Mixing
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Reciprocity of different prime numbers can approximate $1$?
Clash Royale CLAN TAG#URR8PPP
up vote
4
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I want to see if there exist $p_1<p_2<p_3<cdots<p_1000$ different prime numbers such that
$|1/p_1+cdots+1/p_1000-1|le (1over p_1000)^2.$
a) what is my point with this? Nothing. But what is the point of the twin number conjecture? Maybe I can prove that such problem has always a solution, even for more terms.
b) what is my progress? We cannot have $|1/p_1+cdots+1/p_1000-1|=0$ (which I proved), so there must be an error. Also, $|1/2+1/5+1/7+1/11+1/13-1|=0.0107<1/13$, so for five terms there is a solution with power 1, but 1/13^2 cannot be used here. In fact, I proved that the problem has a solution, if the expression on the right is simply $(1over p_1000),$ that is, the second power is just first power. (We can have $k$ terms, in general, too, not just 1000 terms.)
I wish to have second power (or anything bigger than 1 is great as long as we tend to infinity with the number of terms), because of Hurwitz theorem
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)
Remark: the sum of repciprocal of all primes is infinity, that looks good/promising in order not to get a contradiction.
number-theory prime-numbers harmonic-numbers
asked Jul 30 at 22:07
Miranda
543
add a comment |Â
up vote
4
down vote
favorite
I want to see if there exist $p_1<p_2<p_3<cdots<p_1000$ different prime numbers such that
$|1/p_1+cdots+1/p_1000-1|le (1over p_1000)^2.$
a) what is my point with this? Nothing. But what is the point of the twin number conjecture? Maybe I can prove that such problem has always a solution, even for more terms.
b) what is my progress? We cannot have $|1/p_1+cdots+1/p_1000-1|=0$ (which I proved), so there must be an error. Also, $|1/2+1/5+1/7+1/11+1/13-1|=0.0107<1/13$, so for five terms there is a solution with power 1, but 1/13^2 cannot be used here. In fact, I proved that the problem has a solution, if the expression on the right is simply $(1over p_1000),$ that is, the second power is just first power. (We can have $k$ terms, in general, too, not just 1000 terms.)
I wish to have second power (or anything bigger than 1 is great as long as we tend to infinity with the number of terms), because of Hurwitz theorem
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)
Remark: the sum of repciprocal of all primes is infinity, that looks good/promising in order not to get a contradiction.
number-theory prime-numbers harmonic-numbers
asked Jul 30 at 22:07
Miranda
543
5
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I want to see if there exist $p_1<p_2<p_3<cdots<p_1000$ different prime numbers such that
$|1/p_1+cdots+1/p_1000-1|le (1over p_1000)^2.$
a) what is my point with this? Nothing. But what is the point of the twin number conjecture? Maybe I can prove that such problem has always a solution, even for more terms.
b) what is my progress? We cannot have $|1/p_1+cdots+1/p_1000-1|=0$ (which I proved), so there must be an error. Also, $|1/2+1/5+1/7+1/11+1/13-1|=0.0107<1/13$, so for five terms there is a solution with power 1, but 1/13^2 cannot be used here. In fact, I proved that the problem has a solution, if the expression on the right is simply $(1over p_1000),$ that is, the second power is just first power. (We can have $k$ terms, in general, too, not just 1000 terms.)
I wish to have second power (or anything bigger than 1 is great as long as we tend to infinity with the number of terms), because of Hurwitz theorem
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)
Remark: the sum of repciprocal of all primes is infinity, that looks good/promising in order not to get a contradiction.
number-theory prime-numbers harmonic-numbers
asked Jul 30 at 22:07
Miranda
543
I want to see if there exist $p_1<p_2<p_3<cdots<p_1000$ different prime numbers such that
$|1/p_1+cdots+1/p_1000-1|le (1over p_1000)^2.$
a) what is my point with this? Nothing. But what is the point of the twin number conjecture? Maybe I can prove that such problem has always a solution, even for more terms.
b) what is my progress? We cannot have $|1/p_1+cdots+1/p_1000-1|=0$ (which I proved), so there must be an error. Also, $|1/2+1/5+1/7+1/11+1/13-1|=0.0107<1/13$, so for five terms there is a solution with power 1, but 1/13^2 cannot be used here. In fact, I proved that the problem has a solution, if the expression on the right is simply $(1over p_1000),$ that is, the second power is just first power. (We can have $k$ terms, in general, too, not just 1000 terms.)
I wish to have second power (or anything bigger than 1 is great as long as we tend to infinity with the number of terms), because of Hurwitz theorem
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)
Remark: the sum of repciprocal of all primes is infinity, that looks good/promising in order not to get a contradiction.
number-theory prime-numbers harmonic-numbers
asked Jul 30 at 22:07
Miranda
543
edited Jul 30 at 23:02
asked Jul 30 at 22:07
Miranda
543
asked Jul 30 at 22:07
Miranda
543
asked Jul 30 at 22:07
Miranda
543
543
5
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21
add a comment |Â
5
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21
5
5
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :
? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1]," ",vecsort(s)==s," ",Set(s)==s," ",length(s)," ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1 1 1 1000 1
?
The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.
answered Aug 1 at 17:28
Peter
44.9k938119
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :
? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1]," ",vecsort(s)==s," ",Set(s)==s," ",length(s)," ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1 1 1 1000 1
?
The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.
answered Aug 1 at 17:28
Peter
44.9k938119
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
add a comment |Â
up vote
0
down vote
Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :
? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1]," ",vecsort(s)==s," ",Set(s)==s," ",length(s)," ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1 1 1 1000 1
?
The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.
answered Aug 1 at 17:28
Peter
44.9k938119
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :
? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1]," ",vecsort(s)==s," ",Set(s)==s," ",length(s)," ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1 1 1 1000 1
?
The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.
answered Aug 1 at 17:28
Peter
44.9k938119
Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :
? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1]," ",vecsort(s)==s," ",Set(s)==s," ",length(s)," ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1 1 1 1000 1
?
The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.
answered Aug 1 at 17:28
Peter
44.9k938119
answered Aug 1 at 17:28
Peter
44.9k938119
answered Aug 1 at 17:28
Peter
44.9k938119
answered Aug 1 at 17:28
Peter
44.9k938119
44.9k938119
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
add a comment |Â
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
Ingenious! Is it also working for 100,000 terms?
– Dave
Aug 4 at 20:05
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
@Dave Not clear whether this works for every number of primes. I did not find this sequence with a particular method, I just played around and finally succeeded.
– Peter
Aug 6 at 7:40
add a comment |Â
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5
What is my point with this? Nothing. But what is the point of the twin number conjecture? There must be some award somewhere that this quote should win easily.
– Arnaud Mortier
Jul 30 at 22:16
Also $|1/2+1/5+1/7+1/11+1/13-1|=1.0107$ should be $|1/2+1/5+1/7+1/11+1/13-1|colorred< colorred0.0107$.
– Arnaud Mortier
Jul 30 at 22:21