A062052 Numbers with exactly 2 odd integers in their Collatz (or 3x+1) trajectory.

5, 10, 20, 21, 40, 42, 80, 84, 85, 160, 168, 170, 320, 336, 340, 341, 640, 672, 680, 682, 1280, 1344, 1360, 1364, 1365, 2560, 2688, 2720, 2728, 2730, 5120, 5376, 5440, 5456, 5460, 5461, 10240, 10752, 10880, 10912, 10920, 10922, 20480, 21504, 21760, 21824

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

The Collatz (or 3x+1) function is f(x) = x/2 if x is even, 3x+1 if x is odd.

The Collatz trajectory of n is obtained by applying f repeatedly to n until 1 is reached.

The sequence consists of terms of A002450 and their 2^k multiples. The first odd integer in the trajectory is one of the terms of A002450 and the second odd one is the terminal 1. - Antti Karttunen, Feb 21 2006

This sequence looks to appear first in the literature on page 1285 in R. E. Crandall.

Examples

			The Collatz trajectory of 5 is (5,16,8,4,2,1), which contains 2 odd integers.
		

Links

• Reinhard Zumkeller and T. D. Noe, Table of n, a(n) for n = 1..1000 (first 100 terms from Reinhard Zumkeller)
• R. E. Crandall, On the 3x+1 problem, Math. Comp., 32 (1978) 1281-1292.
• J. Shallit and D. Wilson, The "3x+1" Problem and Finite Automata, Bulletin of the EATCS #46 (1992) pp. 182-185.
• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem
• Index entries for 2-automatic sequences.

Crossrefs

Cf. A062053, A062054, A062055, A062056, A062057, A062058, A062059, A062060, A122196, A122197.

Is this a subset of A115774?

Column k=2 of A354236.

Programs

Haskell

import Data.List (elemIndices)
a062052 n = a062052_list !! (n-1)
a062052_list = map (+ 1) $ elemIndices 2 a078719_list
-- Reinhard Zumkeller, Oct 08 2011

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; countOdd[lst_] := Length[Select[lst, OddQ]]; Select[Range[22000], countOdd[Collatz[#]] == 2 &] (* T. D. Noe, Dec 03 2012 *)

PARI

for(n=2,100000,s=n; t=0; while(s!=1,if(s%2==0,s=s/2,s=3*s+1; t++); if(s*t==1,print1(n,","); ); ))

Python

def a(n):
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        if n not in l:
            l.append(n)
            if n<2: break
        else: break
    return len([i for i in l if i % 2])
print([n for n in range(1, 22001) if a(n)==2]) # Indranil Ghosh, Apr 14 2017

Formula

A078719(a(n)) = 2; A006667(a(n)) = 1.

a(n) = 2^x * (4^y - 1)/3 where x = A122196(n) - 1 and y = A122197(n) + 1. - Alan Michael Gómez Calderón, Jan 16 2025 after Antti Karttunen