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A261931 a(0)=1, a(1)=2, for n>=1, a(2*n) = 6*n - 2; a(2*n+1) = min((a(2*n)-1)/3)^, (2*a(2*n))^, (2*a(2*n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3*k nor of the form 6*m-2. 1

%I

%S 1,2,4,8,10,20,16,5,22,7,28,14,34,11,40,13,46,26,52,17,58,19,64,38,70,

%T 23,76,25,82,50,88,29,94,31,100,62,106,35,112,37,118,74,124,41,130,43,

%U 136,86,142,47,148,49,154,98,160,53,166,55,172,110,178,59,184

%N a(0)=1, a(1)=2, for n>=1, a(2*n) = 6*n - 2; a(2*n+1) = min((a(2*n)-1)/3)^, (2*a(2*n))^, (2*a(2*n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3*k nor of the form 6*m-2.

%C Let n>=7. If n == 1 or 3 (mod 6), then a(n) = n-2; if n == 5 (mod 6), then a(n) = 2*(n-4). This could be proved by induction similar to the theorem in A261728.

%C The sequence is a permutation of the positive integers not divisible by 3 which are not of the form 12*s+8, s>=2.

%C This sequence is connected with Collatz's (3*n+1)-conjecture. For example, if n=29, then, by the formulas, 29 = a(31) => 88 = a(30) => 11 = a(13) => 34 = a(12) => 17 = a(19) => 52 = a(18) => 13 = a(15) => 40 = a(14) => 5 = a(7) => 16 = a(6) => 1 = a(0).

%H Peter J. C. Moses, <a href="/A261931/b261931.txt">Table of n, a(n) for n = 0..9999</a>

%F For n>=1, a(2*n) = 6*n - 2.

%F For t>=1, a(6*t+1) = 6*t - 1; a(6*t+3) = 6*t+1; a(6*t+5) = 12*t + 2.

%F And from the name, for n>=3, a(2*n+1) = min((a(2*n)-1)/3)^, (2*a(2*n-1))^).

%e At n=3, since a(1)=2 and a(2)=6*1-2=4, a(3) should be either (4-1)/3=1 or 2*a(2)=8 or 2*a(1)=4; 1 and 4 have already appeared, so a(3)=8.

%e At n=5, since a(3)=8 and a(4)=6*2-2=10, a(5) should be either (10-1)/3=3 or 2*a(4)=20 or 2*a(3)=16; 3 is divisible by 3, and 16 is of the form 6*t-2, so a(5)=20.

%Y Cf. A261690, A261728, A261831.

%K nonn

%O 0,2

%A _Vladimir Shevelev_, Sep 06 2015

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Last modified December 8 00:04 EST 2021. Contains 349590 sequences. (Running on oeis4.)