

A160348


Minimal recursive sequence such that if a(n)>0 then always a(n)>a((f(2n+1)1)/2), where f is defined as in A159885


2



0, 2, 1, 6, 7, 5, 3, 11, 4, 13, 14, 10, 15, 52, 12, 50, 53, 9, 54, 59, 51, 62, 63, 49, 60, 65, 8, 68, 69, 58, 16, 75, 61, 56, 76, 48, 77, 80, 64, 84, 85, 67, 78, 88, 57, 44
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

If the (3x+1)Collatz conjecture is true, then this sequence is a permutation of the nonnegative integers.


LINKS

Table of n, a(n) for n=0..45.


EXAMPLE

Put a(0)=0. Let m=3. Then f(m)=5, f^2(m)=1. The corresponding numbers n=(m1)/2 are 1,2,0. By the condition, a(1)>a(2)>a(0)=0. Therefore put a(2)=1, a(1)=2. Furthermore, consider m=7. Then f(m)=11, f^2(m)=17, f^3(m)=13, f^4(m)=5. The corresponding numbers n=(m1)/2 are 3,5,8,6,2 and, by the condition, a(3)>a(5)>a(8)>a(6)>a(2)=1. Therefore put a(6)=3 (the minimal value which yet did not appear), a(8)=4, a(5)=5, a(3)=6, etc.


CROSSREFS

Cf. A259667 A159945 A160198 A006519 A122458
Sequence in context: A196554 A244647 A324037 * A047708 A256277 A252746
Adjacent sequences: A160345 A160346 A160347 * A160349 A160350 A160351


KEYWORD

nonn,uned


AUTHOR

Vladimir Shevelev, May 10 2009; corrected May 13 2009, May 19 2009


STATUS

approved



