A219696 Numbers k such that the trajectory of 3k + 1 under the '3x + 1' map reaches k.

1, 2, 4, 8, 10, 14, 16, 20, 22, 26, 40, 44, 52, 106, 184, 206, 244, 274, 322, 526, 650, 668, 790, 866, 976, 1154, 1300, 1438, 1732, 1780, 1822, 2308, 2734, 3238, 7288

Offset: 1

Author

Robert C. Lyons, Nov 25 2012

Keywords

• nonn
• nice
• more

Comments

This sequence seems complete; there are no other terms <= 10^9. - T. D. Noe, Dec 03 2012

If the 3x+1 step is replaced with (3x+1)/2, the sequence becomes {1, 2, 4, 8, 10, 14, 20, 22, 26, 40, 44, 206, 244, 650, 668, 866, 1154, 1822, 2308, ...}. - Robert G. Wilson v, Jan 13 2015

From Andrew Slattery, Aug 03 2023: (Start)

For most terms k, the trajectory of 3k + 1 reaches 310 or the trajectory of 310 reaches k.

For the rest of the terms k, the trajectory of 3k + 1 reaches 22 or the trajectory of 22 reaches k.

With the exception of k = 1, k is reached after S steps,

where S = c*8 + d*13 + e*44 + f*75, with c, d, e and f in {0, 1, 2}; in particular, S is in {8, 13, 8+8, 8+13, 13+13, 44, 75, 44+44, 75+13+13, 75+44, 75+75}. (End)

Examples

			For k = 4, the Collatz trajectory of 3k + 1 is (13, 40, 20, 10, 5, 16, 8, 4, 2, 1), which includes 4; thus, 4 is in the sequence.
For k = 5, the Collatz trajectory of 3k + 1 is (16, 8, 4, 2, 1), which does not include 5; thus, 5 is not in the sequence.
		

Links

• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A014682, A070991, A006370, A070165.

Programs

Haskell

a219696 n = a219696_list !! (n-1)
a219696_list = filter (\x -> collatz'' x == x) [1..] where
   collatz'' x = until (`elem` [1, x]) a006370 (3 * x + 1)
-- Reinhard Zumkeller, Aug 11 2014

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Select[Range[10000], MemberQ[Collatz[3 # + 1], #] &] (* T. D. Noe, Dec 03 2012 *)

PARI

a006370(n) = if(n%2==0, n/2, 3*n+1)
is(n) = my(x=3*n+1); while(1, x=a006370(x); if(x==n, return(1), if(x==1, return(0)))) \\ Felix Fröhlich, Jun 10 2021

Python

def ok(n):
    if n==1: return [1]
    N=3*n + 1
    l=[N, ]
    while True:
        if N%2==1: N = 3*N + 1
        else: N/=2
        l+=[N, ]
        if N<2: break
    if n in l: return 1
    return 0 # Indranil Ghosh, Apr 22 2017

Extensions

Initial 1 from Clark R. Lyons, Dec 02 2012