Yet another question on sums of the reciprocals of the primes
$begingroup$
I recall reading once that the sum $$sum_{p ,, small{mbox{is a known prime}}} frac{1}{p}$$
is less than $4$.
Does anybody here know what the ultimate source of this claim is?
Please, let me thank you in advance for your insightful replies...
nt.number-theory reference-request analytic-number-theory computational-number-theory
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
$endgroup$
add a comment |
$begingroup$
I recall reading once that the sum $$sum_{p ,, small{mbox{is a known prime}}} frac{1}{p}$$
is less than $4$.
Does anybody here know what the ultimate source of this claim is?
Please, let me thank you in advance for your insightful replies...
nt.number-theory reference-request analytic-number-theory computational-number-theory
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
$endgroup$
$begingroup$
How do you define a "known" prime?
$endgroup$
– Wojowu
6 hours ago
$begingroup$
I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
$endgroup$
– José Hdz. Stgo.
5 hours ago
4
$begingroup$
Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
$endgroup$
– Wojowu
5 hours ago
$begingroup$
My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
$endgroup$
– Gerry Myerson
1 hour ago
add a comment |
$begingroup$
I recall reading once that the sum $$sum_{p ,, small{mbox{is a known prime}}} frac{1}{p}$$
is less than $4$.
Does anybody here know what the ultimate source of this claim is?
Please, let me thank you in advance for your insightful replies...
nt.number-theory reference-request analytic-number-theory computational-number-theory
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
$endgroup$
I recall reading once that the sum $$sum_{p ,, small{mbox{is a known prime}}} frac{1}{p}$$
is less than $4$.
Does anybody here know what the ultimate source of this claim is?
Please, let me thank you in advance for your insightful replies...
nt.number-theory reference-request analytic-number-theory computational-number-theory
nt.number-theory reference-request analytic-number-theory computational-number-theory
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
edited 6 hours ago
José Hdz. Stgo.
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
asked 6 hours ago
José Hdz. Stgo.José Hdz. Stgo.
5,23734877
5,23734877
$begingroup$
How do you define a "known" prime?
$endgroup$
– Wojowu
6 hours ago
$begingroup$
I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
$endgroup$
– José Hdz. Stgo.
5 hours ago
4
$begingroup$
Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
$endgroup$
– Wojowu
5 hours ago
$begingroup$
My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
$endgroup$
– Gerry Myerson
1 hour ago
add a comment |
$begingroup$
How do you define a "known" prime?
$endgroup$
– Wojowu
6 hours ago
$begingroup$
I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
$endgroup$
– José Hdz. Stgo.
5 hours ago
4
$begingroup$
Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
$endgroup$
– Wojowu
5 hours ago
$begingroup$
My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
$endgroup$
– Gerry Myerson
1 hour ago
$begingroup$
How do you define a "known" prime?
$endgroup$
– Wojowu
6 hours ago
$begingroup$
How do you define a "known" prime?
$endgroup$
– Wojowu
6 hours ago
$begingroup$
I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
$endgroup$
– José Hdz. Stgo.
5 hours ago
$begingroup$
I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
$endgroup$
– José Hdz. Stgo.
5 hours ago
4
4
$begingroup$
Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
$endgroup$
– Wojowu
5 hours ago
$begingroup$
Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
$endgroup$
– Wojowu
5 hours ago
$begingroup$
My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
$endgroup$
– Gerry Myerson
1 hour ago
$begingroup$
I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
$endgroup$
– Gerry Myerson
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's well known that the sum of the reciprocals of the primes below $n$ tends to $log log n + M$, where $M$ is a small constant (the Meissel-Mertens constant). That is to say:
$$ sumlimits_{small{mbox{prime}} , p , < , n} frac{1}{p} = log log n + M + o(1) $$
This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $4$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:
$$ e^{e^{4 - M}} approxeq 1.80 times 10^{18}$$
assuming the $o(1)$ term can be neglected.
Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $pm 1 mod 6$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $10^{16}$, and computing power has advanced considerably since then.
To give a comparison, the first SHA1 collision involved $9.2 times 10^{18}$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $1.80 times 10^{18}$.
edited 4 hours ago
José Hdz. Stgo.
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
$endgroup$
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's well known that the sum of the reciprocals of the primes below $n$ tends to $log log n + M$, where $M$ is a small constant (the Meissel-Mertens constant). That is to say:
$$ sumlimits_{small{mbox{prime}} , p , < , n} frac{1}{p} = log log n + M + o(1) $$
This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $4$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:
$$ e^{e^{4 - M}} approxeq 1.80 times 10^{18}$$
assuming the $o(1)$ term can be neglected.
Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $pm 1 mod 6$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $10^{16}$, and computing power has advanced considerably since then.
To give a comparison, the first SHA1 collision involved $9.2 times 10^{18}$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $1.80 times 10^{18}$.
edited 4 hours ago
José Hdz. Stgo.
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
$endgroup$
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
add a comment |
$begingroup$
It's well known that the sum of the reciprocals of the primes below $n$ tends to $log log n + M$, where $M$ is a small constant (the Meissel-Mertens constant). That is to say:
$$ sumlimits_{small{mbox{prime}} , p , < , n} frac{1}{p} = log log n + M + o(1) $$
This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $4$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:
$$ e^{e^{4 - M}} approxeq 1.80 times 10^{18}$$
assuming the $o(1)$ term can be neglected.
Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $pm 1 mod 6$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $10^{16}$, and computing power has advanced considerably since then.
To give a comparison, the first SHA1 collision involved $9.2 times 10^{18}$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $1.80 times 10^{18}$.
edited 4 hours ago
José Hdz. Stgo.
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
$endgroup$
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
add a comment |
$begingroup$
It's well known that the sum of the reciprocals of the primes below $n$ tends to $log log n + M$, where $M$ is a small constant (the Meissel-Mertens constant). That is to say:
$$ sumlimits_{small{mbox{prime}} , p , < , n} frac{1}{p} = log log n + M + o(1) $$
This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $4$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:
$$ e^{e^{4 - M}} approxeq 1.80 times 10^{18}$$
assuming the $o(1)$ term can be neglected.
Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $pm 1 mod 6$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $10^{16}$, and computing power has advanced considerably since then.
To give a comparison, the first SHA1 collision involved $9.2 times 10^{18}$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $1.80 times 10^{18}$.
edited 4 hours ago
José Hdz. Stgo.
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
$endgroup$
It's well known that the sum of the reciprocals of the primes below $n$ tends to $log log n + M$, where $M$ is a small constant (the Meissel-Mertens constant). That is to say:
$$ sumlimits_{small{mbox{prime}} , p , < , n} frac{1}{p} = log log n + M + o(1) $$
This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $4$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:
$$ e^{e^{4 - M}} approxeq 1.80 times 10^{18}$$
assuming the $o(1)$ term can be neglected.
Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $pm 1 mod 6$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $10^{16}$, and computing power has advanced considerably since then.
To give a comparison, the first SHA1 collision involved $9.2 times 10^{18}$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $1.80 times 10^{18}$.
edited 4 hours ago
José Hdz. Stgo.
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
edited 4 hours ago
José Hdz. Stgo.
5,23734877
edited 4 hours ago
José Hdz. Stgo.
5,23734877
edited 4 hours ago
José Hdz. Stgo.
5,23734877
5,23734877
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
answered 4 hours ago
Adam P. GoucherAdam P. Goucher
6,72022958
6,72022958
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
add a comment |
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$.
$endgroup$
– Wojowu
4 hours ago
1
1
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
$begingroup$
Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about.
$endgroup$
– Wojowu
4 hours ago
1
1
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
$begingroup$
See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4cdot 10^{18}$.
$endgroup$
– Eleanor
3 hours ago
add a comment |
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How do you define a "known" prime?
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– Wojowu
6 hours ago
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I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 , 589 , 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 , 112 , 609}-1,2^{57 , 885 , 161}-1)$, as far as I know there are no "known primes" in that range.
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– José Hdz. Stgo.
5 hours ago
4
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Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result.
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– Wojowu
5 hours ago
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My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes.
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– Wojowu
4 hours ago
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I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so.
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– Gerry Myerson
1 hour ago