Two different scenarios when the Collatz Conjecture fails

Abstract

In this article, we give two different proofs of why the Collatz Conjecture is false.

Full text

Two different scenarios when the Collatz
Conjecture fails
Author: Maya Mohsin Ahmed
Email: maya.ahmed@gmail.com
Abstract
In this article, we give two different proofs of why the Collatz Conjecture is false.
1 Introduction.
Given a positive integer A, construct the sequence cias follows:
ci=Aif i= 0;
= 3ci−1+ 1 if ci−1is odd;
=ci−1/2 if ci−1is even.
The sequence ciis called a Collatz sequence with starting number A. The Collatz
Conjecture says that this sequence will eventually reach the number 1, regardless of which
positive integer is chosen initially. The sequence gets in to an infinite cycle of 4, 2, 1 after
reaching 1.
Example 1.1. The Collatz sequence of 911 is:
911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,
1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,
70,35,106,53,160,80,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1, . . .
For the rest of this article, we will ignore the infinite cycle of 4,2,1, and say that a
Collatz sequence converges to 1, if it reaches 1. A comprehensive study of the Collatz
Conjecture can be found in [4], [5], and [6].
In this article, as we did before in [1], [2], and [3], we focus on the subsequence of odd
numbers of a Collatz sequence. This is because every even number in a Collatz sequence
has to reach an odd number after a finite number of steps. Observe that the Collatz
Conjecture implies that the subsequence of odd numbers of a Collatz sequence converges
to 1.
In Section 2, we use networks of Collatz sequences [2] to prove that the Collatz Conjec-
ture fails. In Section 3, we use the notion of Reverse Collatz sequences [1] to give another
proof of the collapse of the Collatz Conjecture.
1

2 Networking to prove that the Collatz Conjecture
is false.
In this section, we use an array of Collatz sequences to demonstrate how the Collatz
Conjecture fails. In [2], we proved the following theorem which showed that the Collatz
sequence of an odd number Amerges either with the Collatz sequence of (A−1)/2 or
2A+ 1.
Theorem 2.1. (Theorem 2.1, [2]) Let Nbe an odd number. Let n0=N,m0= 2n0+1 =
2N+ 1, and l0= 2m0+ 1 = 4N+3 . Let ni, mi, and lidenote the subsequence of odd
numbers in the Collatz sequence of n0, m0,and l0, respectively. Then, for some integer
r,nr+1 = (3nr+ 1)/2ksuch that k > 1. Let rbe the smallest such integer. Then,
mr+2 = (3mr+1 + 1)/2jfor some j > 1, and
mi= 2ni+ 1,for i≤r,
mr+1 = 2knr+1 + 1
li= 2mi+ 1 for i≤r+ 1,
lr+2 = 2jmr+2 + 1
If k= 2, then mi=nifor i>r+ 1. Otherwise, if k > 2then lr+2 = 4mr+2 + 1 and
li=mifor i > r + 2.
For some integer u0, consider the sequence ui= 2ui−1+1, for i≥1. Let vj,i denote the
sequence of odd integers in the Collatz sequence of ui. Theorem 2.1 tells us that for every
i > 0, there is some rsuch that for j≤r,vj,i = 2vj,i−1+ 1 and vr+1,i = 2kvr,i−1+ 1 where
k > 1. This fact motivates us to construct the array of Collatz sequences in Theorem 2.3.
Let Abe an odd number. Write Ain its binary form,
A= 2i1+ 2i2+· · · + 2im+ 2n+ 2n−1+ 2n−2+· · · + 22+ 2 + 1,
such that i1> i2>· · · > im> n + 1.
The tail of Ais defined as 2n+ 2n−1+ 2n−2+· · · + 22+ 2 + 1. We call nthe length of the
tail of A(See [1]).
Example 2.1. The tail of 27 = 24+ 23+ 2 + 1 is 2 + 1 and hence has length 1, the tail
of 161 = 27+ 25+ 1 is 1 and hence is of length 0, and the tail of 31 = 24+ 23+ 22+ 2 + 1
is the entire number 24+ 23+ 22+ 2 + 1 and therefore has length 4.
In [1] we proved the following theorem about the odd numbers in a Collatz sequence.
Theorem 2.2. (Theorem 2 [1]) Let Abe an odd number and let ndenote the length of
the tail of A. Let aidenote the sequence of odd numbers in the Collatz sequence of Awith
a0=A.
1. If n≥1, then for some i1> i2>· · · > im> n + 1,
ai=3ai−1+ 1
2=3i
2i(2i1+ 2i2+···+ 2im+ 2n+1 )−1,for i= 1, . . . n.
2

The length of the tail of aiis n−i. Hence the length of the tail of the n-th odd
number after Ais 0.
2. If n= 0, then
a1=3A+ 1
2k, k ≥2.
Corollary 2.1. Let Abe an odd number. If A̸≡ 1mod 4, then the next odd term in the
Collatz sequence of Ais (3A+ 1)/2.
Proof. Since Ais odd and A̸≡ 1 mod 4, A≡3 mod 4. This implies that the tail of A
has length greater than zero. Hence the proof follows from Part 1 of Theorem 2.2.
Theorem 2.3. (Theorem 5.1, [2]) For n̸≡ 1mod 3, define a diagonal array as follows.
Let u0= 4n+ 1 and ui= 2ui−1+ 1, for i≥1. For j≥0, let v0,j =uj, and for k≥1, let
vk,k = 3vk−1,k−1+ 2. Finally, for j > i, let vi,j = 2vi,j −1+ 1. We get an array
u0u1u2u3u4u5u6u7. . .
v1,1v1,2v1,3v1,4v1,5v1,6v1,7. . .
v2,2v2,3v2,4v2,5v2,6v2,7. . .
v3,3v3,4v3,5v3,6v3,7. . .
. . . . . .
with the following properties:
1. uk̸≡ 2mod 3for all k≥0, whereas, vi,j ≡2mod 3, if i > 0and j > 0.
2. u0≡1mod 4and vi,i ≡1mod 4for all i.ui̸≡ 1mod 4for i > 0and vi,j ̸≡ 1mod
4if i̸=j.
3. For j≥1, the j-th column is the first few odd numbers in the Collatz sequence of
uj.
4. For j≥i > 0,vi,j = 3vi−1,j−1+ 2.
Proof. 1. Since n̸≡ 1 mod 3, u0= 4n+ 1 ̸≡ 2 mod 3. If ui̸≡ 2 mod 3, then
ui+1 = 2ui+ 1 ̸≡ 2 mod 3. Consequently, uk̸≡ 2 mod 3 for all k≥0.
For i > 0, vi,i ≡2 mod 3, by definition. If vi,j ≡2 mod 3, then vi,j+1 = 2vi,j + 1 ≡2
mod 3. Thus, it follows that vi,j ≡2 mod 3, if i > 0 and j > 0.
2. u0≡1 mod 4 by definition. vi,i = 3vi−1,i−1+ 2 ≡1 mod 4, if vi−1,i−1≡1 mod 4.
Since v0,0=u0, it follows that vi,i ≡1 mod 4 for i≥0. For i > 0, ui= 2ui−1+1 ̸≡ 1
mod 4 since ui−1is an odd number. Similarly, vi,j = 2vi,j−1+ 1 ̸≡ 1 mod 4, when
i̸=j.
3

3. When j≥1, v0,j =uj. We know from Part 2 that vi,j ≡3 mod 4 if i̸=j. Therefore,
by Corollary 2.1, the odd number that comes after vi,j in the Collatz sequence of vi,j
is (3vi,j + 1)/2. We do not compute the odd numbers that come after vi,i . Hence,
assume that j > i.
For k≥1, let zkbe defined as follows:
zk= 2k+ 2k−1+ 2k−2+· · · + 2 + 1 =
k

i=0
2i=2k+1 −1
2−1= 2k+1 −1.
Then, by definition, for j > i,vi,j = 2j−ivii +zj−i−1, and
vi+1,j = 2j−i−1vi+1,i+1 +zj−i−2= 2j−i−1(3vi,i + 2) + zj−i−2.
3vi,j + 1
2=3×2j−ivii + 3 ×zj−i−1+ 1
2= 3 ×2j−i−1vii +3zj−i−1+ 1
2.
3zj−i−1+ 1
2=3(2j−i−1) + 1
2= 3×2j−i−1−1 = 2×2j−i−1+(2j−i−1−1) = 2j−i+zj−i−2.
Consequently, 3vi,j + 1
2=vi+1,j .
This implies vi+1,j is the odd number that comes after vi,j in the Collatz sequence
of vi,j . Hence, for j≥1, the j-th column is the first few odd numbers in the Collatz
sequence of uj.
4. For j≥i > 0, when i=j, we have that vi,j = 3vi−1,j−1+ 2, by definition. Let i̸=j,
then vi,j = 2vi,j−1+ 1. But vi−1,j −1̸≡ 1 mod 4, when i̸=j. Hence, by Corollary
2.1, vi,j−1= (3vi−1,j−1+ 1)/2. Consequently, vi,j = 3vi−1,j−1+ 2.
Theorem 2.3 can be used to construct divergent Collatz sequences as shown in the
example below. Observe that the Collatz sequence of uiis strictly increasing till vi,i.
Our choice of u0and Theorem 2.1 makes sure this is the case. Check that vi+1,i will be
smaller than vi,i. We see that as iincreases, a subsequence of the Collatz sequence of ui
is increasing indefinitely. Thus creating divergent Collatz sequences. This implies that
the Collatz Conjecture is false. Different values of nprovide different arrays that lead to
different divergent Collatz sequences.
Example 2.2. Let uiand vi,j be defined as in Theorem 2.3.
n= 0 :
4

ui: 1 3 7 15 31 63 127 255 511 1023 2047 4095 8191
v1,i : 5 11 23 47 95 191 383 767 1535 3071 6143 12287
v2,i : 17 35 71 143 287 575 1151 2303 4607 9215 18431
v3,i : 53 107 215 431 863 1727 3455 6911 13823 27647
v4,i : 161 323 647 1295 2591 5183 10367 20735 41471
v5,i 485 971 1943 3887 7775 15551 31103 62207
v6,i : 1457 2915 5831 11663 23327 46655 93311
v7,i : 4373 8747 17495 34991 69983 139967
n= 3 :
ui: 13 27 55 111 223 447 895 1791 3583 7167 14335 28671
v1,i : 41 83 167 335 671 1343 2687 5375 10751 21503 43007
v2,i : 125 251 503 1007 2015 4031 8063 16127 32255 64511
v3,i : 377 755 1511 3023 6047 12095 24191 48383 96767
v4,i : 1133 2267 4535 9071 18143 36287 72575 145151
v5,i : 3401 6803 13607 27215 54431 108863 217727
v6,i : 10205 20411 40823 81647 163295 326591
3 Reversing to prove that the Collatz Conjecture is
false
In this section, we provide a different proof of how the Collatz Conjecture fails.
Let Abe an odd integer. We say Ais a jump, if A= 4n+ 1 from some odd number
n. If A= 4i×P+ 4i−1+ 4i−2+· · · + 4 + 1, such that i≥1 and Pis an odd number,
then we say Ais a jump from Pof height i.
Example 3.1. 13 = 4 ×3+1 is a jump from 3 of height 1. 53 = 4 ×13 + 1 = 42×3+4+1
is a jump from 13 of height 1 and a jump from 3 of height 2.
Jumps are studied in great detail in [1] and [2]. We say two Collatz sequences are
equivalent if the second odd number occurring in the sequences are same.
Example 3.2. The Collatz sequence of 3 is
3,10,5,16,8,4,2,1,1, . . .
The Collatz sequence of 13 is
13,40,20,10,5,16,8,4,2,1,1, . . .
Observe that the two sequences merge at the odd number 5. Hence the Collatz sequences
of 3 and 13 are equivalent.
Lemma 3.1 (Corollary 2, Section 2, [1]).Let Abe an odd number and let c0=Aand
ci= 4ci−1+ 1, that is, ciare jumps from A. Then, for any i, the Collatz sequence of A
and ciare equivalent.
5

The Reverse Collatz sequence, ri, of a positive integer Awas defined in [1] as follows.
ri=






Aif i= 0;
ri−1−1
3if ri−1≡1 mod 3 and ri−1is even;
2ri−1if ri−1̸≡ 1 mod 3 and ri−1is even;
2ri−1if ri−1is odd.
We say that a Reverse Collatz sequence converges if the subsequence of odd numbers
of the sequence converges to a multiple of 3.
Example 3.3.
The Reverse Collatz sequence with starting number 121 is :
121,242,484,161,322,107,214,71,142,47,94,31,62,124,41,82,27,54,108,216, . . .
The Reverse Collatz sequence of 121 converges because its subsequence of odd numbers
121,161,107,71,47,31,41,27
converges to 27.
Let pidenote the subsequence of odd numbers in the Reverse Collatz sequence of A.
Then, in [1], it was proved that, if pi≡0 mod 3, then pi+1 do not exist. Otherwise, pi+1
is the smallest odd number before piin any Collatz sequence and
pi+1 =


2pi−1
3if pi≡2 mod 3
4pi−1
3if pi≡1 mod 3
(1)
It was conjectured in [1], that, the Reverse Collatz sequence converges to a multiple
of 3 for every number greater than one. See [1] and [2] for more details about Reverse
Collatz sequences.
Lemma 3.2. Let Abe an odd number such that A≡2 mod 3. Let u= (A−2)/3. If
ridenotes subsequence of odd numbers in the Reverse Collatz sequence of Awith r0=A,
then r1= 2u+ 1. Consequently,
u≡


0 mod 3,=⇒r1≡1 mod 3
1 mod 3,=⇒r1≡0 mod 3
2 mod 3,=⇒r1≡2 mod 3
Observe that r1< r0. Moreover, if u≡2 mod 3, let tirepresent the subsequence of odd
numbers in the Reverse Collatz sequence of u, then t1= (r1−2)/3.
Proof. Since A≡2 mod 3, by definition of Reverse Collatz sequence, r1= (2A−1)/3.
Now
2u+ 1 = 2 A−2
3+ 1 = 2A−1
3.
6

Therefore, r1= 2u+ 1. Consequently,
r1−2
3=(2u+ 1) −2
3=2u−1
3.
If u≡2 mod 3, then by definition of Reverse Collatz sequence, t1= (2u−1)/3. Thus,
t1= (r1−2)/3.
Theorem 3.1. Let Abe an odd number and let A≡2 mod 3. Define a sequence of odd
numbers, v0,j, such that, v0,0=A, and for j > 0,v0,j = (v0,j −1−2)/3. Then, for some
integer n≥0,v0,j ≡2 mod 3 for j < n, and v0,n ̸≡ 2 mod 3. Moreover, there are at
least n+ 1 terms in the subsequence of odd numbers in the Reverse Collatz sequence of
v0,0(by Part 1). Let vi,0,i= 0, . . . , n, denote the first n+ 1 terms of the subsequence of
odd numbers in the Reverse Collatz sequence of v0,0. Then, for each i= 1, . . . , n, we can
form an array vi,j = (vi,j−1−2)/3, where j= 1, . . . , n −i,
v0,0v0,1v0,2v0,3. . . v0,n−2v0,n−1v0,n
v1,0v1,1v1,2v1,3. . . v1,n−2v1,n−1
v2,0v2,1v2,2v2,3. . . v2,n−2
.
.
..
.
..
.
..
.
.
vn−3,0vn−3,1vn−3,2vn−3,3
vn−2,0vn−2,1vn−2,2
vn−1,0vn−1,1
vn,0
with the following properties.
1. For each j= 0, . . . n,vi,j,i= 0,1, . . . , n −j, are the first n+ 1 −jterms of the
subsequence of odd numbers in the Reverse Collatz sequence of v0,j . For i= 0, . . . , n,
vi,j ≡2 mod 3 whenever j̸=n−i, and vi,n−i̸≡ 2 mod 3.
2. For i > 0,vi,j = 2 ∗vi−1,j+1 + 1. Moreover, if v0,n ≡1 mod 3 then,
vi,n−i≡0 mod 3 if iis odd;
1 mod 3 if iis even.
On the other hand, if v0,n ≡0 mod 3 then,
vi,n−i≡1 mod 3 if iis odd;
0 mod 3 if iis even.
3. v0,0= 3n(v0,n + 1) −1and vn,0= 2n(v0,n + 1) −1. Moreover, for i= 1, . . . , n,
vi,0= 3n−i×2i×(v0,n + 1) −1.
Proof.
7

1. By Lemma 3.2, since v0,1≡2 mod 3, v1,0≡2 mod 3, and v1,1is the next odd term
in the subsequence of odd numbers in the Reverse Collatz sequence of v0,1. Now,
because v1,0≡2 mod 3, v2,0is well defined as the next term in the Reverse Collatz
sequence of v1,0, by Equation 1. Since, v0,2≡2 mod 3, we apply Lemma 3.2, again,
to conclude that v1,1≡2 mod 3, and v1,2is the next odd term in the subsequence
of odd numbers in the Reverse Collatz sequence of v0,2.
Applying this argument, repeatedly, we derive that for each j= 1, . . . , n −1, v1,j is
the first odd term that comes after v0,j in the Reverse Collatz sequence of v0,j . We
also get that v1,j ≡2 mod 3, for j= 1, . . . , n −2. Since v0,n ̸≡ 2 mod 3, v1,n−1̸≡ 2
mod 3, by Lemma 3.2.
Now, since we established that v1,1≡2 mod 3, we will repeat the above argument
to derive that for each j= 1, . . . , n −2, v2,j is the first odd term that comes after
v1,j in the Reverse Collatz sequence of v1,j. We also get that v2,j ≡2 mod 3, for
j= 1, . . . , n −3, and v2,n−2̸≡ 2 mod 3.
Continuing thus, we get for each j= 0,...n,vi,j ,i= 0,1, . . . , n −j, are the first
n+ 1 −jterms of the subsequence of odd numbers in the Reverse Collatz sequence
of v0,j . For i= 0, . . . , n,vi,j ≡2 mod 3 whenever j̸=n−i, and vi,n−i̸≡ 2 mod 3.
2. This result follows from Lemma 3.2.
3. Rewriting vi,j = (vi,j−1−2)/3, we get vi,j−1= 3vi,j + 2. Thus,
vi,0= 3vi,1+ 2 = 3(3(vi,2+ 2) + 2 = 32vi,2+ 3 ×2 + 2.
Continuing thus we get vi,0= 3n−i×vi,n−i+ 2 ×n−i−1
s=0 3s. Since
2×
n−i−1

s=0
3s= 3n−i−1,
we get
vi,0= 3n−i(vi,n−i+ 1) −1.
By Part 2, vi,j = 2 ∗vi−1,j+1 + 1, for i > 0. Therefore,
vi,n−i= 2 ∗vi−1,n−i+1 + 1 = 2 ∗(2 ∗vi−2,n−i+2 +1)+1
Continuing thus we get
vi,n−i= 2i×v0,n +
i−1

s=0
2s.
Substituting i−1
s=0 2s= 2i−1, we get vi,n−i= 2i(v0,n + 1) −1. In particular,
v0,0= 3n(v0,n + 1) −1 and vn,0= 2n(v0,n + 1) −1.
Since v0,0= 3n(v0,n + 1) −1 and v0,0≡2 mod 3, v1,0= 3n−1×2×(v0,n + 1) −1, by
Equation 1. By Part 1, v1,0≡2 mod 3. Therefore, again, by Equation 1, we derive
v2,0= 3n−2×22×(v0,n + 1) −1. Repeating this argument, we get, for i= 1, . . . , n,
vi,0= 3n−i×2i×(v0,n + 1) −1.
8

Example 3.4. In this example, we apply Theorem 3.1 to A= 2429 ≡2 mod 3. Here,
n= 5, v0,5= 9, The columns are the first odd terms in the Reverse Collatz sequence of
v0,j . Observe that v0,5= 9, v2,3= 39, v4,1= 159 are ≡0 mod 3 and v1,4= 19, v3,2= 79,
and v5,0= 319 are ≡1 mod 3. v0,0= 2429 = 35×10 −1, and v5,0= 25×10 −1 = 319.
v0,j : 2429 809 269 89 29 9
v1,j : 1619 539 179 59 19
v2,j : 1079 359 119 39
v3,j : 719 239 79
v4,j : 479 159
v5,j : 319
Lemma 3.3. Let A≡1 mod 3 be an odd number. Then we can write A= 3nB+ 1 such
that Bis not divisible by 3. If ridenotes the subsequence of odd numbers in the Reverse
Collatz sequence of A, then, for i= 0, . . . n,ri= 4i×3n−i×B+ 1. Thus, rn= 4n×B+ 1.
For i= 0, . . . n −1,ri≡1 mod 3, and
rn≡2 mod 3,if B≡1 mod 3,
0 mod 3,if B≡2 mod 3.
Observe that ri> ri−1for i= 1, . . . , n.
Proof. Since, A≡1 mod 3, by Equation 1, we get r1= (4A−1)/3 = 4 ×3n−1B+ 1.
If n > 1 then r1≡1 mod 3. Again, by Equation 1, r2= 42×3n−2B+ 1. If n > 2 then
r2≡1 mod 3. Continuing this argument, we get, for i= 0, . . . n−1, ri= 4i×3n−i×B+1,
such that, ri≡1 mod 3. Now since rn−1≡1 mod 3, we get rn= 4n×B+ 1. Since
B̸≡ 0 mod 3, rn̸≡ 1 mod 3. In fact, rn≡2 mod 3, if B≡1 mod 3, and rn≡0
mod 3, if B≡2 mod 3.
Example 3.5. We apply Lemma 3.3 to A= 91. We can write A= 32×10 + 1. Let ri
denote the subsequence of odd numbers in the Reverse Collatz sequence of A. Then,
r0= 91 = 32×10 + 1
r1= 121 = 4 ×3×10 + 1
r2= 161 = 42×10 + 1 ≡2 mod 3.
Since 10 ≡1 mod 3, r2≡2 mod 3.
Example 3.6. In this example, we demonstrate the convergence of the Reverse Collatz
sequence of 2429. By Part 3 of Theorem 3.1, v0,0= 3n(v0,n + 1) −1. Rewriting, we get
(v0,0+ 1)/3n=v0,n + 1. Since (2429 + 1)/35= 10, and 10 ̸≡ 0 mod 3, we get v0,5= 9. By
Part 2 of Theorem 3.1, Since 9 ≡0 mod 3, v5,0≡1 mod 3. Now v5,0= 319 = 3×106+1,
therefore. v6,0= 4 ×106 + 1 = 425, by Lemma 3.3. Also, since, 106 ≡1 mod 3, v6,0≡2
mod 3. Since 426/3 = 142, we get v6,0= 3 ×142 −1, by Part 3 of Theorem 3.1. Since
142 ≡1 mod 3, v7,0≡2 mod 3, by Lemma 3.3. Continuing this argument, we see that
subsequence of odd integers of the Reverse Collatz sequence of 2429 fluctuates between
numbers that are ≡2 mod 3 and 1 mod 3, till it reaches 111 ≡0 mod 3.
9

v0,0= 2429 = 35×10 −1
v1,0= 1619 = 34×2×10 −1
v2,0= 1079 = 33×22×10 −1
v3,0= 719 = 32×23×10 −1
v4,0= 479 = 3 ×24×10 −1
v5,0= 319 = 25×10 −1 = 3 ×106 + 1
v6,0= 425 = 4 ×106 + 1 = 3 ×142 −1
v7,0= 283 = 4 ×142 −1 = 3 ×94 −1
v8,0= 377 = 4 ×94 −1 = 33×14 −1
v9,0= 251 = 32×2×14 −1
v10,0= 167 = 3 ×22×14 −1
v11,0= 111 = 23×14 −1
Thus, we see that the odd numbers of a Reverse Collatz sequence, keep alternating
between numbers that are congruent to 1 mod 3 and 2 mod 3, till it reaches a number
that is divisible by 3. Which also means the sequence increases and decreases at regular
intervals. Does this sequence converge? Or does it alternate forever? We cannot answer
this question yet.
A Reverse Collatz sequence will continue till it reaches a number Athat is a multiple
of 3. Now if Ais a multiple of 3, then 4A+ 1 ≡1 mod 3. So the Reverse Collatz sequence
of 4A+ 1 is non trivial. Moreover, the Collatz sequences of Aand 4A+ 1 are equivalent.
Thus, a Collatz sequence can be extended backwards forever using jumps as in Example
3.7. A sequence containing infinite terms is divergent.
10

Example 3.7. A Collatz sequence can be extended backwards forever using jumps!
204729
153547
230321
4×43185 + 1 = 172741
⇑
43185
8097 ⇒4×8097 + 1 = 32389
6073
4555
6833
1281 ⇒4×1281 + 1 = 5125
961
721
4×135 + 1 = 541
⇑
135
203
305
4×57 + 1 = 229
⇑
57
43
65
49
9⇒4×9 + 1 = 37
7
11
17
4×3 + 1 = 13
⇑
3
1⇒4×1 + 1 = 5
Theorem 3.2. For any odd integer A, there are infinite Collatz sequences that do not
converge.
Proof. Given an odd integer A, consider the sequence of jumps bi= 4bi−1+ 1 with
b0=A. This is an infinite sequence with equivalent Collatz sequences by Lemma 3.1.
If for any i,bi≡0 mod 3, then bi+1 ≡1 mod 3, bi+2 ≡2 mod 3, and bi+3 ≡0 mod 3.
Which implies there are infinite jumps for any number Awhich are not multiples of 3.
The reverse Collatz sequences of these jumps are non trivial. Hence, there are infinite
ways to go backwards. Moreover, as in Example 3.7, these sequences can be extended
backwards infinitely. All these sequences have infinite terms and hence are divergent.
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By Theorem 3.2, the Collatz Conjecture is false. End of story.
References
[1] Ahmed Maya M, A window to the convergence of the Collatz sequence,
arXiv:1510.01274 [math.GM].
[2] Ahmed Maya M, The intricate labyrinth of Collatz sequences, arXiv:1602.01617
[math.CO].
[3] Jeffrey C. Lagarias, The 3x+1 Problem: An Annotated Bibliography, II (2000-2009),
arXiv:math/0608208v5 [math.NT].
[4] Jeffrey C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly
92 (1985) 3-23.
[5] Wirsching, G¨unther J., The Dynamical System Generated by the 3n+1 Function,
Lecture Notes in Mathematics, Vol. 1681, Springer Verlag, 1998.
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