A116641 A116623 sorted, without duplicates.

1, 5, 7, 11, 19, 23, 29, 31, 35, 37, 47, 49, 53, 65, 67, 73, 79, 85, 89, 97, 101, 103, 119, 121, 125, 131, 133, 143, 149, 151, 157, 161, 169, 175, 179, 185, 197, 205, 211, 215, 221, 223, 227, 233, 239, 251, 259, 269, 271, 275, 277, 283, 287, 289, 313, 319, 323

Offset: 0

Author

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Keywords

• nonn

Comments

Related to the parity vectors of Terras and Collatz trajectories.

From Bob Selcoe, Sep 14 2019: (Start)

Let R_s be the reduced Collatz sequence starting with s and let R_s(i), i >= 0 be the i-th term in R_s. Then any term in R_s can be described as (3*s^i + k)/2^j, where j is the total number of halving steps from R_s(0) to R_s(i) i >= 1, and k is some term in A116641. k=1 when i=1; when i > 1, k is determined by the specific order of halving steps in R_s.

Ignoring duplicates, terms in A116641 > 1 can be generated by a series of subsequences:

1. Start with subsequence a(m) = 3+2^m, m >= 1; i.e., a(m) = {5,7,11,19,35,67,...}.

2. For fixed m, generate new subsequences b(n) = 3*a(m) + 2^(m+n), n >= 1; so:

m=1, a(1)=5, b(n) = 3*5 + {4,8,16,32,...} = {19,23,31,47,...};

m=2, a(2)=7, b(n) = 3*7 + {8,16,32,64,...} = {29,37,53,85,...};

m=3, a(3)=11, b(n) = 3*11 + {16,32,64,128,...} = {49,65,97,161,...}; etc.

3. Let 2^y be the summand used to find terms (t) in any previously-generated subsequence. (For instance, in m=2, b(3)=53: y=5 because t=53 = 3*7 + 32.) Continue generating new subsequences p(q) = 3*t + 2^(y+z) {z=1..inf} for all t. So in this example, from t=53 we get p(q) = 3*53 + {64,128,256,512,...} = {223,287,415,671,...}; from t=671 we get p(q) = 3*671 + {1024,2048,4096,...} = {3037,4061,6109,...}, etc. (End)

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A116642 gives the same sequence in binary.