A246007 Length of pseudo-Collatz cycles '3*n - 1' of prime numbers.

2, 5, 3, 6, 7, 7, 19, 5, 4, 11, 7, 15, 10, 9, 14, 17, 12, 8, 21, 20, 16, 15, 11, 33, 36, 36, 18, 10, 14, 31, 26, 22, 21, 13, 26, 34, 16, 12, 21, 42, 25, 16, 16, 37, 20, 29, 19, 24, 32, 90, 28, 28, 19, 19, 85, 23, 40, 14, 36, 27, 22, 49, 17, 31, 31, 40, 13, 44, 43, 26, 66, 43, 25, 25, 25, 30, 21, 30, 30, 51, 20, 25, 25, 33, 47, 16, 47, 91, 46, 46, 29, 46, 28

Offset: 1

Author

Freimut Marschner, Aug 10 2014

Keywords

• nonn

Comments

Define a pseudo-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} depending of the starting point c(1). If c(1) = prime(n) then c(z) might be

(1) finite convergent to c(z) = 1 or

(2) infinite periodic from c(z) = 7 or from c(z) = 17 or

(3) no cycle if c(z) = -1.

The case (3) is not observed out of 10^5 prime numbers. So a(n) = z is the length of the C(prime(n)) up to the stoppping point, where c(z) = 1 or up to the periodical point, where c(z) = 7 or c(z) = 17 or c(z) = c(1). See Table for examples of cases (1) and (2). The longest sequence here is a(99147) = 560 with starting point c(1) = prime(99147) = 1287511 up to the periodical point c(560) = 17.

Examples

			a(1) = {c(1) = prime(1) = 2, 2 mod 2 = 0, c(2) = 2/2 = 1, z=2} = 2.
Table for cases (1) and (2):
case (1)
c(1) = prime(2) = 3
z    1 2 3 4 5
c(z) 3 8 4 2 1
a(2) = 5
c(1) = prime(3) = 5
z    1  2 3
c(z) 5 14 7
a(3) = 3
c(1) = prime(10) = 29
z     1  2  3   4  5  6  7 8 9 10 11
c(z) 29 86 43 128 64 32 16 8 4  2  1
a(10) = 11
case (2)
c(1) = prime(4) = 7
z    1  2  3 4  5 6  7 ...
c(z) 7 20 10 5 14 7 20 ...
a(4) = 6
c(1) = prime(7) = 17
z     1  2  3  4  5   6  7   8  9 10  11 12  13
c(z) 17 50 25 74 37 110 55 164 82 41 122 61 182
z    14  15  16 17 18 19 20 ...
c(z) 91 272 136 68 34 17 50 ...
a(7) = 19
		

Links

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

Crossrefs

A003627 (Primes of form 3n-1), A006370 (Image of n under the '3x+1' map), A014682 (The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2), A006577(Number of halving and tripling steps to reach 1 in '3x+1' problem), A016789({3n+2, n >=0} = {3n-1, n >= 1}).

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}.