

A116641


A116623 sorted, without duplicates.


6



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
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OFFSET

0,2


COMMENTS

Related to the parity vectors of Terras and Collatz trajectories.
Let R_s be the reduced Collatz sequence starting with s and let R_s(i), i >= 0 be the ith 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 previouslygenerated 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



CROSSREFS

Cf. A116642 gives the same sequence in binary.


KEYWORD

nonn


AUTHOR

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


STATUS

approved



