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, 2, 4, 8, 10, 20, 16, 5, 22, 7, 28, 14, 34, 11, 40, 13, 46, 26, 52, 17, 58, 19, 64, 38, 70, 23, 76, 25, 82, 50, 88, 29, 94, 31, 100, 62, 106, 35, 112, 37, 118, 74, 124, 41, 130, 43, 136, 86, 142, 47, 148, 49, 154, 98, 160, 53, 166, 55, 172, 110, 178, 59, 184

Offset: 0

Author

Vladimir Shevelev, Sep 06 2015

Keywords

• nonn

Comments

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.

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

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).

Examples

			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.
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.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 0..9999

Crossrefs

Cf. A261690, A261728, A261831.

Formula

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

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

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