A160348 Minimal recursive sequence such that if a(n) > 0 then always a(n) > a((f(2n+1)-1)/2), where f is defined by f(2n+1) = (3n+2)/A006519(3n+2) for n>=1, that is f(m) = A075677(2*m-1) for odd m.

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

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

a(0)=0. Let m=3. Then f(m)=5, f^2(m)=1. The corresponding numbers n=(m-1)/2 are 1,2,0. By the condition, a(1) > a(2) > a(0)=0. Therefore let 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=(m-1)/2 are 3,5,8,6,2 and, by the condition, a(3) > a(5) > a(8) > a(6) > a(2)=1. Therefore set a(6)=3 (the minimal value which yet did not appear), a(8)=4, a(5)=5, a(3)=6, etc.

Crossrefs

Cf. A006519, A075677, A159885, A159945, A160198, A122458, A259667.

Sequence in context: A196554 A244647 A324037 * A047708 A256277 A252746

Adjacent sequences: A160345 A160346 A160347 * A160349 A160350 A160351

Keyword

nonn,more

Author

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

Extensions

Name edited by Michel Marcus, Feb 01 2021

Status

approved