Prime factorization formalism
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I was hoping someone might know the name for this representation of the positive integers. The idea is similar to Peano's arithmetic but applied to multiplication. So we have $1$, multiplication and a function $p$. We map onto the normal integers by having p map to the nth prime. To illustrate the first few integers would be $1, p(1), p(p(1)), p(1)p(1), p(p(p(1)))$.
This representation has some unusual properties. Computing whether a number is prime is trivial, as is factorization and multiplication.
Addition however is rather tricky since you have to convert back to a normal representation, perform the addition then recursively factorized the result!
prime-numbers prime-factorization
edited Nov 23 at 11:34
amWhy
191k27223439
asked Sep 9 at 22:31
Mumrah
11
add a comment |
up vote
0
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favorite
I was hoping someone might know the name for this representation of the positive integers. The idea is similar to Peano's arithmetic but applied to multiplication. So we have $1$, multiplication and a function $p$. We map onto the normal integers by having p map to the nth prime. To illustrate the first few integers would be $1, p(1), p(p(1)), p(1)p(1), p(p(p(1)))$.
This representation has some unusual properties. Computing whether a number is prime is trivial, as is factorization and multiplication.
Addition however is rather tricky since you have to convert back to a normal representation, perform the addition then recursively factorized the result!
prime-numbers prime-factorization
edited Nov 23 at 11:34
amWhy
191k27223439
asked Sep 9 at 22:31
Mumrah
11
Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was hoping someone might know the name for this representation of the positive integers. The idea is similar to Peano's arithmetic but applied to multiplication. So we have $1$, multiplication and a function $p$. We map onto the normal integers by having p map to the nth prime. To illustrate the first few integers would be $1, p(1), p(p(1)), p(1)p(1), p(p(p(1)))$.
This representation has some unusual properties. Computing whether a number is prime is trivial, as is factorization and multiplication.
Addition however is rather tricky since you have to convert back to a normal representation, perform the addition then recursively factorized the result!
prime-numbers prime-factorization
edited Nov 23 at 11:34
amWhy
191k27223439
asked Sep 9 at 22:31
Mumrah
11
I was hoping someone might know the name for this representation of the positive integers. The idea is similar to Peano's arithmetic but applied to multiplication. So we have $1$, multiplication and a function $p$. We map onto the normal integers by having p map to the nth prime. To illustrate the first few integers would be $1, p(1), p(p(1)), p(1)p(1), p(p(p(1)))$.
This representation has some unusual properties. Computing whether a number is prime is trivial, as is factorization and multiplication.
Addition however is rather tricky since you have to convert back to a normal representation, perform the addition then recursively factorized the result!
prime-numbers prime-factorization
prime-numbers prime-factorization
edited Nov 23 at 11:34
amWhy
191k27223439
asked Sep 9 at 22:31
Mumrah
11
edited Nov 23 at 11:34
amWhy
191k27223439
asked Sep 9 at 22:31
Mumrah
11
edited Nov 23 at 11:34
amWhy
191k27223439
edited Nov 23 at 11:34
amWhy
191k27223439
edited Nov 23 at 11:34
amWhy
191k27223439
191k27223439
asked Sep 9 at 22:31
Mumrah
11
asked Sep 9 at 22:31
Mumrah
11
asked Sep 9 at 22:31
Mumrah
11
11
Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00
add a comment |
Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00
Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00
add a comment |
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Do you have a finite length recipe (i.e. without pre-specifying the value of every prime in a long list) for performing addition in this setting? Even an indirect description of the form '+ is the only binary operation with this this and this property' leaving all the hard work to the reader would be fine. If such a thing exists, that is. I am not so sure here...
– Vincent
Nov 23 at 12:52
I don't have any such mechanism no. You can show its related to a hard problem. Suppose we could efficiently (in polynomial time) compute addition and divide-and-remainder in this representation. It is trivial to convert a binary number into this form since 2^n is simply p(1)^n and we can therefore use addition to covert from binary into this form efficiently. But now we can use repeatedly perform divide and remainder by 2 on the factorized form to compute the binary representation of the prime components. This would allow us to extract the prime factors.
– Mumrah
Nov 27 at 16:00