%I A061641
%S A061641 0,1,3,6,7,9,12,15,18,19,21,24,25,27,30,33,36,37,39,42,43,45,48,51,54,
%T A061641 55,57,60,63,66,69,72,73,75,78,79,81,84,87,90,93,96,97,99,102,105,108,
%U A061641 109,111,114,115,117,120,123,126,127,129,132,133,135,138,141,144,145
%N A061641 Pure numbers in the Collatz (3x+1) iteration. Also called pure hailstone
numbers.
%C A061641 Let {f(k,N), k=0,1,2,...} denote the (3x+1)-sequence with starting value
N; a(n) denotes the smallest positive integer which is not contained
in the union of f(k,0),...,f(k,a(n-1)).
%C A061641 In other words a(n) is the starting value of the next '3x+1'-sequences
in the sense that a(n) is not a value in any sequence f(k,N) with
N < a(n).
%C A061641 f(0,N)=N, f(k+1,N)=f(k,N)/2 if f(k,N) is even and f(k+1,N)=3*f(k,N)+1
if f(k,N) is odd.
%C A061641 For all n, a(n) mod 6 is 0, 1 or 3. I conjecture that a(n)/n -> C=constant
for n->oo, where C=2.311...
%C A061641 The Collatz conjecture says that for all positive n, there exists k such
that C_k(n) = 1. Shaw states [p. 195] that "A positive integer n
is pure if its entire tree of preimages under the Collatz function
C are greater than or equal to it; otherwise n is impure. Equivalently,
a positive integer n is impure if there exists r<n such that C_k(r)
= n for some k." Theorem 2.1: If n == 0 (mod 3), then n is pure.
If n == 2 (mod 3), then n is impure, reducing the field to 1 (mod
3). (mod 18), the following congruences are pure: n == 0 (mod 18);
also 3, 6, 9, 12, 15; while n == 7 (mod 18) may be pure or impure.
n == [2, 4, 5, 8, 10, 11, 13, 14, 16, or 17] (mod 18) are all impure.
The asymptotic density d of the set of impure numbers I is such that
d <= 2/3. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 28 2007
%C A061641 Pure numbers remaining after deleting the impure numbers in the hailstone
(Collatz) problem; where the operation C(n) = {3n+1, n odd; n/2,
n even}. Add the 0 mod 3 terms in order, among the terms of A127633,
since all 0 mod 3 numbers are pure. - Gary W. Adamson (qntmpkt(AT)yahoo.com),
Jan 28 2007
%C A061641 After computing all a(n) < 10^9, the ratio a(n)/n appears to be converging
to 2.31303... Hence it appears that the numbers in this sequence
have a density of about 1/3 (due to all multiples of 3) + 99/1000.
- T. D. Noe, Oct 12 2007
%C A061641 A016945 is a subsequence. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com),
Apr 17 2008
%D A061641 Douglas J. Shaw, "The Pure Numbers Generated by the Collatz Sequence",
Fibonacci Quarterly, Vol. 44, Number 3, August 2006, p. 194.
%H A061641 T. D. Noe, <a href="b061641.txt">Table of n, a(n) for n=1..10000</a>
%H A061641 <a href="Sindx_3.html#3x1">Index entries for sequences related to 3x+1
(or Collatz) problem</a>
%e A061641 Consider n=3: C(n), C_2(n), C_3(n)...; the iterates are 10, 5, 16, 8,
4, 2, 1, 4, 2, 1; where 4, 5, 8, 10 and 16 have appeared in the orbit
of 3 and are thus impure.
%e A061641 a(1)=1 since Im(f(k,0))={0} for all k and so 1 is not a value of f(k,
0). a(2)=3 since Im(f(k,0)) union Im(f(k,1))={0,1,2,4} and 3 is the
smallest positive integer not contained in this set.
%Y A061641 Cf. A070165 (Collatz trajectories), A127633.
%Y A061641 Sequence in context: A120684 A026227 A026232 this_sequence A085359 A087916
A023982
%Y A061641 Adjacent sequences: A061638 A061639 A061640 this_sequence A061642 A061643
A061644
%K A061641 nice,nonn
%O A061641 1,3
%A A061641 Frederick Magata (fmagata(AT)mi.uni-koeln.de), Jun 14 2001
%E A061641 Edited by T. D. Noe and N. J. A. Sloane (njas(AT)research.att.com), Oct
16 2007
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