Distance from One Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Find a high-scoring 53-digit prime number chainDo better than chanceMaking π from 1 2 3 4 5 6 7 8 9Eight distinct numbers in the tableMy time-travelling adventureLeast amount of (n)x(n+1) tiles to perfectly cover a 7x7 floor with two irregular tilesI am not sure if this is related to magic squares but is something that is unanswered in my mind since I was a kidMore than equilibriumAn odd way to subtract digitsHappy Pi-Day! Try to solve this “PiDoku”
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Distance from One
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Find a high-scoring 53-digit prime number chainDo better than chanceMaking π from 1 2 3 4 5 6 7 8 9Eight distinct numbers in the tableMy time-travelling adventureLeast amount of (n)x(n+1) tiles to perfectly cover a 7x7 floor with two irregular tilesI am not sure if this is related to magic squares but is something that is unanswered in my mind since I was a kidMore than equilibriumAn odd way to subtract digitsHappy Pi-Day! Try to solve this “PiDoku”
$begingroup$
For random natural number $N_0 > 1$, we find $N_1$ by the rule:
$N_i+1 = fracN_i2$ if $N_i$ is even
$N_i+1 = 3N_i + 1$ if $N_i$ is odd
If $N_1$ isn't equal to $1$, we find $N_2$ by applying the same rule on $N_1$.
$N_0$ is at a distance $X$ from $1$, when after exactly $X$ applications of the rule, we get $1$ for the first time.
What's the amount of natural numbers at distance 61 from 1 ?
mathematics
edited Apr 11 at 11:15
jafe
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
$endgroup$
add a comment |
$begingroup$
For random natural number $N_0 > 1$, we find $N_1$ by the rule:
$N_i+1 = fracN_i2$ if $N_i$ is even
$N_i+1 = 3N_i + 1$ if $N_i$ is odd
If $N_1$ isn't equal to $1$, we find $N_2$ by applying the same rule on $N_1$.
$N_0$ is at a distance $X$ from $1$, when after exactly $X$ applications of the rule, we get $1$ for the first time.
What's the amount of natural numbers at distance 61 from 1 ?
mathematics
edited Apr 11 at 11:15
jafe
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
$endgroup$
$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago
add a comment |
$begingroup$
For random natural number $N_0 > 1$, we find $N_1$ by the rule:
$N_i+1 = fracN_i2$ if $N_i$ is even
$N_i+1 = 3N_i + 1$ if $N_i$ is odd
If $N_1$ isn't equal to $1$, we find $N_2$ by applying the same rule on $N_1$.
$N_0$ is at a distance $X$ from $1$, when after exactly $X$ applications of the rule, we get $1$ for the first time.
What's the amount of natural numbers at distance 61 from 1 ?
mathematics
edited Apr 11 at 11:15
jafe
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
$endgroup$
For random natural number $N_0 > 1$, we find $N_1$ by the rule:
$N_i+1 = fracN_i2$ if $N_i$ is even
$N_i+1 = 3N_i + 1$ if $N_i$ is odd
If $N_1$ isn't equal to $1$, we find $N_2$ by applying the same rule on $N_1$.
$N_0$ is at a distance $X$ from $1$, when after exactly $X$ applications of the rule, we get $1$ for the first time.
What's the amount of natural numbers at distance 61 from 1 ?
mathematics
mathematics
edited Apr 11 at 11:15
jafe
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
edited Apr 11 at 11:15
jafe
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
edited Apr 11 at 11:15
jafe
26k475255
edited Apr 11 at 11:15
jafe
26k475255
edited Apr 11 at 11:15
jafe
26k475255
26k475255
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
asked Apr 11 at 11:00
Kradec na kysmetKradec na kysmet
918
918
$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago
add a comment |
$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago
$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know if this is strictly a puzzle but more of a computational programming exercise.
Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an entry on OEIS which describes the number of numbers that are exactly $n$ steps away from $1$, see here.
The value for $61$ is not listed here but they do give functions in a few different languages to obtain the number of steps for arbitrary $n$, and, saying that, it's not terribly difficult to write one yourself, for which the result is
$$1040490$$
Some other interesting things
This is a nice graph from Wikipedia showing the numbers which are within $20$ steps of $1$:
Also, the general pattern of the sequence tends to settle into exponential growth with ratio between adjacent terms thought to approach $frac3 + sqrt216$, which is something I didn't know before.
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
$endgroup$
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know if this is strictly a puzzle but more of a computational programming exercise.
Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an entry on OEIS which describes the number of numbers that are exactly $n$ steps away from $1$, see here.
The value for $61$ is not listed here but they do give functions in a few different languages to obtain the number of steps for arbitrary $n$, and, saying that, it's not terribly difficult to write one yourself, for which the result is
$$1040490$$
Some other interesting things
This is a nice graph from Wikipedia showing the numbers which are within $20$ steps of $1$:
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
$endgroup$
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
add a comment |
$begingroup$
I don't know if this is strictly a puzzle but more of a computational programming exercise.
Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an entry on OEIS which describes the number of numbers that are exactly $n$ steps away from $1$, see here.
The value for $61$ is not listed here but they do give functions in a few different languages to obtain the number of steps for arbitrary $n$, and, saying that, it's not terribly difficult to write one yourself, for which the result is
$$1040490$$
Some other interesting things
This is a nice graph from Wikipedia showing the numbers which are within $20$ steps of $1$:
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
$endgroup$
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
add a comment |
$begingroup$
I don't know if this is strictly a puzzle but more of a computational programming exercise.
Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an entry on OEIS which describes the number of numbers that are exactly $n$ steps away from $1$, see here.
The value for $61$ is not listed here but they do give functions in a few different languages to obtain the number of steps for arbitrary $n$, and, saying that, it's not terribly difficult to write one yourself, for which the result is
$$1040490$$
Some other interesting things
This is a nice graph from Wikipedia showing the numbers which are within $20$ steps of $1$:
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
$endgroup$
I don't know if this is strictly a puzzle but more of a computational programming exercise.
Many people on this site will already be familiar with the iterative function in the set-up to this problem as that from the famous Collatz conjecture. As you would expect, much research has been conducted into the properties of this function and there is even an entry on OEIS which describes the number of numbers that are exactly $n$ steps away from $1$, see here.
The value for $61$ is not listed here but they do give functions in a few different languages to obtain the number of steps for arbitrary $n$, and, saying that, it's not terribly difficult to write one yourself, for which the result is
$$1040490$$
Some other interesting things
This is a nice graph from Wikipedia showing the numbers which are within $20$ steps of $1$:
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
answered Apr 11 at 11:42
hexominohexomino
47.1k4143221
47.1k4143221
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
add a comment |
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
$begingroup$
The link you shared is for just 3*x + 1
$endgroup$
– Kradec na kysmet
Apr 11 at 11:51
2
2
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
@Kradecnakysmet It's sometimes called the "3x+1" problem but the set-up is as you've described.
$endgroup$
– hexomino
Apr 11 at 11:53
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
How did you count it so fast, is there formula for it ?
$endgroup$
– Kradec na kysmet
Apr 11 at 12:23
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
$begingroup$
Here's one interesting way to look at this issue: once you hit 2^X, you are X steps away from 1. The only way to get to 1 is to be at 2, then 4, etc. So, really your somewhat random walk ends on any power of 2.
$endgroup$
– Jim
Apr 11 at 18:40
add a comment |
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$begingroup$
This stems from the $3n+1$ conjecture! $(+1)$ :D
$endgroup$
– user477343
Apr 11 at 11:48
$begingroup$
Clarification needed: is the final question "what is the sum of all the natural numbers that are a distance of 61 from 1?", or does this mean "how many natural numbers are a distance of 61 from 1?"
$endgroup$
– Hugh
Apr 11 at 21:03
$begingroup$
It's asking how many (count)
$endgroup$
– Kradec na kysmet
2 days ago