Find a formula for the collatz branch numbers
Clash Royale CLAN TAG#URR8PPP
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Context: Collatz conjecture
What I call a 'branch number', is a number accessible by 2 different routes.
Example :
- 24 is not a branch number, it can be accessed only from 48 (division by 2)
- 16 is a branch number, it can be accessed from 32 (divison by 2) or 5 (3x+1)
Is it possible to find a formula that generates these numbers or is this tied to the problem itself - so solving this would resolve the problem?
Thanks
Update
I'm talking about finding a function that generates these numbers with this sequence :
[10, 16, 22, 28, 34, 40, 46, ...]
collatz
asked Aug 21 at 13:20
toto
1214
add a comment |Â
up vote
0
down vote
favorite
Context: Collatz conjecture
What I call a 'branch number', is a number accessible by 2 different routes.
Example :
- 24 is not a branch number, it can be accessed only from 48 (division by 2)
- 16 is a branch number, it can be accessed from 32 (divison by 2) or 5 (3x+1)
Is it possible to find a formula that generates these numbers or is this tied to the problem itself - so solving this would resolve the problem?
Thanks
Update
I'm talking about finding a function that generates these numbers with this sequence :
[10, 16, 22, 28, 34, 40, 46, ...]
collatz
asked Aug 21 at 13:20
toto
1214
7
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
1
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
1
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
@JaapScherphuis thanks
– toto
Aug 22 at 9:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Context: Collatz conjecture
What I call a 'branch number', is a number accessible by 2 different routes.
Example :
- 24 is not a branch number, it can be accessed only from 48 (division by 2)
- 16 is a branch number, it can be accessed from 32 (divison by 2) or 5 (3x+1)
Is it possible to find a formula that generates these numbers or is this tied to the problem itself - so solving this would resolve the problem?
Thanks
Update
I'm talking about finding a function that generates these numbers with this sequence :
[10, 16, 22, 28, 34, 40, 46, ...]
collatz
asked Aug 21 at 13:20
toto
1214
Context: Collatz conjecture
What I call a 'branch number', is a number accessible by 2 different routes.
Example :
- 24 is not a branch number, it can be accessed only from 48 (division by 2)
- 16 is a branch number, it can be accessed from 32 (divison by 2) or 5 (3x+1)
Is it possible to find a formula that generates these numbers or is this tied to the problem itself - so solving this would resolve the problem?
Thanks
Update
I'm talking about finding a function that generates these numbers with this sequence :
[10, 16, 22, 28, 34, 40, 46, ...]
collatz
asked Aug 21 at 13:20
toto
1214
edited Aug 22 at 9:19
asked Aug 21 at 13:20
toto
1214
asked Aug 21 at 13:20
toto
1214
asked Aug 21 at 13:20
toto
1214
1214
7
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
1
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
1
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
@JaapScherphuis thanks
– toto
Aug 22 at 9:17
add a comment |Â
7
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
1
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
1
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
@JaapScherphuis thanks
– toto
Aug 22 at 9:17
7
7
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
1
1
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
1
1
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
@JaapScherphuis thanks
– toto
Aug 22 at 9:17
@JaapScherphuis thanks
– toto
Aug 22 at 9:17
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
[10,$qquad$ 16,$quad$ 22,$qquad$ 28,$quad$ 34,$qquad$ 40,$qquad$ 46, ...] can be rewritten as
[1*6+4, 2*6+4, 3*6+4, 4*6+4, 5*6+4, ...] . You surely see the pattern - and the general formula?
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
add a comment |Â
up vote
0
down vote
We know that odd numbers cannot be branch numbers, because they can only be generated in one way (i.e., dividing by 2). Even numbers can only be branch numbers if they are of the form $3x+1$ for some $x$. This means that all positive even numbers $e$ that satisfy $eequiv1mod3$ are branch numbers.
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
[10,$qquad$ 16,$quad$ 22,$qquad$ 28,$quad$ 34,$qquad$ 40,$qquad$ 46, ...] can be rewritten as
[1*6+4, 2*6+4, 3*6+4, 4*6+4, 5*6+4, ...] . You surely see the pattern - and the general formula?
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
add a comment |Â
up vote
0
down vote
accepted
[10,$qquad$ 16,$quad$ 22,$qquad$ 28,$quad$ 34,$qquad$ 40,$qquad$ 46, ...] can be rewritten as
[1*6+4, 2*6+4, 3*6+4, 4*6+4, 5*6+4, ...] . You surely see the pattern - and the general formula?
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
[10,$qquad$ 16,$quad$ 22,$qquad$ 28,$quad$ 34,$qquad$ 40,$qquad$ 46, ...] can be rewritten as
[1*6+4, 2*6+4, 3*6+4, 4*6+4, 5*6+4, ...] . You surely see the pattern - and the general formula?
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
[10,$qquad$ 16,$quad$ 22,$qquad$ 28,$quad$ 34,$qquad$ 40,$qquad$ 46, ...] can be rewritten as
[1*6+4, 2*6+4, 3*6+4, 4*6+4, 5*6+4, ...] . You surely see the pattern - and the general formula?
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
answered Aug 22 at 12:21
Gottfried Helms
22.7k24195
22.7k24195
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
add a comment |Â
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
I was missing one number in my sequence, thus making it more complex than it was :/
– toto
Aug 22 at 13:21
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
@toto : hmm, sh** happens :)
– Gottfried Helms
Aug 22 at 13:47
add a comment |Â
up vote
0
down vote
We know that odd numbers cannot be branch numbers, because they can only be generated in one way (i.e., dividing by 2). Even numbers can only be branch numbers if they are of the form $3x+1$ for some $x$. This means that all positive even numbers $e$ that satisfy $eequiv1mod3$ are branch numbers.
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
add a comment |Â
up vote
0
down vote
We know that odd numbers cannot be branch numbers, because they can only be generated in one way (i.e., dividing by 2). Even numbers can only be branch numbers if they are of the form $3x+1$ for some $x$. This means that all positive even numbers $e$ that satisfy $eequiv1mod3$ are branch numbers.
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We know that odd numbers cannot be branch numbers, because they can only be generated in one way (i.e., dividing by 2). Even numbers can only be branch numbers if they are of the form $3x+1$ for some $x$. This means that all positive even numbers $e$ that satisfy $eequiv1mod3$ are branch numbers.
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
We know that odd numbers cannot be branch numbers, because they can only be generated in one way (i.e., dividing by 2). Even numbers can only be branch numbers if they are of the form $3x+1$ for some $x$. This means that all positive even numbers $e$ that satisfy $eequiv1mod3$ are branch numbers.
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
answered Aug 21 at 13:25
Rushabh Mehta
1,430217
1,430217
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
add a comment |Â
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
So they are all $4 mod 6$, i.e. of the form $6k+4$ for some $k$.
– Jaap Scherphuis
Aug 21 at 13:39
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
My mistake, eyes aren't working.
– Daniel Littlewood
Aug 21 at 13:41
add a comment |Â
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7
Unsolicited advice: Stay away from the horrific time-sink which is this conjecture.
– Matt
Aug 21 at 13:22
1
What @Matt said above. This isn't something that can be resolved with simple algebraic manipulation.
– anomaly
Aug 21 at 13:27
How would it help resolve the problem? Can you elaborate a little more in your question? Also what @Matt said, Paul Erdos (a famous mathematician of the 20th century) once said "Mathematics is not yet ready for such problems", and in my very humble opinion I don't think much has changed since he said that.
– Logan Toll
Aug 21 at 13:32
1
Your numbers are simply $6k+4$ for $k=1,2,3,4...$.
– Jaap Scherphuis
Aug 22 at 9:10
@JaapScherphuis thanks
– toto
Aug 22 at 9:17