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

Author

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

Keywords

• nonn
• more

Comments

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

Examples

			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.

Extensions

Name edited by Michel Marcus, Feb 01 2021