Mixing
Is there a maximum number of consecutive decreasing steps a Collatz cycle can have?
$begingroup$
If we take a look into the (known) cycles of the Collatz Conjecture when all integers are included, we get 4 cycles:
1 → 4 → 2 → 1 …
−1 → −2 → −1 …
−5 → −14 → −7 → −20 → −10 → −5 …
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …
The 1 and -5 cycles have at most 2 consecutive decreasing steps (n/2), the -1 cycle has just 1, and the -17 cycle has 4, which brings me to the question:
If there was any other cycle, would there be a maximum of consecutive decreasing steps it can have? And if so, would this maximum be different between a negative and a positive cycle?
Also, do we know for example why the -17 cycle reaches a point with 4 consecutive decreasing steps? As in, do we know why any given number reaches for example 4 consecutive decreasing steps?
And lastly, would having a maximum number of consecutive decreasing steps higher than 2 for a positive cycle prove the existence of another cycle different than the trivial 4:2:1 ?
Sorry for the amount of followup questions.
number-theory collatz
asked Jan 20 at 20:25
j jj j
91
$endgroup$
add a comment |
$begingroup$
If we take a look into the (known) cycles of the Collatz Conjecture when all integers are included, we get 4 cycles:
1 → 4 → 2 → 1 …
−1 → −2 → −1 …
−5 → −14 → −7 → −20 → −10 → −5 …
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …
The 1 and -5 cycles have at most 2 consecutive decreasing steps (n/2), the -1 cycle has just 1, and the -17 cycle has 4, which brings me to the question:
If there was any other cycle, would there be a maximum of consecutive decreasing steps it can have? And if so, would this maximum be different between a negative and a positive cycle?
Also, do we know for example why the -17 cycle reaches a point with 4 consecutive decreasing steps? As in, do we know why any given number reaches for example 4 consecutive decreasing steps?
And lastly, would having a maximum number of consecutive decreasing steps higher than 2 for a positive cycle prove the existence of another cycle different than the trivial 4:2:1 ?
Sorry for the amount of followup questions.
number-theory collatz
asked Jan 20 at 20:25
j jj j
91
$endgroup$
$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23
add a comment |
$begingroup$
If we take a look into the (known) cycles of the Collatz Conjecture when all integers are included, we get 4 cycles:
1 → 4 → 2 → 1 …
−1 → −2 → −1 …
−5 → −14 → −7 → −20 → −10 → −5 …
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …
The 1 and -5 cycles have at most 2 consecutive decreasing steps (n/2), the -1 cycle has just 1, and the -17 cycle has 4, which brings me to the question:
If there was any other cycle, would there be a maximum of consecutive decreasing steps it can have? And if so, would this maximum be different between a negative and a positive cycle?
Also, do we know for example why the -17 cycle reaches a point with 4 consecutive decreasing steps? As in, do we know why any given number reaches for example 4 consecutive decreasing steps?
And lastly, would having a maximum number of consecutive decreasing steps higher than 2 for a positive cycle prove the existence of another cycle different than the trivial 4:2:1 ?
Sorry for the amount of followup questions.
number-theory collatz
asked Jan 20 at 20:25
j jj j
91
$endgroup$
If we take a look into the (known) cycles of the Collatz Conjecture when all integers are included, we get 4 cycles:
1 → 4 → 2 → 1 …
−1 → −2 → −1 …
−5 → −14 → −7 → −20 → −10 → −5 …
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …
The 1 and -5 cycles have at most 2 consecutive decreasing steps (n/2), the -1 cycle has just 1, and the -17 cycle has 4, which brings me to the question:
If there was any other cycle, would there be a maximum of consecutive decreasing steps it can have? And if so, would this maximum be different between a negative and a positive cycle?
Also, do we know for example why the -17 cycle reaches a point with 4 consecutive decreasing steps? As in, do we know why any given number reaches for example 4 consecutive decreasing steps?
And lastly, would having a maximum number of consecutive decreasing steps higher than 2 for a positive cycle prove the existence of another cycle different than the trivial 4:2:1 ?
Sorry for the amount of followup questions.
number-theory collatz
number-theory collatz
asked Jan 20 at 20:25
j jj j
91
asked Jan 20 at 20:25
j jj j
91
asked Jan 20 at 20:25
j jj j
91
asked Jan 20 at 20:25
j jj j
91
asked Jan 20 at 20:25
j jj j
91
91
$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23
add a comment |
$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23
$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23
$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.
A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1over 2^A } tag 1$$ then, if continued, $c = {3b+1over 2^B }$. One more, $d = {3c+1over 2^C }$. One could write the whole concatenated trajectory as $$ d=mathcal C(a;[A,B,C]) tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.
Then we can introduce even longer trajectories, using indexes, $$ a_N=mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])qquad text{with} quad S=sum_{k=0}^{N-1}A_k tag 3$$
If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $mathcal C()$.
Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=mathcal (a,[1,1,1,1]) approx 1.5^4 cdot a implies e approx 5.06 cdot a$ and we can have at best $a = mathcal C(e,[4])$ which means $a ={ 3 e+1over 16} approx {15 a+1 over 16} approx a $.
So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.
You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form
$$ a_N= a_0 = mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) tag 4$$
and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.
But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$).
(This is by John Simons & Benne de Weger in 2000 and later)
However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...
(For references to Steiner/Simons see external links in the wikipedia-article)
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
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add a comment |
$begingroup$
The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.
Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
add a comment |
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2 Answers
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$begingroup$
The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.
A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1over 2^A } tag 1$$ then, if continued, $c = {3b+1over 2^B }$. One more, $d = {3c+1over 2^C }$. One could write the whole concatenated trajectory as $$ d=mathcal C(a;[A,B,C]) tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.
Then we can introduce even longer trajectories, using indexes, $$ a_N=mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])qquad text{with} quad S=sum_{k=0}^{N-1}A_k tag 3$$
If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $mathcal C()$.
Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=mathcal (a,[1,1,1,1]) approx 1.5^4 cdot a implies e approx 5.06 cdot a$ and we can have at best $a = mathcal C(e,[4])$ which means $a ={ 3 e+1over 16} approx {15 a+1 over 16} approx a $.
So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.
You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form
$$ a_N= a_0 = mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) tag 4$$
and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.
But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$).
(This is by John Simons & Benne de Weger in 2000 and later)
However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...
(For references to Steiner/Simons see external links in the wikipedia-article)
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.
A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1over 2^A } tag 1$$ then, if continued, $c = {3b+1over 2^B }$. One more, $d = {3c+1over 2^C }$. One could write the whole concatenated trajectory as $$ d=mathcal C(a;[A,B,C]) tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.
Then we can introduce even longer trajectories, using indexes, $$ a_N=mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])qquad text{with} quad S=sum_{k=0}^{N-1}A_k tag 3$$
If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $mathcal C()$.
Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=mathcal (a,[1,1,1,1]) approx 1.5^4 cdot a implies e approx 5.06 cdot a$ and we can have at best $a = mathcal C(e,[4])$ which means $a ={ 3 e+1over 16} approx {15 a+1 over 16} approx a $.
So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.
You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form
$$ a_N= a_0 = mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) tag 4$$
and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.
But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$).
(This is by John Simons & Benne de Weger in 2000 and later)
However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...
(For references to Steiner/Simons see external links in the wikipedia-article)
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.
A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1over 2^A } tag 1$$ then, if continued, $c = {3b+1over 2^B }$. One more, $d = {3c+1over 2^C }$. One could write the whole concatenated trajectory as $$ d=mathcal C(a;[A,B,C]) tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.
Then we can introduce even longer trajectories, using indexes, $$ a_N=mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])qquad text{with} quad S=sum_{k=0}^{N-1}A_k tag 3$$
If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $mathcal C()$.
Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=mathcal (a,[1,1,1,1]) approx 1.5^4 cdot a implies e approx 5.06 cdot a$ and we can have at best $a = mathcal C(e,[4])$ which means $a ={ 3 e+1over 16} approx {15 a+1 over 16} approx a $.
So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.
You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form
$$ a_N= a_0 = mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) tag 4$$
and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.
But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$).
(This is by John Simons & Benne de Weger in 2000 and later)
However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...
(For references to Steiner/Simons see external links in the wikipedia-article)
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.
A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1over 2^A } tag 1$$ then, if continued, $c = {3b+1over 2^B }$. One more, $d = {3c+1over 2^C }$. One could write the whole concatenated trajectory as $$ d=mathcal C(a;[A,B,C]) tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.
Then we can introduce even longer trajectories, using indexes, $$ a_N=mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])qquad text{with} quad S=sum_{k=0}^{N-1}A_k tag 3$$
If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $mathcal C()$.
Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=mathcal (a,[1,1,1,1]) approx 1.5^4 cdot a implies e approx 5.06 cdot a$ and we can have at best $a = mathcal C(e,[4])$ which means $a ={ 3 e+1over 16} approx {15 a+1 over 16} approx a $.
So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.
You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form
$$ a_N= a_0 = mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) tag 4$$
and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.
But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$).
(This is by John Simons & Benne de Weger in 2000 and later)
However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...
(For references to Steiner/Simons see external links in the wikipedia-article)
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
edited Jan 21 at 12:24
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
answered Jan 21 at 11:52
Gottfried HelmsGottfried Helms
23.5k24599
23.5k24599
add a comment |
add a comment |
$begingroup$
The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.
Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
add a comment |
$begingroup$
The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.
Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
add a comment |
$begingroup$
The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.
Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
$endgroup$
The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.
Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
answered Jan 20 at 20:31
Ross MillikanRoss Millikan
298k24200373
298k24200373
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
add a comment |
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
Thanks for the answer. What I meant with how do we know it has 4 consecutive decreasing steps was if by just having any given number, without iterating the sequence, we can tell what is the max number of decreasing steps it will have; if by any characteristics of our given number we can deduct the most consecutive decreasing steps it will eventually reach just looking at the number and not knowing the next iterations. Hope it makes sense
$endgroup$
– j j
Jan 20 at 20:36
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
$begingroup$
I haven't read much about the massive amount of work done on this. My impression is that you can't tell much where a number will lead without doing the iteration.
$endgroup$
– Ross Millikan
Jan 20 at 20:40
add a comment |
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$begingroup$
Most likely would there be a different maximum between a negative and positive cycle. This is because of the $+1$ in $3n+1$ equation. This addition gives different behaviour for negatives than positives. adding $1$ to a negative means that this local is approaching zero, while adding $1$ to a positive does not. One question raises, would $3n-1$ for negatives and $3n+1$ for positives give same maximum?
$endgroup$
– Natural Number Guy
Jan 21 at 22:23