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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. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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).

LINKS

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

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.

EXAMPLE

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

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 *)

CROSSREFS

Cf. A109732, A261690, A261728.

Sequence in context: A167837 A222228 A263449 * A260458 A275958 A275957

Adjacent sequences:  A261828 A261829 A261830 * A261832 A261833 A261834

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Sep 02 2015

STATUS

approved

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Last modified December 4 13:00 EST 2021. Contains 349526 sequences. (Running on oeis4.)