Mixing
Is every integer the sum of distinct prime numbers?
8
$begingroup$
Let $Bbb{P}$ be the set of prime numbers and $Q=Bbb{P}cup{1}$. Is it true that every natural numbers ($neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
elementary-number-theory prime-numbers
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
$endgroup$
add a comment |
8
$begingroup$
Let $Bbb{P}$ be the set of prime numbers and $Q=Bbb{P}cup{1}$. Is it true that every natural numbers ($neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
elementary-number-theory prime-numbers
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
$endgroup$
3
$begingroup$
Heard about Goldbach's Conjecture? Still not proven.
$endgroup$
– SchrodingersCat
Nov 18 '15 at 15:36
$begingroup$
Goldbach's conjecture? No one proved it yet.
$endgroup$
– Atvin
Nov 18 '15 at 15:37
$begingroup$
@whatever see weak Goldbach's conj.
$endgroup$
– martin
Nov 18 '15 at 15:37
1
$begingroup$
If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
$endgroup$
– barak manos
Nov 18 '15 at 15:38
1
$begingroup$
FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
$endgroup$
– John Omielan
2 days ago
add a comment |
8
8
8
4
$begingroup$
Let $Bbb{P}$ be the set of prime numbers and $Q=Bbb{P}cup{1}$. Is it true that every natural numbers ($neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
elementary-number-theory prime-numbers
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
$endgroup$
Let $Bbb{P}$ be the set of prime numbers and $Q=Bbb{P}cup{1}$. Is it true that every natural numbers ($neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
edited Nov 18 '15 at 20:02
Solomonoff's Secret
3,65211233
3,65211233
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
asked Nov 18 '15 at 15:35
BPPBPP
2,166927
2,166927
3
$begingroup$
Heard about Goldbach's Conjecture? Still not proven.
$endgroup$
– SchrodingersCat
Nov 18 '15 at 15:36
$begingroup$
Goldbach's conjecture? No one proved it yet.
$endgroup$
– Atvin
Nov 18 '15 at 15:37
$begingroup$
@whatever see weak Goldbach's conj.
$endgroup$
– martin
Nov 18 '15 at 15:37
1
$begingroup$
If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
$endgroup$
– barak manos
Nov 18 '15 at 15:38
1
$begingroup$
FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
$endgroup$
– John Omielan
2 days ago
add a comment |
3
$begingroup$
Heard about Goldbach's Conjecture? Still not proven.
$endgroup$
– SchrodingersCat
Nov 18 '15 at 15:36
$begingroup$
Goldbach's conjecture? No one proved it yet.
$endgroup$
– Atvin
Nov 18 '15 at 15:37
$begingroup$
@whatever see weak Goldbach's conj.
$endgroup$
– martin
Nov 18 '15 at 15:37
1
$begingroup$
If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
$endgroup$
– barak manos
Nov 18 '15 at 15:38
1
$begingroup$
FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
$endgroup$
– John Omielan
2 days ago
3
3
$begingroup$
Heard about Goldbach's Conjecture? Still not proven.
$endgroup$
– SchrodingersCat
Nov 18 '15 at 15:36
$begingroup$
Heard about Goldbach's Conjecture? Still not proven.
$endgroup$
– SchrodingersCat
Nov 18 '15 at 15:36
$begingroup$
Goldbach's conjecture? No one proved it yet.
$endgroup$
– Atvin
Nov 18 '15 at 15:37
$begingroup$
Goldbach's conjecture? No one proved it yet.
$endgroup$
– Atvin
Nov 18 '15 at 15:37
$begingroup$
@whatever see weak Goldbach's conj.
$endgroup$
– martin
Nov 18 '15 at 15:37
$begingroup$
@whatever see weak Goldbach's conj.
$endgroup$
– martin
Nov 18 '15 at 15:37
1
1
$begingroup$
If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
$endgroup$
– barak manos
Nov 18 '15 at 15:38
$begingroup$
If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
$endgroup$
– barak manos
Nov 18 '15 at 15:38
1
1
$begingroup$
FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
$endgroup$
– John Omielan
2 days ago
$begingroup$
FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
$endgroup$
– John Omielan
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
27
$begingroup$
By Bertrand's postulate, you can find a prime satisfying $lfloor n/2rfloor <p< n$. Proceed by induction.
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
|
show 5 more comments
1
$begingroup$
Note that we can not write $2$ in required form.
Take any integer (say) $x$ with $3le x.$
Let $y$ be the largest prime strictly less than $x.$ If $x-yin Q$ we are already done.
Suppose $x-ynotin Q.$ Then, take the largest prime strictly less than $x-y.$
And continue this..
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
27
$begingroup$
By Bertrand's postulate, you can find a prime satisfying $lfloor n/2rfloor <p< n$. Proceed by induction.
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
|
show 5 more comments
27
$begingroup$
By Bertrand's postulate, you can find a prime satisfying $lfloor n/2rfloor <p< n$. Proceed by induction.
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
|
show 5 more comments
27
27
27
$begingroup$
By Bertrand's postulate, you can find a prime satisfying $lfloor n/2rfloor <p< n$. Proceed by induction.
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
By Bertrand's postulate, you can find a prime satisfying $lfloor n/2rfloor <p< n$. Proceed by induction.
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
answered Nov 18 '15 at 15:46
Thomas AndrewsThomas Andrews
131k12147298
131k12147298
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
|
show 5 more comments
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
1
1
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
$begingroup$
So, for example, if you want to solve $n=64$, find a prime between $32$ and $64$, say, $p_1=37$. Then solve for $64-37=27$ - take $p_2=17$ between $27/2$ and $27$. Then solve for $27-17=10$ with $p_3=7$ between $10/2$ and $10$. Now you are down to $10-7=3$ which is prime, so you are done. So $64=3+7+17+37$. (We could have obviously gone faster by picking bigger primes at each step.)
$endgroup$
– Thomas Andrews
Nov 18 '15 at 15:55
2
2
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
$begingroup$
Sure, I left out a big step - which is the initial range of the induction. You have to treat the small values on a case-by-case basis. Not meant to be a complete answer. @RamirodelaVega
$endgroup$
– Thomas Andrews
Nov 18 '15 at 16:00
2
2
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
$begingroup$
$2,3,11$ are the sum of sets of one prime. $6=5+1$ - OP specifically includes $1$ in the set, and $11=7+3+1$. @gnasher729
$endgroup$
– Thomas Andrews
Nov 18 '15 at 21:00
1
1
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
$begingroup$
And the linked Wikipedia article even mentions this explicitly in the subsection Consequences.
$endgroup$
– Jeppe Stig Nielsen
Nov 18 '15 at 21:36
1
1
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
$begingroup$
Would you consider integrating the content of your comments into your answer? Right now someone reading this answer needs to sift through the comments to find all the useful information. Thanks.
$endgroup$
– Najib Idrissi
Dec 1 '15 at 8:58
|
show 5 more comments
1
$begingroup$
Note that we can not write $2$ in required form.
Take any integer (say) $x$ with $3le x.$
Let $y$ be the largest prime strictly less than $x.$ If $x-yin Q$ we are already done.
Suppose $x-ynotin Q.$ Then, take the largest prime strictly less than $x-y.$
And continue this..
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
add a comment |
1
$begingroup$
Note that we can not write $2$ in required form.
Take any integer (say) $x$ with $3le x.$
Let $y$ be the largest prime strictly less than $x.$ If $x-yin Q$ we are already done.
Suppose $x-ynotin Q.$ Then, take the largest prime strictly less than $x-y.$
And continue this..
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
add a comment |
1
1
1
$begingroup$
Note that we can not write $2$ in required form.
Take any integer (say) $x$ with $3le x.$
Let $y$ be the largest prime strictly less than $x.$ If $x-yin Q$ we are already done.
Suppose $x-ynotin Q.$ Then, take the largest prime strictly less than $x-y.$
And continue this..
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
$endgroup$
Note that we can not write $2$ in required form.
Take any integer (say) $x$ with $3le x.$
Let $y$ be the largest prime strictly less than $x.$ If $x-yin Q$ we are already done.
Suppose $x-ynotin Q.$ Then, take the largest prime strictly less than $x-y.$
And continue this..
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
answered Dec 1 '15 at 8:57
BumblebeeBumblebee
9,74912551
9,74912551
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
add a comment |
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
$begingroup$
After I post my answer, I feel that, It would be easy to use strong induction than continue above process.
$endgroup$
– Bumblebee
Dec 1 '15 at 8:58
add a comment |
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3
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Heard about Goldbach's Conjecture? Still not proven.
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– SchrodingersCat
Nov 18 '15 at 15:36
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Goldbach's conjecture? No one proved it yet.
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– Atvin
Nov 18 '15 at 15:37
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@whatever see weak Goldbach's conj.
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– martin
Nov 18 '15 at 15:37
1
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If every even integer can be expressed as a sum of $2$ prime numbers, then every integer can be expressed as a sum of at most $3$ prime numbers. The "if" part of this statement has yet to be proven.
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– barak manos
Nov 18 '15 at 15:38
1
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FYI, what you are asking about is an example of a Complete sequence. The linked Wikipedia article provides a definition, conditions, several examples (starting with the primes & 1), and applications where this may be used.
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– John Omielan
2 days ago