

A109732


a(1) = 1; for n>1, a(n) is the smallest number not already present which is entailed by the rules (i) k present => 2k+1 present; (ii) 3k present => k present.


11



1, 3, 7, 15, 5, 11, 23, 31, 47, 63, 21, 43, 87, 29, 59, 95, 119, 127, 175, 191, 239, 255, 85, 171, 57, 19, 39, 13, 27, 9, 55, 79, 111, 37, 75, 25, 51, 17, 35, 71, 103, 115, 143, 151, 159, 53, 107, 207, 69, 139, 215, 223, 231, 77, 155, 279, 93, 187, 287, 303, 101, 203
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OFFSET

1,2


COMMENTS

Van der Poorten asks if every odd number is in the sequence. This seems very likely.
Comment from Max Alekseyev, Aug 28 2015 (Start):
The question of whether every odd number is present in this sequence can be reformulated as follows. Can every odd number m be transformed into 1 using the maps: m > (m1)/2 (only if the result is integer) and m > 3m, applied in some order? It is clear that even numbers cannot appear in such a transformation, since they would remain even and thus not reach 1.
Replacing m by n = (m+1)/2, we get an equivalent question: Can any number n be transformed into 1 using the maps: n > n/2 (only if n is even) and n > 3n1 applied in some order?
An affirmative answer to this question would follow from the 3x1 variation of Collatz conjecture. This states that the maps x > x/2 (for even x) and x > 3x1 (for odd x) eventually reach one of the three cycles: (1,2), (5, ...) of length 5  see A003079  or (17, ,..) of length 17  see A003124.
However, in our problem, we have the freedom of choosing either of the two maps at each stage (the only restriction being that n > n/2 can be used only if n is even). With this freedom, we can transform 5 and 17 from the nontrivial cycles of the 3x1 problem to 1: (5, 14, 7, 20, 10, 29, 86, 43, 128, 64, 32, 16, 8, 4, 2, 1) or (17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, 136, 68, 203, 608, 304, 152, 76, 38, 19, 56, 28, 14, ... as before).
That is, under the 3x1 variation of Collatz conjecture, we can transform any number either to 1, 5, or 17, and in the latter two cases we can proceed further as explained above and still reach 1. (End)
In short, the question of showing that every odd number occurs is likely to be very difficult.  N. J. A. Sloane, Aug 29 2015
Odd numbers of the form 2^k+1 take a long time to appear; e.g., 2^12+1 appears at a(64607).  T. D. Noe, Aug 10 2005. A109734, A261412, A261413, A261414 are related to this question.  N. J. A. Sloane, Aug 27 2015


LINKS

T. D. Noe and N. J. A. Sloane, Table of n, a(n) for n = 1..20000 (First 1000 terms from T. D. Noe)
T. D. Noe, Graph of first 1000 terms


MAPLE

with(LinearAlgebra);
hit:=Array(1..200000); a:=[1, 3, 7]; hit[1]:=1; hit[3]:=1; hit[7]:=1; S:={15}; L:=7;
for n from 4 to 20000 do
if (L mod 3 = 0) and hit[L/3]=0 then
L:=L/3; a:=[op(a), L]; hit[L]:=1; S:= S minus {L};
if hit[2*L+1]=0 then S:=S union {2*L+1}; fi;
else L:=min(S); a:=[op(a), L]; hit[L]:=1; S:=S minus {L};
if hit[2*L+1]=0 then S:=S union {2*L+1}; fi;
fi; od: a; # From N. J. A. Sloane, Aug 25 2015


MATHEMATICA

maxVal=1000; f[n_]:=Module[{lst={}, x=n}, While[x=2x+1; x<maxVal, AppendTo[lst, x]]; lst]; M={1}; pending=f[1]; While[Length[pending]>0, next=First[pending]; pending=Rest[pending]; If[ !MemberQ[M, next], AppendTo[M, next]; While[Mod[next, 3]==0 && !MemberQ[M, next/3], next=next/3; AppendTo[M, next]; pending=Union[pending, f[next]]]]]; M (Noe)


CROSSREFS

Cf. A109734 (inverse), A261412 and A261413 (records), A261414 (where 2^k+1 appears),
A261690 (an analog connected with (3n+1)problem).
See also A003124, A003079.
Sequence in context: A234042 A001203 A154883 * A114396 A102032 A086517
Adjacent sequences: A109729 A109730 A109731 * A109733 A109734 A109735


KEYWORD

nonn,easy,look


AUTHOR

N. J. A. Sloane, prompted by a posting by Alf van der Poorten to the Number Theory List, Aug 10 2005


EXTENSIONS

More terms from T. D. Noe, Aug 10 2005


STATUS

approved



