# Proof to the Collatz conjecture

Made a quick app in Python.
Brilliant!

I happen to have Python installed. Maybe I'll run some submission requests from Michael345.

While I'm learning Python

Code said:
X = input("Enter a seed number X: ")
seednumX = int(X)
steps = 0
evensteps = 0
oddsteps = 0
Xhi = 0

while X != 1:
if (int(X) % 2) == 0:
X = int(X) / 2
if int(X) > int(Xhi):
Xhi = int(X)
print(int(X))
evensteps += 1
steps += 1
else:
X = 3 * int(X) + 1
if int(X) > int(Xhi):
Xhi = int(X)
print(int(X))
oddsteps += 1
steps += 1

print("Seed Number X = " + str(seednumX) + " | X Seed Length = " + str(len(str(seednumX))))
print("X Highest = " + str(Xhi) + " | X Highest Length = " + str(len(str(Xhi))))
print("Even Steps = " + str(evensteps))
print("Odd Steps = " + str(oddsteps))
print("Total Steps = " + str(steps))

Output:

Enter a seed number X: 9663
28990
14495
...
9038141
27114424
13557212
6778606
...
4
2
1
Seed Number X = 9663 | X Seed Length = 4
X Highest = 27114424 | X Highest Length = 8
Even Steps = 118
Odd Steps = 66
Total Steps = 184

While I'm learning Python

Output:

Enter a seed number X: 9663
28990
14495
...
9038141
27114424
13557212
6778606
...
4
2
1
Seed Number X = 9663 | X Seed Length = 4
X Highest = 27114424 | X Highest Length = 8
Even Steps = 118
Odd Steps = 66
Total Steps = 184
Cool. Bonus marks if you can plot the output as a graph!

This one is actually done in Python!

https://ultimatetheorem.blogspot.com/2020/08/visualization-of-collatz-conjecture_10.html

Last edited:
BdS
James R;

This seems unlikely, because it would make sense to first publish any such proof in a peer-reviewed journal, rather than on an internet forum like this one.

dictionary:
peer-reviewed) vt
evaluate something as an expert: to assess an article, piece of work, or research as an expert on the subject

There are no experts, only people with more experience than others.
Peer reviewed journals are typically biased, assuming only people associated with eduational or scientific organizations are capable of critical thinking. Many expect to gain copyrights to the submitted material.

If anyone is interested in a solution to this 'recreational math' problem, here is a pdf.

#### Attachments

• The Collatz Conjecture - a proof.pdf
108.1 KB · Views: 9
If anyone is interested in a solution to this 'recreational math' problem, here is a pdf.
The paper is dated only five weeks ago.

Are we expected to accept that this is has been reviewed and accepted as a formal proof?

Peer reviewed journals are typically biased, assuming only people associated with eduational or scientific organizations are capable of critical thinking.
This is an opinion. Hard to claim what others are thinking without telepathy.

Still, regardless of whether the opinion is warranted, it's not justification for not submitting one's work to a peer-review, since a review would certainly be capable of finding errors if there were some.

Not submitting one's work to a peer review process is - at the very least - a tacit admission that the paper probably has holes in it big enough for one's peers to find (otherwise, what's the harm?).

Last edited:
The paper is dated only five weeks ago.

Are we expected to accept that this is has been reviewed and accepted as a formal proof?

This is an opinion. Hard to claim what others are thinking without telepathy.

Still, regardless of whether the opinion is warranted, it's not justification for not submitting one's work to a peer-review, since a review would certainly be capable of finding errors if there were some.

Not submitting one's work to a peer review process is - at the very least - a tacit admission that the paper probably has holes in it big enough for one's peers to find (otherwise, what's the harm?).
The paper has been revised a few times over the years, to make it comprehensible for the average reader with basic math skills, and correcting errors.
It's an opinion based on attempted submissions!
If the 'experts' have the ability to critique submissions, why don't they solve the problems?
I'm open to opinions by readers who understand the problem.

Code v1.3 said:
X = input("Enter a seed number X: ")
seednumX = int(X)
steps = 0
evensteps = 0
consecutiveevensteps = 0
consecutiveevenstepstring = ""
consecutivestepseven = False
oddsteps = 0
consecutiveoddsteps = 0
consecutiveoddstepstring = ""
consecutivestepsodd = False
Xhi = 0

while X != 1:

if (int(str(int(X))[-1]) % 2) == 0:
X = int(X) / 2
if int(consecutiveoddsteps) > 1:
consecutiveoddstepstring = consecutiveoddstepstring + str(consecutiveoddsteps) + ", "

consecutivestepsodd = False
consecutiveoddsteps = 0

if consecutivestepseven == True:
consecutiveevensteps += 1
else:
consecutivestepseven = True
consecutiveevensteps += 1

if int(X) > int(Xhi):
Xhi = int(X)

if (int(str(int(X))[-1]) % 2) == 0:
print(str(int(X)) + " EVEN")
else:
print(str(int(X)) + " ODD")

evensteps += 1
steps += 1
else:
X = 3 * int(X) + 1
if int(consecutiveevensteps) > 1:
consecutiveevenstepstring = consecutiveevenstepstring + str(consecutiveevensteps) + ", "
consecutivestepseven = False
consecutiveevensteps = 0

if consecutivestepsodd == True:
consecutiveoddsteps += 1
else:
consecutivestepsodd = True
consecutiveoddsteps += 1

oddsteps += 1
if int(X) > int(Xhi):
Xhi = int(X)

if (int(str(int(X))[-1]) % 2) == 0:
print(str(int(X)) + " EVEN")
else:
print(str(int(X)) + " ODD")
steps += 1

if int(consecutiveevensteps) > 1:
consecutiveevenstepstring = consecutiveevenstepstring + str(consecutiveevensteps)
if int(consecutiveoddsteps) > 1:
consecutiveoddstepstring = consecutiveoddstepstring + str(consecutiveoddsteps)

print("Seed Number X = " + str(seednumX) + " | X Seed Length = " + str(len(str(seednumX))))
print("X Highest = " + str(Xhi) + " | X Highest Length = " + str(len(str(Xhi))))
print("Even Steps = " + str(evensteps))
print("Odd Steps = " + str(oddsteps))
print("Total Steps = " + str(steps))
print("Consecutive Even Steps = " + str(consecutiveevenstepstring))
print("Consecutive Odd Steps = " + str(consecutiveoddstepstring))

output:

Enter a seed number X: 69
208 EVEN
104 EVEN
52 EVEN
26 EVEN
13 ODD
40 EVEN
20 EVEN
10 EVEN
5 ODD
16 EVEN
8 EVEN
4 EVEN
2 EVEN
1 ODD
Seed Number X = 69 | X Seed Length = 2
X Highest = 208 | X Highest Length = 3
Even Steps = 11
Odd Steps = 3
Total Steps = 14
Consecutive Even Steps = 4, 3, 4
Consecutive Odd Steps =

There is never any consecutive odd steps. Because as soon as X is odd then you times by 3, X always becomes odd again, then when you add 1 you make it even... ?

Seed Number X = 6325874123658987 ... 88745566321456987 | X Seed Length = 206
X Highest = 1897762237097 ... 36698964370962 | X Highest Length = 207
Even Steps = 928
Odd Steps = 154
Total Steps = 1082
Consecutive Even Steps = 634, 6, 2, 5, 2, 5, 2, 2, 2, 2, 5, 2, 2, 2, 2, 4, 2, 2, 4, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 5, 2, 2, 3, 5, 2, 4, 3, 4, 2, 2, 4, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 5, 5, 2, 2, 7, 2, 3, 2, 2, 6, 2, 2, 6, 3, 3, 3, 3, 4, 3, 2, 3, 4
Consecutive Odd Steps =

Not sure why when I enter very long numbers its always starts with a high amount of consecutive even steps? the longer the X number length the greater the first even consecutive steps value. Maybe a limitation to handling large numbers?

There is never any consecutive odd steps. Because as soon as X is odd then you times by 3, X always becomes odd again, then when you add 1 you make it even... ?

Yes, that sounds correct.

Seed Number X = 6325874123658987 ... 88745566321456987 | X Seed Length = 206
X Highest = 1897762237097 ... 36698964370962 | X Highest Length = 207
Even Steps = 928
Odd Steps = 154
Total Steps = 1082
Consecutive Even Steps = 634, 6, 2, 5, 2, 5, 2, 2, 2, 2, 5, 2, 2, 2, 2, 4, 2, 2, 4, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 5, 2, 2, 3, 5, 2, 4, 3, 4, 2, 2, 4, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 5, 5, 2, 2, 7, 2, 3, 2, 2, 6, 2, 2, 6, 3, 3, 3, 3, 4, 3, 2, 3, 4
Consecutive Odd Steps =

Not sure why when I enter very long numbers its always starts with a high amount of consecutive even steps? the longer the X number length the greater the first even consecutive steps value. Maybe a limitation to handling large numbers?

Interesting. Also, it appears that 7 is the maximum number of even consecutive steps after the first large number of even consecutive steps is done. At least that is how it looks from your examples.

BdS,

Just for fun, try multiplying the odd numbers by 1.5 and then adding 0.5. That will allow some odd numbers to lead to other odd numbers, which will give you some results in the Consecutive Odd Steps = results. It may take forever though, so be careful not to melt your cpu!

Since the rule (odd * 3) + 1 always produces an even outcome, we know the next step will be to divide by 2. So...

((odd * 3) + 1) / 2
(odd * 1.5) + 0.5

So the rule (odd * 1.5) + 0.5 will ultimately result in the same outcome. But it allows us to track how many consecutive odd steps happen, just for reference.

Interesting.
Not sure whether it is interesting or just a technical limitation in calculating long numbers.
At first I thought this line "if (int(X) % 2) == 0:" used to check if X is currently odd or even was having a problem parsing the long numbers. I changed it to "if (int(str(int(X))[-1]) % 2) == 0:" to only look at the last digit of X and determine if X is odd or even from the last digit. It produces the same results with both lines, so now I'm not sure whats causing it.

Just for fun, try multiplying the odd numbers by 1.5 and then adding 0.5.
It's working with natural integer numbers so to change it to use decimals is too much effort
load python copy paste the code, you'll just need to get the indentations sorted out, because the website is trimming the leading spaces/tabs on the beginning of every line.

calculating long numbers.

Don't know how many phone numbers on telephone communication systems

Indonesia has started to register sim card to phone's IMEI number so it will not work in another phone

Asked friend over in Bali to buy sim card for me and gave photo of passport

Friend went to sim card selling shop. When my details were checked shop gave friend new sim card and friend sent to me

I always put own phone number in the phone. If it gets mislaid call own number and might hear it to track down

Number on sim card friend given was / is same as number I had / used 5 years agol

Guessing shop recreated my old number registered it to new device (my upgraded phone) and since NOW REGISTERED will not work in any other phone

Think of the number of locals with phones all the tourist who use a local number on holiday and then keeping check no phone numbers are duplicated and where it was last used