Caribb. J. Math. Comput. Sci.
13, 2006, 1-11
A Solution to the 3x + 1 Problem
Charles Cadogan
Abstract
This paper presents a solution to the 3x + 1 problem. The method employs the convergence of the trajectories of the odd positive integers by exploiting the role of the positive integers of the form 1 + 4n, where n
is a non-negative integer. Equivalence relations also contribute to the analysis.
AMS 2000 Classification: 11A25, 11Y55, 11Y70
Key words: trajectory, coalesce, equivalence relation.
1. Introduction
The references at [5,6,7] present an excellent bibliography of the literature resulting from research on this
problem. No formal proof of the conjecture associated with the problem has been previously achieved. This
paper presents a formal proof of the conjecture.
The following basic definitions, descriptions and terminology will be used. Our discussion is confined to
non-negative integers.
Let N= {1,2,3,....} denote the set of positive integers and let E and O be, respectively, the subsets of
even integers and of odd integers in N. Let N0 = N ∪ {0} = {0, 1, 2, ....}.
For any x ∈ N, let thešfunction f : N → N be given by,
1 + 3x, if x ∈ O,
f (x) =
x
if x ∈ E.
2,
(1.1)
The 3x + 1 conjecture asserts that in the sequence of iterates x, f (x), f 2 (x), f 3 (x),... for any x ∈ N,
∃ k ∈ N0 , such that f k (x) = 1.
By convention f 0 (x) = x, ∀ x ∈ N, and the least value of k satisfying f k (x) = 1 is what is being sought.
In [4] an alternative formulation of the conjecture was provided by means of the function h : O → O
given by:
h(x) = 21+3x
(1.2)
m(x) , x ∈ O,
where 2m(x) is the maximum power of 2 dividing 1 + 3x. The conjecture then reduces to showing that for
each x ∈ O there is an integer k ∈ N0 such that hk (x) = 1.
Definition 1.1.
The trajectory, under f , of x ∈ N is the sequence Lf (x) : x, f (x), f 2 (x), f 3 (x), . . ..
Definition 1.2.
The trajectory, under h, of x ∈ O is the sequence Lh (x) : x, h(x), h2 (x), h3 (x), . . ..
1
Charles Cadogan
The trajectory Lh (x) of any x ∈ O is obtained from Lf (x) by selecting the odd integers in Lf (x).
Example 1.1. Lf (9) : 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 and Lh (9) : 9, 7, 11, 17, 13, 5, 1.
Definition 1.3.
Two trajectories are said to coalesce if they have a common element.
Example 1.2. Lf (15) : 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, so that Lf (15) and Lf (9)
coalesce at 40. Also, Lh (15) : 15, 23, 35, 53, 5, 1, so that Lh (15) and Lh (9) coalesce at 5.
Definition 1.4. Let ∼ be the relation defined on N as follows: for x, y ∈ N, x ∼ y iff the trajectories Lf (x)
and Lf (y) coalesce.
Henceforth, we shall write x ∼ y to mean that the trajectories of x, y ∈ N coalesce.
Example 1.3. From Example 1.2, 15 ∼ 9.
Remark 1: If Lf (x) and Lf (y) coalesce, then so do Lh (x) and Lh (y) for x, y ∈ O, and x = y ⇒ x ∼ y.
Further, if Lf (x) and Lf (y) coalesce at u ∈ N, then they do not separate again and, thereafter, the common
trajectory of Lf (x) and Lf (y) may be represented by Lf (u).
From [7], we adopt the notion of a predecessor in a trajectory defined as follows.
Definition 1.5. Let u, v ∈ Lf (x), x ∈ N. Then u is said to be a predecessor of v iff ∃ r, s ∈ N0 such that
0 ≤ r < s and u = f r (x), v = f s (x).
Example 1.4. In Example 1.2, in Lf (15), 23 is a predecessor of 53; so also is 15.
Lemma 1.1. The relation ∼ is an equivalence relation on N and partitions N into ∼-classes.
Proof: Let x, y, z ∈ N.
(a) x ∼ x for all x ∈ N since Lf (x) coalesces with itself at each u ∈ Lf (x), hence ∼ is reflexive on N.
(b) x ∼ y ⇒ Lf (x) and Lf (y) coalesce at some u ∈ N ⇒ Lf (y) and Lf (x) coalesce at u ⇒ y ∼ x ⇒∼ is
symmetric on N.
(c) Let x ∼ y so that Lf (x) and Lf (y) coalesce at u, and let y ∼ z so that Lf (y) and Lf (z) coalesce at v.
Now, u, v ∈ Lf (y) so we have to consider the following three cases, viz, (i) u is a predecessor of v, (ii) u = v,
(iii) v is a predecessor of u. In case (i), Lf (u) represents the common trajectory of Lf (x) and Lf (y), and
since Lf (y), Lf (z) coalesce at v it follows that Lf (u) and Lf (z) coalesce at v so that all three trajectories
Lf (x), Lf (y) and Lf (z) coalesce at v. By similar arguments, we obtain that in case (ii), Lf (x), Lf (y) and
Lf (z) coalesce at u = v, and in case (iii), Lf (x), Lf (y) and Lf (z) coalesce at u, that is, in each case x ∼ y ∼ z.
Thus, x ∼ y and y ∼ z ⇒ x ∼ y ∼ z ⇒ x ∼ z and ∼ is transitive on N.
By (a),(b) and (c), ∼ is an equivalence relation on N and thus partitions N into ∼ - classes.
Let Φ(x) denote the ∼-class generated by x ∈ N. Our aim is to prove that Φ(x) = Φ(1) for all x ∈ N.
2
A Solution to the 3X+1 Problem
2. Previous Results
In [1,2,3] several results were established using the function f . Any e ∈ E is of the form e = 2i q, where
i ∈ N and q ∈ O, so that f i (e) = q, that is, e ∼ q. As a consequence, we concentrate our attention on O.
We start with the partition of O as given in [1].
The non-empty subsets of O given by:
Ri = {x ∈ O : x ≡ 2i − 1 (mod 2i+1 )}, i ∈ N,
form a partition of O.
Proof: Any x ∈ O is of the form x = 2i q − 1 for some i ∈ N, q ∈ O. Also, q ∈ O ⇒ q = 2j + 1 and
j ∈ N0 , so that,
x = 2i+1 j + 2i − 1 ≡ 2i − 1 (mod 2i+1 ),
Lemma 2.1.
i ∈ N, j ∈ N0 , that is, x ∈ Ri for some i ∈ N, so that Ri 6= ∅, ∀ i ∈ N. Thus,
O = ∪i Ri , i ∈ N.
Now, suppose that x ∈ Ri ∩ Rk , i 6= k, with i, k ∈ N. Without loss of generality, we may assume that
i > k. Then, x ∈ Ri , so that x = 2i qi − 1, qi ∈ O, and x ∈ Rk , so that x = 2k qk − 1, qk ∈ O. Thus,
x = 2i qi − 1 = 2k qk − 1 ⇒ 2i−k qi = qk ,
6 k, and the sets Ri , i ∈ N,
which is a contradiction to qk ∈ O since i − k > 0. Hence, 6 ∃ x ∈ Ri ∩ Rk , i =
partition O.
The partition of O given in Lemma 2.1 can be arranged in rows Ri , i ∈ N, and columns Cj , j ∈ N0 as
in Table 1 below which provides the start of a doubly infinite array T with the integer x ∈ O represented in
the form
and with each ti,j
ti,j = 2i+1 j + 2i − 1, i ≥ 1, j ≥ 0,
in row i and column j of T. In particular, when j = 0, we get ti,0 = 2i − 1.
(2.1)
Table 1
R1
R2
R3
R4
R5
R6
R7
·
C0
1
3
7
15
31
63
127
·
C1
5
11
23
47
95
191
383
·
C2
9
19
39
79
159
319
639
·
C3
13
27
55
111
223
447
895
·
C4
17
35
71
143
287
575
1151
·
C5
21
43
87
175
351
703
1407
·
C6
25
51
103
207
415
831
1663
·
C7
29
59
119
239
479
959
1919
·
C8
33
67
135
271
543
1087
2175
·
·
·
·
·
·
·
·
·
·
The binary expansion of x ∈ O may be used to identify uniquely the i and the j associated with the
representation of x by ti,j in (2.1).
3
Charles Cadogan
Example 2.1.
For x = 479 , we have,
479 = 1 + 2 × 239 = 1 + 2(1 + 2 × 119) = 1 + 2(1 + 2(1 + 2 × 59))
= 1 + 2(1 + 2(1 + 2(1 + 2 × 29))) = 1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(2 × 7)))))
= 26 7 + 25 − 1 = 1110111112 = t5,7 .
The 5 ones to the right of the zero give the row number and the binary number 111, or decimal 7, to the
left of the zero gives the column number.
The following result is now immediate from (2.1).
Lemma 2.2. (i) In each row i, ti,j+1 = ti,j + 2i+1 .
(ii) In each column j, ti+1,j = 2ti,j + 1.
Lemma 2.3. For any i ≥ 2, j ≥ 0,
(i) f 2 (ti,j ) = ti−1,3j+1 .
(ii) f 2(i−1) (ti,j ) = t1,ni−1,j , where,
i−1
Pi−2
ni−1,j = 3i−1 j + r=0 3r = 3 (2j+1)−1
.
2
Proof: (i) f (ti,j ) = 1 + 3 · 2i+1 j + 3 · 2i − 3 = 3 · 2i+1 j + 3 · 2i − 2,
⇒ f 2 (ti,j ) = 3 · 2i j + 3 · 2i−1 − 1 = 2i (3j + 1) + 2i−1 − 1 = ti−1,3j+1 .
(ii)
follows by iteration on (i) above.
Theorem 2.4. For any i ≥ 1, j ≥ 0, f 2(i−1) (ti,j ) = t1,ni−1,j .
For i = 1, ni−1,j = n0,j = j, so that,
f 2(i−1) (ti,j ) = f 0 (ti,j ) = t1,n0,j = t1,j .
The result now follows from Lemma 2.3.
Proof:
The typical integer t1,ni−1,j ∈ O in Theorem 2.4 is of the form t1,ni−1,j = 22 ni−1,j + 2 − 1 = 1 + 4ni−1,j
from (2.1) above; this means that the trajectory Lf (ti,j ) of each ti,j , i ≥ 1, j ≥ 0, contains elements of the
form t1,ni−1,j , so that each such trajectory Lf (ti,j ) contains some element in the set R1 in Lemma 2.1 (see
Table 1), a result previously derived in [1,2,3].
In order to determine the complete trajectory Lf (ti,j ) it is necessary to be able to identify the trajectories
of each t1,n , n ∈ N0 . Our approach is to show that for each t1,n , n ∈ N0 , ∃ t1,n1 , n1 ∈ N0 , such that n1 ≤ n
(equality when n = 0) and t1,n ∼ t1,n1 . Such a result would establish that for each ti,j , i ≥ 1, j ≥ 0, ti,j ∼ t1,n
for some n ∈ N0 and would produce a chain t1,n ∼ t1,n1 ∼ t1,n2 ∼ · · · ∼ t1,0 = 1 of elements in R1 in the
array T , with n > n1 > n2 > · · · ≥ 0, where n, n1 , n2 , · · · are elements of N0 ; this development in turn
would determine a crucial aspect of an efficient algorithm to prove that for each q ∈ O, q ∼ 1. Since for
each e ∈ E, e ∼ q for some q ∈ O as seen earlier, we would conclude that y ∼ 1, ∀ y ∈ N, and the 3x + 1
conjecture would be established. To this end we pursue the trajectory Lf (ti,j ) for arbitrary i, j through to
t1,n , and beyond, by means of a series of preliminary results for t1,n , n ∈ N0 .
4
A Solution to the 3X+1 Problem
Lemma 2.5.
Let n ∈ N. Then t1,n = 1 + 4n, and,
(i) f 3 (t1,n ) = 1 + 3n, for all n ∈ N,
(ii) f 3 (t1,n ) = f (n), for all n ∈ O.
Proof: (i) For all n ∈ N, t1,n = 1 + 4n, so
f (t1,n ) = 1 + 3(1 + 4n) = 4(1 + 3n),
⇒ f 3 (t1,n ) = 1 + 3n.
(ii) When n ∈ O, 1 + 3n = f (n), so
f 3 (t1,n ) = f (n), for all n ∈ O.
We note that for n = 0, 1 + 4n = 1 + 3n = 1.
In terms of the equivalence relation ∼ defined on N in Definition 1.4, the results in Lemma 2.5 can be
stated as follows:
Corollary 2.6. For all n ∈ N, t1,n = 1 + 4n ∼ 1 + 3n, and,
Φ(1 + 4n) = Φ(1 + 3n).
For all n ∈ O, t1,n = 1 + 4n ∼ 1 + 3n ∼ n, and,
Φ(1 + 4n) = Φ(1 + 3n) = Φ(n).
We now combine Theorem 2.4 and Lemma 2.5 to obtain,
ti,j ∼ t1,ni−1,j = 1 + 4ni−1,j ∼ f 2i+1 (ti,j ) = 1 + 3ni−1,j ,
with,
i−1
, i ≥ 1, j ≥ 0.
ni−1,j = 3 (2j+1)−1
2
(2.2)
(2.3)
In place of (2.3), we use the typical pair, namely,
n0,j = j, ni,j =
3i (2j+1)−1
,
2
i ≥ 1, j ≥ 0.
(2.4)
The values of the ni,j ∈ N corresponding to the ti,j in array T of Table 1 are displayed in Table 2 below,
providing a second doubly infinite array T 0 of elements of N.
Table 2
R1
R2
R3
R4
R5
R6
R7
·
C0
1
4
13
40
121
364
1093
·
C1
4
13
40
121
364
1093
3280
·
C2
7
22
67
202
607
1822
5467
·
C3
10
31
94
283
850
2551
7654
·
C4
13
40
121
364
1093
3280
9841
·
C5
16
49
148
445
1336
4009
12028
·
C6
19
58
175
526
1579
4738
14215
·
C7
22
67
202
607
1822
5467
16402
·
C8
25
76
229
688
2065
6196
18589
·
·
·
·
·
·
·
·
·
·
The entries ni,j in Table 2 provide a means of identifying each x ∈ O in Table 1 in the trajectory Lf (ti,j )
as far as t1,ni−1,j .
5
Charles Cadogan
Example 2.2. In Table 2, n5,0 = n4,1 = n3,4 = 121. In Table 1, t5,0 = 31, t4,1 = 47 and t3,4 = 71, so
that the first three odd integers in the trajectory Lf (31) are 31, 47 and 71. The next odd integer in Lf (31)
is t2,13 = 107 for which n2,13 = 121; indeed, n1,40 = 121 also, and this value in Table 2 corresponds to
t1,40 = 161 in Table 1.
The pattern of correspondence demonstrated in Example 2.2 may be characterised as follows:
Remark 2: The Ri , i ∈ N, in Table 1 partition O and for any arbitrary ni,j in Table 2, i ≥ 1, j ≥ 0, each
odd integer ti−k,nk,j , 0 ≤ k < i − 1, in Table 1 is a predecessor of t1,ni−1,j in Lf (ti,j ).
Corresponding to Lemma 2.2 (ii) for the array T is the following result for the array T 0 .
Lemma 2.7. In each column j ≥ 0, ni,j = 1 + 3ni−1,j , i ≥ 1.
Proof: From (2.4), ni,j =
=
=
3i (2j+1)−1
,
2
3i−1 (2j+1)−1
),
1 + 3(
2
1 + 3ni−1,j .
Corollary 2.8. For all i ≥ 1 and fixed j ≥ 0,
(i) ni−1,j and ni,j have opposite parity,
(ii) ti,j ∼ ni,j .
Proof: (i) follows from the fact that ni,j = 1 + 3ni−1,j .
(ii) ti,j ∼ 1 + 3ni−1,j by (2.2), hence, ti,j ∼ ni,j .
The parity of ni,j in (2.4) above now assumes enormous significance.
We focus on ti,j and commence consideration of parity with the case ni,j ∈ O.
Lemma 2.9. If ni,j ∈ O, then i + j ∈ O.
Proof: If ni,j ∈ O, then ni,j ≡ 1 (mod 2).
i
Now, ni,j ≡ 1 (mod 2)⇒ 3 (2j+1)−1
≡ 1 (mod 2)⇒ 3i−1 (2j + 1) ≡ 1 (mod 4). There are two cases.
2
Case (i): i − 1 ∈ O ⇒ 3(2j + 1) ≡ 1 (mod 4)⇒ j ≡ 1 (mod 2) ⇒ j ∈ O.
Case (ii): i − 1 ∈ E ⇒ 2j + 1 ≡ 1 (mod 4)⇒ j ≡ 0 (mod 2) ⇒ j ∈ E.
Hence, i + j ∈ O in each case.
Lemma 2.10. If ni,j ∈ E, then i + j ∈ E.
Proof: If ni,j ∈ E, then ni,j ≡ 0 (mod 2).
Now, ni,j ≡ 0 (mod 2) ⇒
3i (2j+1)−1
2
≡ 0 (mod 2) ⇒ 3i (2j + 1) ≡ 1 (mod 4). There are two cases.
Case (i): i ∈ O ⇒ 3(2j + 1) ≡ 1 (mod 4) ⇒ j ≡ 1 (mod 2) ⇒ j ∈ O.
Case (ii): i ∈ E ⇒ 2j + 1 ≡ 1 (mod 4) ⇒ j ≡ 0 (mod 2) ⇒ j ∈ E.
Hence, i + j ∈ E in each case.
Lemma 2.9 and Lemma 2.10 combine to produce the following result.
6
A Solution to the 3X+1 Problem
Theorem 2.11. (i) ni,j ∈ O iff i + j ∈ O.
(ii) ni,j ∈ E iff i + j ∈ E.
Proof: (i) By Lemma 2.9, ni,j ∈ O ⇒ i + j ∈ O. By Lemma 2.10, ni,j ∈ E ⇒ i + j ∈ E, so that, by
the contrapositive, i + j ∈ O ⇒ ni,j ∈ O, hence ni,j ∈ O iff i + j ∈ O.
(ii) By Lemma 2.10, ni,j ∈ E ⇒ i + j ∈ E. By Lemma 2.9, ni,j ∈ O ⇒ i + j ∈ O, so that, by
the contrapositive, i + j ∈ E ⇒ ni,j ∈ E, hence ni,j ∈ E iff i + j ∈ E.
We can now proceed to derive the following important result.
Theorem 2.12. If ni,j ∈ O, then ti+1,j ∼ ti,j .
Proof: ti+1,j ∼ ni+1,j ,
= 1 + 3ni,j ,
∼ ni,j ,
∼ ti,j ,
by Corollary 2.8(ii),
by Lemma 2.7,
by Corollary 2.6,
by Corollary 2.8(ii).
Corollary 2.13. If ni,j ∈ E, then ti,j ∼ ti−1,j for i > 1. When i = 1, ti,j ∼ j and j ∈ O.
Proof: For i > 1, we replace i with i − 1 in Theorem 2.12 and obtain ti,j ∼ ti−1,j . When i = 1, ti,j = t1,j ∼
1 + 3j = n1,j ∈ E, so that the column number j ∈ O.
In terms of the parity of ni,j , the results of Theorem 2.12 and Corollary 2.13 may be summarised as
follows.
Corollary 2.14 (i) If ni,j ∈ O, then ti+1,j ∼ ti,j .
(ii) If ni,j ∈ E, then ti,j ∼ ti−1,j , i > 1 and t1,j ∼ j.
The result of Corollary 2.8(ii) is a consequence of the equation f 2i+1 (ti,j ) = ni,j , i ≥ 1, i ≥ 0. Iteration
on i gives f 2i+3 (ti+1,j ) = ni+1,j , which is equivalent to f 2i+3 (1 + 2ti,j ) = 1 + 3ni,j that is 1 + 2ti,j ∼ 1 + 3ni,j ,
thus, we have,
2i+1
ti,j ∼ ni,j ⇒ 1 + 2ti,j ∼ 1 + 3ni,j .
(2.5)
2i+2
Now, f
(ti,j ) = ni,j with ni,j ∈ O as in Theorem 2.12 implies that f
(ti,j ) = f (ni,j ) = 1 + 3ni,j =
ni+1,j = f 2i+3 (ti+1,j ), so that f 2i+2 (ti,j ) = ni+1,j and, by (2.5) above, we obtain, ti,j ∼ ni+1,j ⇒ ti+1,j =
1 + 2ti,j ∼ 1 + 3ni+1,j = ni+2,j ∼ ti+2,j , that is, for ni,j ∈ O,
(2.6)
ti,j ∼ ti+1,j ∼ ti+2,j .
We now establish the analogue of Theorem 2.12 for ni,j ∈ E.
Theorem 2.15.
If ni,j ∈ E, then ti+1,j ∼ ti,j .
Proof: If ni,j ∈ E, then 1 + 3ni−1,j ∈ E, so that, ni−1,j ∈ O with ni−1,j = j when i = 1, (Corollary 2.13),
thus, ti,j ∼ ni,j ∼ ni−1,j , and by the transitivity of ∼ on O, ti,j ∼ ni−1,j so that, by replacing i with i − 1
in condition (2.6) we obtain that ti+1,j ∼ ti,j .
7
Charles Cadogan
Corollary 2.16.
ti+1,j ∼ ti,j and ni+1,j ∼ ni,j are independent of the parity of ni,j .
Proof: The result follows from Theorems 2.12 and 2.15.
3. Main Results
Our immediate aim is to show that for fixed j ≥ 0, ti,j ∼ t1,j , ∀ i ≥ 1. As in (2.4), ni,j =
i ≥ 1, n0,j = j.
3i (2j+1)−1
,
2
Some important preliminary results follow.
Lemma 3.1. For all x ∈ O, 1 + 2x ∼ x.
Proof: If x ∈ O, then x = ti,j in T for some i ≥ 1, j ≥ 0, so that 1 + 2x = ti+1,j . By Theorems 2.12 and
Theorem 2.15, ti+1,j ∼ ti,j , hence, 1 + 2x ∼ x, for all x ∈ O.
Example 3.1. The Appendix contains the results for 1 ≤ x ≤ 103, x ∈ O, and shows the point A of
coalescence of Lf (x) and Lf (1 + 2x) for each such x.
The following Corollary is now immediate:
Corollary 3.2. If x, y ∈ O such that x ∼ y, then x ∼ y ∼ 1 + 2x ∼ 1 + 2y ∼ 1 + 3x ∼ 1 + 3y.
4
Example 3.2. From Tables 1 and 2, 31 ∼ 47 ∼ 71 ∼ 107 ∼ 161 with 31 = t5,0 and n5,0 = 3 2−1 = 40. For
any pair x, y of integers in the set {31, 47, 71, 107, 161}, x ∼ y and 1 + 2x ∼ 121 = 1 + 3 × 40. We note that
40 is the column number of the integer 161. In fact, 161 = t1,40 .
We now have,
Theorem 3.3. For fixed j ≥ 0, ti,j ∼ t1,j , ∀ i ≥ 1.
Proof: By Lemma 3.1, x ∼ 1 + 2x ∀ x ∈ O. But, for fixed j ≥ 0 and ∀ i ≥ 1, ti,j ∈ O, hence,
ti,j ∼ 1 + 2ti,j = ti+1,j , ∀ i ≥ 1, for fixed j ≥ 0,
thus, inductively, by the transitive property of the equivalence relation ∼,
ti,j ∼ t1,j , ∀ i ≥ 1, and for fixed j ≥ 0.
Now, for any fixed j ≥ 0, t1,j = 1 + 4j ∼ 1 + 3j, and by Lemma 2.2 we obtain that t1,j+1 = t1,j + 4 which
implies that t1,j+r = t1,j +4r > t1,j for all r ∈ N while ti,j = 2ti−1,j +1 implies that ti,j = 2i−1 t1,j +2i−1 −1 >
t1,j , i ≥ 2, so that, we have the following result.
Lemma 3.4. For any arbitrary j ≥ 1 there is an l ≥ 0 such that l < j and t1,l ∼ t1,j .
Proof: For any j ≥ 1, 1 + 3j < 1 + 4j = t1,j , and 1 + 3j ∼ t1,j by Corollary 2.6. But 1 + 3j < t1,j ⇒
1 + 3j < t1,j+r and 1 + 3j < t1+r,j for r ≥ 1. If j ∈ E then 1 + 3j ∈ O, and if j ∈ O then 1 + 3j ∼ j
by Corollary 2.6 and j < 1 + 3j, so that j is in the array T with 1 + 3j ∼ j. Hence, independently of the
8
A Solution to the 3X+1 Problem
parity of j, 1 + 3j ∼ tk,l for some k ≥ 1 and l < j. Also, by Theorem 3.3, tk,l ∼ t1,l , hence 1 + 3j ∼ t1,j and
1 + 3j ∼ tk,l ∼ t1,l , that is, t1,j ∼ t1,l with 0 ≤ l < j.
By repeated application of the result in Lemma 3.4 we obtain:
Lemma 3.5. For each j ≥ 0, t1,j ∼ t1,0 = 1.
By combining the results in Theorem 3.3 through to Lemma 3.5, we obtain that for i ≥ 1, j ≥ 0, the
trajectory Lf (ti,j ) contains ti,j , t1,j and t1,0 = 1, hence, we have the following important result.
Theorem 3.6. For any i ≥ 1, j ≥ 0, ti,j ∼ t1,0 = 1.
We observed earlier in Section 2 that any e ∈ E can be expressed in the form e = 2i q with i ∈ N and
q ∈ O so that e ∼ q with q = ti,j for some i ≥ 1, j ≥ 0. Thus, we conclude with the following Major
Theorem.
Theorem 3.7. For any x ∈ N, ∃ k ∈ N0 such that,
f k (x) = 1, that is, Φ(x) = Φ(1) for all x ∈ N.
Remark 3: Theorem 3.7 establishes the validity of the 3x + 1 Conjecture.
Alternative methods of proof are already being investigated, based on the results in Lemma 3.1 and
Theorem 3.3.
4. Conclusion
The method of identifying integers with coalescing trajectories provided an effective strategy to tackle
the 3x + 1 problem since two trajectories, having coalesced, can never separate again; in each such case
the smaller integer is used to represent the common trajectory from the point of coalescence onwards. This
approach enhanced the possibility of reaching the integer 1 in the shortest number of steps and was used
repeatedly in proving this monumental result. Several techniques were combined in this work to achieve the
goal of proving the 3x + 1 Conjecture; among these are basic results on partitions of positive odd integers
and equivalence relations.
Acknowledgements
The author wishes to thank all those who contributed to the success of this research, in particular, Dr.
Tane Ray of the Department of Computer Science, Mathematics and Physics of the University of the West
Indies in Barbados, and Professor Larry Cummings of the Department of Pure Mathematics at the University of Waterloo in Canada.
9
Charles Cadogan
REFERENCES
[1 ] Cadogan, C.C., A Note on the 3x + 1 Problem, Caribb. J. Math. 3,(2)(1984)67-72.
[2 ] Cadogan, C.C., Trajectories in the 3x + 1 Problem, Jour.Comb.Math. Comb.Comp. 44, Feb.(2003)
177-187.
[3 ] Cadogan, C.C., The 3x + 1 Problem: Towards a Solution, Caribb. J. Math. Comput. Sci. 10, (1 &
2) (2000) (Electronically Available).
[4 ] Crandall, R.E., On the “3x + 1” Problem, Math. Comput., Vol. 32, Oct. (1978), 1281-1292, MR 58
# 494.
[5 ] Lagarias, J.C., The 3x + 1 Problem and Its Generalisations, Amer. Math. Monthly, Vol 92, No. 1
(1985), 3-23.
[6 ] Lagarias, J.C., 3x + 1 Problem Annotated Bibliography, September 22, 1997.
[7 ] Wirsching, G.J., The Dynamical System Generated by the 3n + 1 Function, Springer, Lecture Notes
in Mathematics 1681, (1998).
Department of Computer Science, Mathematics & Physics
University of the West Indies
Cave Hill Campus
Barbados, West Indies
e-mail:
[email protected]
10
A Solution to the 3X+1 Problem
APPENDIX
A = point at which Lf (x) and Lf (1 + 2x) coalesce.
N (x) = number of iterations for x ∈ O to reach A.
x
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
39
41
43
45
47
49
51
1 + 2x
3
7
11
15
19
23
27
31
35
39
43
47
51
55
59
63
67
71
75
79
83
87
91
95
99
103
N (x)
1
1
1
8
4
6
1
1
4
1
3
1
4
6
1
12
4
1
17
8
4
6
8
10
4
16
N (1 + 2x)
5
10
10
9
5
7
103
90
5
15
25
90
5
7
15
13
5
90
10
9
5
7
84
11
5
79
A
4
10
16
40
22
40
40
46
40
58
16
70
58
94
88
364
76
106
16
202
94
148
40
364
112
40
x
53
55
57
59
61
63
65
67
69
71
73
75
77
79
81
83
85
87
89
91
93
95
97
99
101
103
11
1 + 2x
107
111
115
119
123
127
131
135
139
143
147
151
155
159
163
167
171
175
179
183
187
191
195
199
203
207
N (x)
1
44
4
6
11
99
4
12
3
8
4
6
14
27
4
43
5
22
4
6
9
97
4
17
10
8
N (1 + 2x)
90
1
5
7
38
38
5
26
30
9
5
7
77
46
5
0
120
72
5
7
36
36
5
111
24
9
A
160
334
130
202
40
40
148
22
52
364
166
256
40
40
184
167
16
40
202
310
40
40
220
40
22
526