

A261831


a(2*n1) = 2*n1; 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
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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*n2; if n == 2(mod 6), then a(n) = 2*n; if n == 4(mod 6), then a(n) = n2.
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*n1) = 2*n1, for n>=1.
a(6*k) = 12*k2, 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, 2n2, 2, 2n, 4, n2]]; Array[a, 81] (* JeanFrançois 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



