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%I #15 Feb 01 2021 13:29:45
%S 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,
%T 8,68,69,58,16,75,61,56,76,48,77,80,64,84,85,67,78,88,57,44
%N 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.
%C If the (3x+1)-Collatz conjecture is true, then this sequence is a permutation of the nonnegative integers.
%e 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.
%Y Cf. A006519, A075677, A159885, A159945, A160198, A122458, A259667.
%K nonn,more
%O 0,2
%A _Vladimir Shevelev_, May 10 2009; corrected May 13 2009, May 19 2009
%E Name edited by _Michel Marcus_, Feb 01 2021
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