A061641 Pure numbers in the Collatz (3x+1) iteration. Also called pure hailstone numbers.

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, 55, 57, 60, 63, 66, 69, 72, 73, 75, 78, 79, 81, 84, 87, 90, 93, 96, 97, 99, 102, 105, 108, 109, 111, 114, 115, 117, 120, 123, 126, 127, 129, 132, 133, 135, 138, 141, 144, 145

Offset: 1

Author

Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 14 2001

Keywords

• nice
• nonn

Comments

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)).

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).

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.

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...

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 rGary W. Adamson, Jan 28 2007

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, Jan 28 2007

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

A016945 is a subsequence. - Reinhard Zumkeller, Apr 17 2008

Examples

			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.
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.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Douglas J. Shaw, The Pure Numbers Generated by the Collatz Sequence, The Fibonacci Quarterly, Vol. 44, Number 3, August 2006, pp. 194-201.
• B. Snapp and M. Tracy, The Collatz Problem and analogues, JIS 11 (2008) 08.4.7.
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A070165 (Collatz trajectories), A127633, A336938, A336938. See A177729 for a variant.

Programs

Mathematica

DoCollatz[n_] := Module[{m = n}, While[m > nn || ! reached[[m]], If[m <= nn, reached[[m]] = True]; If[EvenQ[m], m = m/2, m = 3 m + 1]]]; nn = 200; reached = Table[False, {nn}]; t = {0, 1}; While[DoCollatz[t[[-1]]]; pos = Position[reached, False, 1, 1]; pos != {}, AppendTo[t, pos[[1, 1]]]]; t (* T. D. Noe, Jan 22 2013 *)

PARI

firstMiss(A) = { my(i); if(#A == 0 || A[1] > 0, return(0)); for(i = 1, A[#A] + 1, if(!setsearch(A,i), return(i))); };
iter(A) = { my(a = firstMiss(A)); while(!setsearch(A,a), A = setunion(A, Set([a])); a = if(a % 2, 3*a+1, a/2)); A; };
makeVec(m) = { my(v = [], A = Set([]), i); for(i = 1, m, v = concat(v, firstMiss(A)); if (i < m, A = iter(A))); v; };
makeVec(64) \\ Markus Sigg, Aug 08 2020

Extensions

Edited by T. D. Noe and N. J. A. Sloane, Oct 16 2007