

A232895


Sequence (or tree) S of all positive integers in the order generated by these rules: 1 and 2 are in S; if x is in S then x + 2 and 2*x are in S, where duplicates are deleted as they occur.


2



1, 2, 3, 4, 5, 6, 8, 7, 10, 12, 16, 9, 14, 20, 24, 18, 32, 11, 28, 22, 40, 26, 48, 36, 34, 64, 13, 30, 56, 44, 42, 80, 52, 50, 96, 38, 72, 68, 66, 128, 15, 60, 58, 112, 46, 88, 84, 82, 160, 54, 104, 100, 98, 192, 76, 74, 144, 70, 136, 132, 130, 256, 17, 62
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Let S be the sequence (or tree) of numbers generated by these rules: 1 and 2 are in S; if x is in S then x + 2 and 2*x are in S, where duplicates are deleted as they occur. Every positive integer occurs exactly once in S, so that S is a permutation of the natural numbers. Deleting duplicates as they occur, the generations of S are given by g(1) = (1,2), g(2) = (3,4), g(3) = (5,6,8), g(4) = (7,10,12,16), ... Concatenating gives 1,2,3,4,5,6,8,... Conjecture: the position of the nth odd positive integer in S is the linearly recurrent sequence given by A232896(n) for n>=1.


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..1000


EXAMPLE

To generate S, start with g(1) = (1,2). Then 1 begets 3 and 2, but 2 is deleted as a duplicate, and 2 begets 4 and 4, of which the second 4 is deleted; thus g(2) = (3,4).


MATHEMATICA

x = {1, 2}; dx = 0; Do[x = DeleteDuplicates[Flatten[AppendTo[x, Transpose[{# + 2, 2*#}] &[Drop[x, Length[x]  dx]]]]]; dx = Length[x]  dx, {31}]; x (* A232895 *)
t = Flatten[Position[Denominator[x/2], 2]] (* A232896 conjectured *)
(* Peter J. C. Moses, Dec 02 2013 *)


CROSSREFS

Cf. A232559, A232896.
Sequence in context: A125624 A262388 A297440 * A274607 A262374 A299442
Adjacent sequences: A232892 A232893 A232894 * A232896 A232897 A232898


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Dec 02 2013


STATUS

approved



