Mixing
The orbit of $2^n+1$
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Consider the Collatz function, $$T(n)=frac n2, text if $n$ is even$$ and $$ T(n)=frac 3n+12, text if $n$ is odd$$
Observing the orbit of $2^n+1$, I came up with the following formula.
$$ T^n(2^n+1)= 3^n/2+1, text if $n$ is even$$
$$T^n(2^n+1)= 3^(n+1)/2+2, text if $n$ is odd$$
I came up with the formula by following the orbit.
$$2^n+1to 3(2^n-1)+2 to 3(2^n-2)+1to 3^2(2^n-3)+2to 3^2(2^n-4)+1to...$$
For example $$T^5(2^5+1)=3^3+2=29$$
which is verified by $$33to 50to 25to 38to 19 to 29$$
For larger $n$ this is significant, for example:
$$T^100(2^100+1)= 3^50+1$$
$$T^1000(2^1000+1)= 3^500+1$$
Question:
How do we turn this into a formal proof?
number-theory collatz
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
add a comment |Â
up vote
4
down vote
favorite
Consider the Collatz function, $$T(n)=frac n2, text if $n$ is even$$ and $$ T(n)=frac 3n+12, text if $n$ is odd$$
Observing the orbit of $2^n+1$, I came up with the following formula.
$$ T^n(2^n+1)= 3^n/2+1, text if $n$ is even$$
$$T^n(2^n+1)= 3^(n+1)/2+2, text if $n$ is odd$$
I came up with the formula by following the orbit.
$$2^n+1to 3(2^n-1)+2 to 3(2^n-2)+1to 3^2(2^n-3)+2to 3^2(2^n-4)+1to...$$
For example $$T^5(2^5+1)=3^3+2=29$$
which is verified by $$33to 50to 25to 38to 19 to 29$$
For larger $n$ this is significant, for example:
$$T^100(2^100+1)= 3^50+1$$
$$T^1000(2^1000+1)= 3^500+1$$
Question:
How do we turn this into a formal proof?
number-theory collatz
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
1
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
1
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
1
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider the Collatz function, $$T(n)=frac n2, text if $n$ is even$$ and $$ T(n)=frac 3n+12, text if $n$ is odd$$
Observing the orbit of $2^n+1$, I came up with the following formula.
$$ T^n(2^n+1)= 3^n/2+1, text if $n$ is even$$
$$T^n(2^n+1)= 3^(n+1)/2+2, text if $n$ is odd$$
I came up with the formula by following the orbit.
$$2^n+1to 3(2^n-1)+2 to 3(2^n-2)+1to 3^2(2^n-3)+2to 3^2(2^n-4)+1to...$$
For example $$T^5(2^5+1)=3^3+2=29$$
which is verified by $$33to 50to 25to 38to 19 to 29$$
For larger $n$ this is significant, for example:
$$T^100(2^100+1)= 3^50+1$$
$$T^1000(2^1000+1)= 3^500+1$$
Question:
How do we turn this into a formal proof?
number-theory collatz
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
Consider the Collatz function, $$T(n)=frac n2, text if $n$ is even$$ and $$ T(n)=frac 3n+12, text if $n$ is odd$$
Observing the orbit of $2^n+1$, I came up with the following formula.
$$ T^n(2^n+1)= 3^n/2+1, text if $n$ is even$$
$$T^n(2^n+1)= 3^(n+1)/2+2, text if $n$ is odd$$
I came up with the formula by following the orbit.
$$2^n+1to 3(2^n-1)+2 to 3(2^n-2)+1to 3^2(2^n-3)+2to 3^2(2^n-4)+1to...$$
For example $$T^5(2^5+1)=3^3+2=29$$
which is verified by $$33to 50to 25to 38to 19 to 29$$
For larger $n$ this is significant, for example:
$$T^100(2^100+1)= 3^50+1$$
$$T^1000(2^1000+1)= 3^500+1$$
Question:
How do we turn this into a formal proof?
number-theory collatz
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
edited Jul 15 at 18:46
TheSimpliFire
9,69261951
9,69261951
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
asked Jul 15 at 18:37
Mohammad Riazi-Kermani
27.6k41852
27.6k41852
1
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
1
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
1
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26
add a comment |Â
1
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
1
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
1
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26
1
1
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
1
1
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
1
1
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
By induction.
We have
$$2^a3^b+1tofrac2^a3^b+1+42to2^a-23^b+1+1.$$
Then for even $a$, after $a/2$ steps,
$$2^a3^0+1to2^03^a/2+1$$
and for odd $a$, after $(a-1)/2$ steps,
$$2^a3^0+1to2^13^(a-1)/2+1tofrac2^13^(a+1)/2+42=2^03^(a+1)/2+2.$$
answered Jul 15 at 19:13
Yves Daoust
111k665204
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
By induction.
We have
$$2^a3^b+1tofrac2^a3^b+1+42to2^a-23^b+1+1.$$
Then for even $a$, after $a/2$ steps,
$$2^a3^0+1to2^03^a/2+1$$
and for odd $a$, after $(a-1)/2$ steps,
$$2^a3^0+1to2^13^(a-1)/2+1tofrac2^13^(a+1)/2+42=2^03^(a+1)/2+2.$$
answered Jul 15 at 19:13
Yves Daoust
111k665204
add a comment |Â
up vote
4
down vote
accepted
By induction.
We have
$$2^a3^b+1tofrac2^a3^b+1+42to2^a-23^b+1+1.$$
Then for even $a$, after $a/2$ steps,
$$2^a3^0+1to2^03^a/2+1$$
and for odd $a$, after $(a-1)/2$ steps,
$$2^a3^0+1to2^13^(a-1)/2+1tofrac2^13^(a+1)/2+42=2^03^(a+1)/2+2.$$
answered Jul 15 at 19:13
Yves Daoust
111k665204
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
By induction.
We have
$$2^a3^b+1tofrac2^a3^b+1+42to2^a-23^b+1+1.$$
Then for even $a$, after $a/2$ steps,
$$2^a3^0+1to2^03^a/2+1$$
and for odd $a$, after $(a-1)/2$ steps,
$$2^a3^0+1to2^13^(a-1)/2+1tofrac2^13^(a+1)/2+42=2^03^(a+1)/2+2.$$
answered Jul 15 at 19:13
Yves Daoust
111k665204
By induction.
We have
$$2^a3^b+1tofrac2^a3^b+1+42to2^a-23^b+1+1.$$
Then for even $a$, after $a/2$ steps,
$$2^a3^0+1to2^03^a/2+1$$
and for odd $a$, after $(a-1)/2$ steps,
$$2^a3^0+1to2^13^(a-1)/2+1tofrac2^13^(a+1)/2+42=2^03^(a+1)/2+2.$$
answered Jul 15 at 19:13
Yves Daoust
111k665204
answered Jul 15 at 19:13
Yves Daoust
111k665204
answered Jul 15 at 19:13
Yves Daoust
111k665204
answered Jul 15 at 19:13
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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1
Very nice to see such patterns with that definition of the counting of steps. I'm used to the version $n_k+1=3n+1over2^A$ and analysis of that function would never have shown such a nice pattern. Still new surprises in the Collatz-problem :-)
– Gottfried Helms
Jul 16 at 6:35
@GottfriedHelms The Collatz conjecture has lots of surprises and beauties.
– Mohammad Riazi-Kermani
Jul 16 at 10:21
1
another one: $T^18n(acdot 2048^n-61)= acdot 2187^n-61$. There are some other here: math.stackexchange.com/q/2428208
– Collag3n
Jul 18 at 18:13
1
This is equivalent to $hcdot4^p+1->hcdot3^p+1$ in the above link. For odd $n$ , you just set $h=2$ and use even exponent $n-1$.
– Collag3n
Jul 18 at 18:26