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)).

Sequence in context: A019186 A019187 A019243 * A373411 A100871 A111860

Adjacent sequences: A246006 A246007 A246008 * A246010 A246011 A246012

Keyword

nonn

Author

Freimut Marschner, Aug 12 2014

Status

approved