A261831 a(2*n-1) = 2*n-1; otherwise a(n) is the smallest even number not already present which is obtained from the existing terms by the rules of (3*n+1)-problem.

1, 4, 3, 2, 5, 10, 7, 16, 9, 8, 11, 22, 13, 28, 15, 14, 17, 34, 19, 40, 21, 20, 23, 46, 25, 52, 27, 26, 29, 58, 31, 64, 33, 32, 35, 70, 37, 76, 39, 38, 41, 82, 43, 88, 45, 44, 47, 94, 49, 100, 51, 50, 53, 106, 55, 112, 57, 56, 59, 118, 61, 124, 63, 62, 65, 130, 67, 136, 69, 68, 71, 142, 73, 148, 75, 74, 77, 154, 79, 160, 81, 80

Offset: 1

Author

Vladimir Shevelev, Sep 02 2015

Keywords

• nonn

Comments

By the rules of the (3*n+1)-problem, an even number can appear either by the operation 3*x+1 only when x is an odd number or by the division of a number of the form 4*k by 2.

Using induction as in the proof of the Theorem in A261728, one can prove that if n == 0(mod 6), then a(n) = 2*n-2; if n == 2(mod 6), then a(n) = 2*n; if n == 4(mod 6), then a(n) = n-2.

The sequence is a permutation of the positive integers not divisible by 6 (A047253).

Examples

			Let n=28. Since 28 is of the form 6*k+4 with k=4, then a(28) = 6*4+2 = 26.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Crossrefs

Cf. A109732, A261690, A261728.

Programs

Mathematica

a[n_] := If[OddQ[n], n, Switch[Mod[n, 6], 0, 2n-2, 2, 2n, 4, n-2]]; Array[a, 81] (* Jean-François Alcover, Sep 02 2015, from given formula *)
LinearRecurrence[{0,0,0,0,0,2,0,0,0,0,0,-1},{1,4,3,2,5,10,7,16,9,8,11,22},90] (* Harvey P. Dale, Feb 24 2025 *)

Formula

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

a(6*k) = 12*k-2, a(6*k+2) = 12*k+4 and a(6*k+4) = 6*k+2, for k>=0.

O.g.f.:(1+x (4+x (3+x (2+x (5+x (10+x (5+x (8+x (3+x (4+x (1+2 x)))))))))))/(-1+x^6)^2.