Mixing
An interesting pattern in the differences between prime numbers.
$begingroup$
Once upon a time, I was looking at interesting properties of prime numbers. One thing I noticed was that if we take the absolute values of the differences between each prime, and repeat this process on the differences recursively, the first column turns out to always be $1$ (With the exception of the first row) as demonstrated below for the first $10$ primes:
$$begin{matrix} p_n & 2 & & 3 & & 5 & & 7 & & 11 & & 13 & & 17 & & 19 & & 23 & & 29 & cdots \ 1^{text{st}}text{ difference}&& color{#007777}1 && 2 && 2 && 4 && 2 && 4 && 2 && 4 && 6 \2^{text{nd}}text{ difference}& && color{#007777}1 && 0 && 2 && 2 && 2 && 2 && 2 && 2 && \3^{text{rd}}text{ difference} &&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 && 0 \ 4^{text{th}}text{ difference}&&&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 \ vdots &&&&&& color{#007777}1 && 2 && 0 && 0 && 0 \ vdots&&&&&&&color{#007777}1 && 2 && 0 && 0 \ vdots &&&&&&&&color{#007777}1 && 2 && 0 \ vdots &&&&&&&&&color{#007777}1 && 2 \ vdots &&&&&&&&&&color{#007777}1end{matrix}$$
For simplicity, we denote the terms in the form $a_{m,n}$ where $m$ is the row number and $n$ is the column number, like this:
$$begin{matrix} a_{1,1} & & a_{1,2} & & a_{1,3} & & a_{1,4} && a_{1,5} & cdots \ & a_{2,1} &&a_{2,2} && a_{2,3} && a_{2,4} \ && a_{3,1} && a_{3,2} && a_{3,3} \ &&& a_{4,1} && a_{4,2} \ &&&&a_{5,1} end{matrix}$$
The general form for the differences is given by:
$$a_{m+1,n}=|a_{m,n+1}-a_{m,n}| tag{1}$$
Therefore, I conjectured the following:
Let $m$ and $n$ denote the row and column number respectively. Therefore, $a_{m,1}=1$ is true $forall mgeq 2$ where $m in mathbb{Z}^+$,
I'd like to know whether this conjecture is true. If so, it would be nice if a proof was provided. Therefore, here I show my thoughts on the problem:
The process on $(1)$ of course should be applied recursively to obtain an expression for the elements on the first row. For example, if i put $a_{4,1}$ in terms of the elements of the first row, we obtain:
$$begin{align} a_{4,1} & =|a_{3,2}-a_{3,1}| \ &=||a_{2,3}-a_{2,2}|-|a_{2,2}-a_{2,1}|| \ &=|||a_{1,4}-a_{1,3}|-|a_{1,3}-a_{1,2}||-||a_{1,3}-a_{1,2}|-|a_{1,2}-a_{1,1}||| end{align}tag{2}$$
This does not appear obvious as to why the the conjecture is true due to the absolute value signs.
Therefore, I decided to abandon using this idea. Somebody had suggested that I use Bertrand's Postulate. I will use the weaker form of the theorem, which states that if $p_n$ is the $n$-th prime, for all $ngeq 1$, then:
$$p_{n+1}<2p_n$$
In that case, I figured that if I apply this on the series with increasing powers of $2$, then:
$$begin{matrix} color{#007777}{1} & & color{#777700}2 & & color{green}{4} & & color{orange}8 && 16 & cdots \ & color{#007777}1 && color{#777700}2 && color{green}4 && color{orange}8 \ && color{#007777}1 && color{#777700}2 && color{green}4 \ &&& color{#007777}1 && color{#777700}2 \ &&&& color{#007777}1 end{matrix}$$
Then each of the columns will have identical values as shown by the different colors above. I figured this was quite similar, and since $p_nleq 2^n$ where $nin mathbb{Z}^+$, thus this would be true for the primes as well. However, it seems doubtful because the differences are not strictly increasing as we can see in the $2^{text{nd}} text{ difference}$ in the prime number series and also, we are taking the absolute value of the differences recursively as shown on the example on $(2)$.
Therefore, I do not think Bertrand's Postulate can be applied directly as I have done.
In summary, I'd like to know whether the conjecture is true. If so, a proof would be nice. Otherwise, if the conjecture is false a form of disproof such as a counterexample would be nice.
number-theory prime-numbers
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
$endgroup$
add a comment |
$begingroup$
Once upon a time, I was looking at interesting properties of prime numbers. One thing I noticed was that if we take the absolute values of the differences between each prime, and repeat this process on the differences recursively, the first column turns out to always be $1$ (With the exception of the first row) as demonstrated below for the first $10$ primes:
$$begin{matrix} p_n & 2 & & 3 & & 5 & & 7 & & 11 & & 13 & & 17 & & 19 & & 23 & & 29 & cdots \ 1^{text{st}}text{ difference}&& color{#007777}1 && 2 && 2 && 4 && 2 && 4 && 2 && 4 && 6 \2^{text{nd}}text{ difference}& && color{#007777}1 && 0 && 2 && 2 && 2 && 2 && 2 && 2 && \3^{text{rd}}text{ difference} &&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 && 0 \ 4^{text{th}}text{ difference}&&&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 \ vdots &&&&&& color{#007777}1 && 2 && 0 && 0 && 0 \ vdots&&&&&&&color{#007777}1 && 2 && 0 && 0 \ vdots &&&&&&&&color{#007777}1 && 2 && 0 \ vdots &&&&&&&&&color{#007777}1 && 2 \ vdots &&&&&&&&&&color{#007777}1end{matrix}$$
For simplicity, we denote the terms in the form $a_{m,n}$ where $m$ is the row number and $n$ is the column number, like this:
$$begin{matrix} a_{1,1} & & a_{1,2} & & a_{1,3} & & a_{1,4} && a_{1,5} & cdots \ & a_{2,1} &&a_{2,2} && a_{2,3} && a_{2,4} \ && a_{3,1} && a_{3,2} && a_{3,3} \ &&& a_{4,1} && a_{4,2} \ &&&&a_{5,1} end{matrix}$$
The general form for the differences is given by:
$$a_{m+1,n}=|a_{m,n+1}-a_{m,n}| tag{1}$$
Therefore, I conjectured the following:
Let $m$ and $n$ denote the row and column number respectively. Therefore, $a_{m,1}=1$ is true $forall mgeq 2$ where $m in mathbb{Z}^+$,
I'd like to know whether this conjecture is true. If so, it would be nice if a proof was provided. Therefore, here I show my thoughts on the problem:
The process on $(1)$ of course should be applied recursively to obtain an expression for the elements on the first row. For example, if i put $a_{4,1}$ in terms of the elements of the first row, we obtain:
$$begin{align} a_{4,1} & =|a_{3,2}-a_{3,1}| \ &=||a_{2,3}-a_{2,2}|-|a_{2,2}-a_{2,1}|| \ &=|||a_{1,4}-a_{1,3}|-|a_{1,3}-a_{1,2}||-||a_{1,3}-a_{1,2}|-|a_{1,2}-a_{1,1}||| end{align}tag{2}$$
This does not appear obvious as to why the the conjecture is true due to the absolute value signs.
Therefore, I decided to abandon using this idea. Somebody had suggested that I use Bertrand's Postulate. I will use the weaker form of the theorem, which states that if $p_n$ is the $n$-th prime, for all $ngeq 1$, then:
$$p_{n+1}<2p_n$$
In that case, I figured that if I apply this on the series with increasing powers of $2$, then:
$$begin{matrix} color{#007777}{1} & & color{#777700}2 & & color{green}{4} & & color{orange}8 && 16 & cdots \ & color{#007777}1 && color{#777700}2 && color{green}4 && color{orange}8 \ && color{#007777}1 && color{#777700}2 && color{green}4 \ &&& color{#007777}1 && color{#777700}2 \ &&&& color{#007777}1 end{matrix}$$
Then each of the columns will have identical values as shown by the different colors above. I figured this was quite similar, and since $p_nleq 2^n$ where $nin mathbb{Z}^+$, thus this would be true for the primes as well. However, it seems doubtful because the differences are not strictly increasing as we can see in the $2^{text{nd}} text{ difference}$ in the prime number series and also, we are taking the absolute value of the differences recursively as shown on the example on $(2)$.
Therefore, I do not think Bertrand's Postulate can be applied directly as I have done.
In summary, I'd like to know whether the conjecture is true. If so, a proof would be nice. Otherwise, if the conjecture is false a form of disproof such as a counterexample would be nice.
number-theory prime-numbers
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
$endgroup$
14
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
3
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
3
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
2
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52
add a comment |
$begingroup$
Once upon a time, I was looking at interesting properties of prime numbers. One thing I noticed was that if we take the absolute values of the differences between each prime, and repeat this process on the differences recursively, the first column turns out to always be $1$ (With the exception of the first row) as demonstrated below for the first $10$ primes:
$$begin{matrix} p_n & 2 & & 3 & & 5 & & 7 & & 11 & & 13 & & 17 & & 19 & & 23 & & 29 & cdots \ 1^{text{st}}text{ difference}&& color{#007777}1 && 2 && 2 && 4 && 2 && 4 && 2 && 4 && 6 \2^{text{nd}}text{ difference}& && color{#007777}1 && 0 && 2 && 2 && 2 && 2 && 2 && 2 && \3^{text{rd}}text{ difference} &&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 && 0 \ 4^{text{th}}text{ difference}&&&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 \ vdots &&&&&& color{#007777}1 && 2 && 0 && 0 && 0 \ vdots&&&&&&&color{#007777}1 && 2 && 0 && 0 \ vdots &&&&&&&&color{#007777}1 && 2 && 0 \ vdots &&&&&&&&&color{#007777}1 && 2 \ vdots &&&&&&&&&&color{#007777}1end{matrix}$$
For simplicity, we denote the terms in the form $a_{m,n}$ where $m$ is the row number and $n$ is the column number, like this:
$$begin{matrix} a_{1,1} & & a_{1,2} & & a_{1,3} & & a_{1,4} && a_{1,5} & cdots \ & a_{2,1} &&a_{2,2} && a_{2,3} && a_{2,4} \ && a_{3,1} && a_{3,2} && a_{3,3} \ &&& a_{4,1} && a_{4,2} \ &&&&a_{5,1} end{matrix}$$
The general form for the differences is given by:
$$a_{m+1,n}=|a_{m,n+1}-a_{m,n}| tag{1}$$
Therefore, I conjectured the following:
Let $m$ and $n$ denote the row and column number respectively. Therefore, $a_{m,1}=1$ is true $forall mgeq 2$ where $m in mathbb{Z}^+$,
I'd like to know whether this conjecture is true. If so, it would be nice if a proof was provided. Therefore, here I show my thoughts on the problem:
The process on $(1)$ of course should be applied recursively to obtain an expression for the elements on the first row. For example, if i put $a_{4,1}$ in terms of the elements of the first row, we obtain:
$$begin{align} a_{4,1} & =|a_{3,2}-a_{3,1}| \ &=||a_{2,3}-a_{2,2}|-|a_{2,2}-a_{2,1}|| \ &=|||a_{1,4}-a_{1,3}|-|a_{1,3}-a_{1,2}||-||a_{1,3}-a_{1,2}|-|a_{1,2}-a_{1,1}||| end{align}tag{2}$$
This does not appear obvious as to why the the conjecture is true due to the absolute value signs.
Therefore, I decided to abandon using this idea. Somebody had suggested that I use Bertrand's Postulate. I will use the weaker form of the theorem, which states that if $p_n$ is the $n$-th prime, for all $ngeq 1$, then:
$$p_{n+1}<2p_n$$
In that case, I figured that if I apply this on the series with increasing powers of $2$, then:
$$begin{matrix} color{#007777}{1} & & color{#777700}2 & & color{green}{4} & & color{orange}8 && 16 & cdots \ & color{#007777}1 && color{#777700}2 && color{green}4 && color{orange}8 \ && color{#007777}1 && color{#777700}2 && color{green}4 \ &&& color{#007777}1 && color{#777700}2 \ &&&& color{#007777}1 end{matrix}$$
Then each of the columns will have identical values as shown by the different colors above. I figured this was quite similar, and since $p_nleq 2^n$ where $nin mathbb{Z}^+$, thus this would be true for the primes as well. However, it seems doubtful because the differences are not strictly increasing as we can see in the $2^{text{nd}} text{ difference}$ in the prime number series and also, we are taking the absolute value of the differences recursively as shown on the example on $(2)$.
Therefore, I do not think Bertrand's Postulate can be applied directly as I have done.
In summary, I'd like to know whether the conjecture is true. If so, a proof would be nice. Otherwise, if the conjecture is false a form of disproof such as a counterexample would be nice.
number-theory prime-numbers
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
$endgroup$
Once upon a time, I was looking at interesting properties of prime numbers. One thing I noticed was that if we take the absolute values of the differences between each prime, and repeat this process on the differences recursively, the first column turns out to always be $1$ (With the exception of the first row) as demonstrated below for the first $10$ primes:
$$begin{matrix} p_n & 2 & & 3 & & 5 & & 7 & & 11 & & 13 & & 17 & & 19 & & 23 & & 29 & cdots \ 1^{text{st}}text{ difference}&& color{#007777}1 && 2 && 2 && 4 && 2 && 4 && 2 && 4 && 6 \2^{text{nd}}text{ difference}& && color{#007777}1 && 0 && 2 && 2 && 2 && 2 && 2 && 2 && \3^{text{rd}}text{ difference} &&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 && 0 \ 4^{text{th}}text{ difference}&&&&&color{#007777}1 && 2 && 0 && 0 && 0 && 0 \ vdots &&&&&& color{#007777}1 && 2 && 0 && 0 && 0 \ vdots&&&&&&&color{#007777}1 && 2 && 0 && 0 \ vdots &&&&&&&&color{#007777}1 && 2 && 0 \ vdots &&&&&&&&&color{#007777}1 && 2 \ vdots &&&&&&&&&&color{#007777}1end{matrix}$$
For simplicity, we denote the terms in the form $a_{m,n}$ where $m$ is the row number and $n$ is the column number, like this:
$$begin{matrix} a_{1,1} & & a_{1,2} & & a_{1,3} & & a_{1,4} && a_{1,5} & cdots \ & a_{2,1} &&a_{2,2} && a_{2,3} && a_{2,4} \ && a_{3,1} && a_{3,2} && a_{3,3} \ &&& a_{4,1} && a_{4,2} \ &&&&a_{5,1} end{matrix}$$
The general form for the differences is given by:
$$a_{m+1,n}=|a_{m,n+1}-a_{m,n}| tag{1}$$
Therefore, I conjectured the following:
Let $m$ and $n$ denote the row and column number respectively. Therefore, $a_{m,1}=1$ is true $forall mgeq 2$ where $m in mathbb{Z}^+$,
I'd like to know whether this conjecture is true. If so, it would be nice if a proof was provided. Therefore, here I show my thoughts on the problem:
The process on $(1)$ of course should be applied recursively to obtain an expression for the elements on the first row. For example, if i put $a_{4,1}$ in terms of the elements of the first row, we obtain:
$$begin{align} a_{4,1} & =|a_{3,2}-a_{3,1}| \ &=||a_{2,3}-a_{2,2}|-|a_{2,2}-a_{2,1}|| \ &=|||a_{1,4}-a_{1,3}|-|a_{1,3}-a_{1,2}||-||a_{1,3}-a_{1,2}|-|a_{1,2}-a_{1,1}||| end{align}tag{2}$$
This does not appear obvious as to why the the conjecture is true due to the absolute value signs.
Therefore, I decided to abandon using this idea. Somebody had suggested that I use Bertrand's Postulate. I will use the weaker form of the theorem, which states that if $p_n$ is the $n$-th prime, for all $ngeq 1$, then:
$$p_{n+1}<2p_n$$
In that case, I figured that if I apply this on the series with increasing powers of $2$, then:
$$begin{matrix} color{#007777}{1} & & color{#777700}2 & & color{green}{4} & & color{orange}8 && 16 & cdots \ & color{#007777}1 && color{#777700}2 && color{green}4 && color{orange}8 \ && color{#007777}1 && color{#777700}2 && color{green}4 \ &&& color{#007777}1 && color{#777700}2 \ &&&& color{#007777}1 end{matrix}$$
Then each of the columns will have identical values as shown by the different colors above. I figured this was quite similar, and since $p_nleq 2^n$ where $nin mathbb{Z}^+$, thus this would be true for the primes as well. However, it seems doubtful because the differences are not strictly increasing as we can see in the $2^{text{nd}} text{ difference}$ in the prime number series and also, we are taking the absolute value of the differences recursively as shown on the example on $(2)$.
Therefore, I do not think Bertrand's Postulate can be applied directly as I have done.
In summary, I'd like to know whether the conjecture is true. If so, a proof would be nice. Otherwise, if the conjecture is false a form of disproof such as a counterexample would be nice.
number-theory prime-numbers
number-theory prime-numbers
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
edited Mar 18 '17 at 16:30
projectilemotion
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
asked Mar 18 '17 at 15:26
projectilemotionprojectilemotion
11.5k62241
11.5k62241
14
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
3
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
3
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
2
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52
add a comment |
14
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
3
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
3
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
2
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52
14
14
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
3
3
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
3
3
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
2
2
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have re-discovered Gilbreath's conjecture, namely that creating a sequence in which each term $s_n$ is the difference between the consecutive primes $p_{n+1}-p_n$, and then repeating this process for the newly created sequence, and so forth, will always yield sequences that begin with $s^k_0=1$, for the $k^{th}$ sequence.
Andrew Odlyzko verified the conjecture for $k=3.4 times10^{11}$ iterations, however, there are no proofs of this conjecture as of today.
Your calculations regarding the sequence ${2^n}$ have, unfortunately, nothing to do with Gilbreath's conjecture. It is easily shown that the difference between any consecutive terms $s_n$ and $s_{n+1}$ is:
$$
s_{n+1}-s_n=2^{n+1}-2^n=2^n,
$$
yielding the element in the next sequence $q_n=2^n=s_n$.
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
$endgroup$
add a comment |
$begingroup$
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4
1 1 2 2 2 2 2 4 4 4 2 2 2 2 0 4 4 2 2 4 2 2 2 4
0 1 0 0 0 0 2 0 0 2 0 0 0 2 4 0 2 0 2 2 0 0 2
1 1 0 0 0 2 0 0 2 2 0 0 2 2 4 2 2 2 0 2 0 2
0 1 0 0 2 2 0 2 0 2 0 2 0 2 2 0 0 2 2 2 2
1 1 0 2 0 2 2 2 2 2 2 2 2 0 2 0 2 0 0 0
0 1 2 2 2 0 0 0 0 0 0 0 2 2 2 2 2 0 0
1 1 0 0 2 0 0 0 0 0 0 2 0 0 0 0 2 0
0 1 0 2 2 0 0 0 0 0 2 2 0 0 0 2 2
1 1 2 0 2 0 0 0 0 2 0 2 0 0 2 0
0 1 2 2 2 0 0 0 2 2 2 2 0 2 2
1 1 0 0 2 0 0 2 0 0 0 2 2 0
0 1 0 2 2 0 2 2 0 0 2 0 2
1 1 2 0 2 2 0 2 0 2 2 2
0 1 2 2 0 2 2 2 2 0 0
1 1 0 2 2 0 0 0 2 0
0 1 2 0 2 0 0 2 2
1 1 2 2 2 0 2 0
0 1 0 0 2 2 2
1 1 0 2 0 0
0 1 2 2 0
1 1 0 2
0 1 2
1 1
0
103
2
2
2
0
2
0
0
0
0
2
2
0
0
2
0
0
0
2
0
0
0
0
2
0
1
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
$endgroup$
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2192314%2fan-interesting-pattern-in-the-differences-between-prime-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have re-discovered Gilbreath's conjecture, namely that creating a sequence in which each term $s_n$ is the difference between the consecutive primes $p_{n+1}-p_n$, and then repeating this process for the newly created sequence, and so forth, will always yield sequences that begin with $s^k_0=1$, for the $k^{th}$ sequence.
Andrew Odlyzko verified the conjecture for $k=3.4 times10^{11}$ iterations, however, there are no proofs of this conjecture as of today.
Your calculations regarding the sequence ${2^n}$ have, unfortunately, nothing to do with Gilbreath's conjecture. It is easily shown that the difference between any consecutive terms $s_n$ and $s_{n+1}$ is:
$$
s_{n+1}-s_n=2^{n+1}-2^n=2^n,
$$
yielding the element in the next sequence $q_n=2^n=s_n$.
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
$endgroup$
add a comment |
$begingroup$
You have re-discovered Gilbreath's conjecture, namely that creating a sequence in which each term $s_n$ is the difference between the consecutive primes $p_{n+1}-p_n$, and then repeating this process for the newly created sequence, and so forth, will always yield sequences that begin with $s^k_0=1$, for the $k^{th}$ sequence.
Andrew Odlyzko verified the conjecture for $k=3.4 times10^{11}$ iterations, however, there are no proofs of this conjecture as of today.
Your calculations regarding the sequence ${2^n}$ have, unfortunately, nothing to do with Gilbreath's conjecture. It is easily shown that the difference between any consecutive terms $s_n$ and $s_{n+1}$ is:
$$
s_{n+1}-s_n=2^{n+1}-2^n=2^n,
$$
yielding the element in the next sequence $q_n=2^n=s_n$.
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
$endgroup$
add a comment |
$begingroup$
You have re-discovered Gilbreath's conjecture, namely that creating a sequence in which each term $s_n$ is the difference between the consecutive primes $p_{n+1}-p_n$, and then repeating this process for the newly created sequence, and so forth, will always yield sequences that begin with $s^k_0=1$, for the $k^{th}$ sequence.
Andrew Odlyzko verified the conjecture for $k=3.4 times10^{11}$ iterations, however, there are no proofs of this conjecture as of today.
Your calculations regarding the sequence ${2^n}$ have, unfortunately, nothing to do with Gilbreath's conjecture. It is easily shown that the difference between any consecutive terms $s_n$ and $s_{n+1}$ is:
$$
s_{n+1}-s_n=2^{n+1}-2^n=2^n,
$$
yielding the element in the next sequence $q_n=2^n=s_n$.
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
$endgroup$
You have re-discovered Gilbreath's conjecture, namely that creating a sequence in which each term $s_n$ is the difference between the consecutive primes $p_{n+1}-p_n$, and then repeating this process for the newly created sequence, and so forth, will always yield sequences that begin with $s^k_0=1$, for the $k^{th}$ sequence.
Andrew Odlyzko verified the conjecture for $k=3.4 times10^{11}$ iterations, however, there are no proofs of this conjecture as of today.
Your calculations regarding the sequence ${2^n}$ have, unfortunately, nothing to do with Gilbreath's conjecture. It is easily shown that the difference between any consecutive terms $s_n$ and $s_{n+1}$ is:
$$
s_{n+1}-s_n=2^{n+1}-2^n=2^n,
$$
yielding the element in the next sequence $q_n=2^n=s_n$.
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
edited Aug 18 '17 at 13:52
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
answered Aug 18 '17 at 11:31
KlangenKlangen
1,66311334
1,66311334
add a comment |
add a comment |
$begingroup$
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4
1 1 2 2 2 2 2 4 4 4 2 2 2 2 0 4 4 2 2 4 2 2 2 4
0 1 0 0 0 0 2 0 0 2 0 0 0 2 4 0 2 0 2 2 0 0 2
1 1 0 0 0 2 0 0 2 2 0 0 2 2 4 2 2 2 0 2 0 2
0 1 0 0 2 2 0 2 0 2 0 2 0 2 2 0 0 2 2 2 2
1 1 0 2 0 2 2 2 2 2 2 2 2 0 2 0 2 0 0 0
0 1 2 2 2 0 0 0 0 0 0 0 2 2 2 2 2 0 0
1 1 0 0 2 0 0 0 0 0 0 2 0 0 0 0 2 0
0 1 0 2 2 0 0 0 0 0 2 2 0 0 0 2 2
1 1 2 0 2 0 0 0 0 2 0 2 0 0 2 0
0 1 2 2 2 0 0 0 2 2 2 2 0 2 2
1 1 0 0 2 0 0 2 0 0 0 2 2 0
0 1 0 2 2 0 2 2 0 0 2 0 2
1 1 2 0 2 2 0 2 0 2 2 2
0 1 2 2 0 2 2 2 2 0 0
1 1 0 2 2 0 0 0 2 0
0 1 2 0 2 0 0 2 2
1 1 2 2 2 0 2 0
0 1 0 0 2 2 2
1 1 0 2 0 0
0 1 2 2 0
1 1 0 2
0 1 2
1 1
0
103
2
2
2
0
2
0
0
0
0
2
2
0
0
2
0
0
0
2
0
0
0
0
2
0
1
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
$endgroup$
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
|
show 5 more comments
$begingroup$
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4
1 1 2 2 2 2 2 4 4 4 2 2 2 2 0 4 4 2 2 4 2 2 2 4
0 1 0 0 0 0 2 0 0 2 0 0 0 2 4 0 2 0 2 2 0 0 2
1 1 0 0 0 2 0 0 2 2 0 0 2 2 4 2 2 2 0 2 0 2
0 1 0 0 2 2 0 2 0 2 0 2 0 2 2 0 0 2 2 2 2
1 1 0 2 0 2 2 2 2 2 2 2 2 0 2 0 2 0 0 0
0 1 2 2 2 0 0 0 0 0 0 0 2 2 2 2 2 0 0
1 1 0 0 2 0 0 0 0 0 0 2 0 0 0 0 2 0
0 1 0 2 2 0 0 0 0 0 2 2 0 0 0 2 2
1 1 2 0 2 0 0 0 0 2 0 2 0 0 2 0
0 1 2 2 2 0 0 0 2 2 2 2 0 2 2
1 1 0 0 2 0 0 2 0 0 0 2 2 0
0 1 0 2 2 0 2 2 0 0 2 0 2
1 1 2 0 2 2 0 2 0 2 2 2
0 1 2 2 0 2 2 2 2 0 0
1 1 0 2 2 0 0 0 2 0
0 1 2 0 2 0 0 2 2
1 1 2 2 2 0 2 0
0 1 0 0 2 2 2
1 1 0 2 0 0
0 1 2 2 0
1 1 0 2
0 1 2
1 1
0
103
2
2
2
0
2
0
0
0
0
2
2
0
0
2
0
0
0
2
0
0
0
0
2
0
1
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
$endgroup$
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
|
show 5 more comments
$begingroup$
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4
1 1 2 2 2 2 2 4 4 4 2 2 2 2 0 4 4 2 2 4 2 2 2 4
0 1 0 0 0 0 2 0 0 2 0 0 0 2 4 0 2 0 2 2 0 0 2
1 1 0 0 0 2 0 0 2 2 0 0 2 2 4 2 2 2 0 2 0 2
0 1 0 0 2 2 0 2 0 2 0 2 0 2 2 0 0 2 2 2 2
1 1 0 2 0 2 2 2 2 2 2 2 2 0 2 0 2 0 0 0
0 1 2 2 2 0 0 0 0 0 0 0 2 2 2 2 2 0 0
1 1 0 0 2 0 0 0 0 0 0 2 0 0 0 0 2 0
0 1 0 2 2 0 0 0 0 0 2 2 0 0 0 2 2
1 1 2 0 2 0 0 0 0 2 0 2 0 0 2 0
0 1 2 2 2 0 0 0 2 2 2 2 0 2 2
1 1 0 0 2 0 0 2 0 0 0 2 2 0
0 1 0 2 2 0 2 2 0 0 2 0 2
1 1 2 0 2 2 0 2 0 2 2 2
0 1 2 2 0 2 2 2 2 0 0
1 1 0 2 2 0 0 0 2 0
0 1 2 0 2 0 0 2 2
1 1 2 2 2 0 2 0
0 1 0 0 2 2 2
1 1 0 2 0 0
0 1 2 2 0
1 1 0 2
0 1 2
1 1
0
103
2
2
2
0
2
0
0
0
0
2
2
0
0
2
0
0
0
2
0
0
0
0
2
0
1
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
$endgroup$
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4
1 1 2 2 2 2 2 4 4 4 2 2 2 2 0 4 4 2 2 4 2 2 2 4
0 1 0 0 0 0 2 0 0 2 0 0 0 2 4 0 2 0 2 2 0 0 2
1 1 0 0 0 2 0 0 2 2 0 0 2 2 4 2 2 2 0 2 0 2
0 1 0 0 2 2 0 2 0 2 0 2 0 2 2 0 0 2 2 2 2
1 1 0 2 0 2 2 2 2 2 2 2 2 0 2 0 2 0 0 0
0 1 2 2 2 0 0 0 0 0 0 0 2 2 2 2 2 0 0
1 1 0 0 2 0 0 0 0 0 0 2 0 0 0 0 2 0
0 1 0 2 2 0 0 0 0 0 2 2 0 0 0 2 2
1 1 2 0 2 0 0 0 0 2 0 2 0 0 2 0
0 1 2 2 2 0 0 0 2 2 2 2 0 2 2
1 1 0 0 2 0 0 2 0 0 0 2 2 0
0 1 0 2 2 0 2 2 0 0 2 0 2
1 1 2 0 2 2 0 2 0 2 2 2
0 1 2 2 0 2 2 2 2 0 0
1 1 0 2 2 0 0 0 2 0
0 1 2 0 2 0 0 2 2
1 1 2 2 2 0 2 0
0 1 0 0 2 2 2
1 1 0 2 0 0
0 1 2 2 0
1 1 0 2
0 1 2
1 1
0
103
2
2
2
0
2
0
0
0
0
2
2
0
0
2
0
0
0
2
0
0
0
0
2
0
1
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
edited Jan 26 at 10:59
Gottfried Helms
23.6k245101
23.6k245101
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
answered Jan 26 at 3:35
Brian StorieBrian Storie
1
1
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
|
show 5 more comments
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
Go figure A-Z +1
$endgroup$
– Brian Storie
Jan 26 at 3:36
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
01010101... Repeated
$endgroup$
– Brian Storie
Jan 26 at 3:45
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
$begingroup$
... was not too difficult to format for readability...
$endgroup$
– Gottfried Helms
Jan 26 at 11:00
2
2
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
The differences in second row are wrong. You should delete this
$endgroup$
– Gottfried Helms
Jan 26 at 11:18
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
$begingroup$
It's the 2 changed values that that make this work. when adding the columns and subtracting them from the corresponding prime and running the differences again you get 13 prime doing a count down 5,4,3,2,1,0,1,0,1..... Try and find the 2 incorrect values and explain why changing 2 values gives a repeating 1,0,1,0... Pattern.
$endgroup$
– Brian Storie
Feb 19 at 22:24
|
show 5 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2192314%2fan-interesting-pattern-in-the-differences-between-prime-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
14
$begingroup$
See en.wikipedia.org/wiki/Gilbreath%27s_conjecture
$endgroup$
– Barry Cipra
Mar 18 '17 at 15:30
3
$begingroup$
@BarryCipra: Short as it is, I think your comment should be an answer.
$endgroup$
– user21820
Mar 18 '17 at 15:42
3
$begingroup$
If someone knows an heuristic why the conjecture should be true (or not) ?
$endgroup$
– reuns
Mar 18 '17 at 15:55
2
$begingroup$
See there paris8.free.fr/ODLYZKO-gilbreath_conjecture.pdf it seems there is an heuristic applying to many sequences whose first element is a 1, the others even, and where the gaps between consecutive elements are not too large and are sufficiently random
$endgroup$
– reuns
Mar 18 '17 at 16:22
$begingroup$
what happens if you started at 5 instead of 2? Can you get a simple formula for the sequence when you start at 5?
$endgroup$
– user25406
Mar 20 '17 at 23:52