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%I #18 Oct 21 2019 02:19:21
%S 2,8,6,17,15,10,13,21,16,19,107,22,110,30,105,12,33,20,28,103,116,36,
%T 111,31,119,26,88,101,114,13,47,29,91,42,24,16,37,24,68,32,32,19,45,
%U 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
%N Length of Collatz cycles '3*n + 1' of prime numbers.
%C 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.
%C The longest C(prime(n)) out of 10^5 prime numbers is C(prime(96648)) = C(1252663) with a(96648) = 510.
%C 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.
%H Freimut Marschner, <a href="/A246009/b246009.txt">Table of n, a(n) for n = 1..100000</a>
%F 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.
%e a(1) = {c(1) = prime(1) = 2, 2 mod 2 = 0, c(2) = 2/2 = 1, z=2} = 2;
%e 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.
%o (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
%Y 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)).
%K nonn
%O 1,1
%A _Freimut Marschner_, Aug 12 2014
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