

A246007


Length of pseudoCollatz cycles '3*n  1' of prime numbers.


2



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
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OFFSET

1,1


COMMENTS

Define a pseudoCollatz 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.


LINKS



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}.


EXAMPLE

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


CROSSREFS

A003627 (Primes of form 3n1), 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} = {3n1, n >= 1}).


KEYWORD

nonn


AUTHOR



STATUS

approved



