Arbitrarily large sequence of numbers with a property
$begingroup$
We say that a positive integer is good if it has an even number of prime factors. Does there exist an arbitrarily large sequence of consecutive good numbers?
In fact, this problem came from another one: Show that there exists an infinite set $S$ of positive integers such that the sum of any two distinct elements of S has an even number of distinct prime factors.
If the first statement about long sequences of good numbers is true, then I can finish the second problem as follows: Choose $s_{1}in mathbb{N}$ randomly. Then, we will define the increasing sequence $s_{n}$ inductively. Take a sequence of $s_{n-1}+1$ consecutive good numbers. If $a$ is the smallest one in this sequence, take $s_{n}=a$. Then set $S={s_{n}}^{infty}_{n=1}$.
combinatorics number-theory prime-numbers prime-factorization
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
$endgroup$
|
show 11 more comments
$begingroup$
We say that a positive integer is good if it has an even number of prime factors. Does there exist an arbitrarily large sequence of consecutive good numbers?
In fact, this problem came from another one: Show that there exists an infinite set $S$ of positive integers such that the sum of any two distinct elements of S has an even number of distinct prime factors.
If the first statement about long sequences of good numbers is true, then I can finish the second problem as follows: Choose $s_{1}in mathbb{N}$ randomly. Then, we will define the increasing sequence $s_{n}$ inductively. Take a sequence of $s_{n-1}+1$ consecutive good numbers. If $a$ is the smallest one in this sequence, take $s_{n}=a$. Then set $S={s_{n}}^{infty}_{n=1}$.
combinatorics number-theory prime-numbers prime-factorization
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
$endgroup$
1
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
1
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
1
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
1
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
1
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29
|
show 11 more comments
$begingroup$
We say that a positive integer is good if it has an even number of prime factors. Does there exist an arbitrarily large sequence of consecutive good numbers?
In fact, this problem came from another one: Show that there exists an infinite set $S$ of positive integers such that the sum of any two distinct elements of S has an even number of distinct prime factors.
If the first statement about long sequences of good numbers is true, then I can finish the second problem as follows: Choose $s_{1}in mathbb{N}$ randomly. Then, we will define the increasing sequence $s_{n}$ inductively. Take a sequence of $s_{n-1}+1$ consecutive good numbers. If $a$ is the smallest one in this sequence, take $s_{n}=a$. Then set $S={s_{n}}^{infty}_{n=1}$.
combinatorics number-theory prime-numbers prime-factorization
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
$endgroup$
We say that a positive integer is good if it has an even number of prime factors. Does there exist an arbitrarily large sequence of consecutive good numbers?
In fact, this problem came from another one: Show that there exists an infinite set $S$ of positive integers such that the sum of any two distinct elements of S has an even number of distinct prime factors.
If the first statement about long sequences of good numbers is true, then I can finish the second problem as follows: Choose $s_{1}in mathbb{N}$ randomly. Then, we will define the increasing sequence $s_{n}$ inductively. Take a sequence of $s_{n-1}+1$ consecutive good numbers. If $a$ is the smallest one in this sequence, take $s_{n}=a$. Then set $S={s_{n}}^{infty}_{n=1}$.
combinatorics number-theory prime-numbers prime-factorization
combinatorics number-theory prime-numbers prime-factorization
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
edited Jan 26 at 14:46
Alex Ravsky
43.6k32584
43.6k32584
asked Jan 15 at 18:02
alexp9alexp9
459314
asked Jan 15 at 18:02
alexp9alexp9
459314
asked Jan 15 at 18:02
alexp9alexp9
459314
459314
1
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
1
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
1
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
1
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
1
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29
|
show 11 more comments
1
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
1
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
1
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
1
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
1
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29
1
1
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
1
1
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
1
1
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
1
1
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
1
1
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29
|
show 11 more comments
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1
$begingroup$
Is 12 good? (That is, are you taking into account multiplicity of prime factors?)
$endgroup$
– Michael Lugo
Jan 15 at 18:10
1
$begingroup$
Heuristically, which numbers are "good" is essentially coin-flipping, and we should expect arbitrarily long runs of heads in a sequence of coin flips. But these sort of things are notoriously hard to prove.
$endgroup$
– Michael Lugo
Jan 15 at 18:19
1
$begingroup$
Peter's tests are roughly in line with my coin-flipping heuristic (that is, that we should have a run of length $n$ by around $2^n$).
$endgroup$
– Michael Lugo
Jan 15 at 18:47
1
$begingroup$
I just want to solve the original problem. If this strstegy is hard to proof, I will find another one.
$endgroup$
– alexp9
Jan 15 at 18:48
1
$begingroup$
The original question has a nice non-constructive proof using Ramsey's theorem, see math.stackexchange.com/questions/7527/…
$endgroup$
– Dap
Jan 18 at 11:29