Mixing
For $n ge 20$, there are at least $0.6 cdot frac{2^n}{n}$ primes in $[2^{n-1},2^n - 1]$.
$begingroup$
From the prime number theorem, one can deduce the following inequality:
For $x ge 355991$, if $pi(x) = |{p le x:p text{ is prime}}|$
then we have
$frac{x}{ln(x)}Big(1+frac{1}{ln(x)}Big) < pi(x) <
frac{x}{ln(x)}Big(1+frac{1}{ln(x)}+frac{2.51}{(ln(x))^2}Big)$
Form here I have to deduce that for $n ge 20$ one has that there are at least $0.6 cdot frac{2^n}{n}$ primes in $[2^{n-1},2^n - 1]$.
My try
Observation: $pi(2^n -1) = pi(2^n)$ since $2^n$ is not prime.
So the number of primes in $[2^{n-1},2^n - 1]$ is given by $pi(2^n)-pi(2^{n-1})$.
We have:
$pi(2^{n-1}) <frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) implies$
$-frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < -pi(2^{n-1}) $
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) < pi(2^n)$
adding up these two:
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) - frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < pi(2^n) - pi(2^{n-1})$
At this point I could do various simplifications but I cannot find the right approximation. Any ideas?
abstract-algebra number-theory cryptography
asked Jan 9 at 13:38
JavierJavier
2,01821133
$endgroup$
add a comment |
$begingroup$
From the prime number theorem, one can deduce the following inequality:
For $x ge 355991$, if $pi(x) = |{p le x:p text{ is prime}}|$
then we have
$frac{x}{ln(x)}Big(1+frac{1}{ln(x)}Big) < pi(x) <
frac{x}{ln(x)}Big(1+frac{1}{ln(x)}+frac{2.51}{(ln(x))^2}Big)$
Form here I have to deduce that for $n ge 20$ one has that there are at least $0.6 cdot frac{2^n}{n}$ primes in $[2^{n-1},2^n - 1]$.
My try
Observation: $pi(2^n -1) = pi(2^n)$ since $2^n$ is not prime.
So the number of primes in $[2^{n-1},2^n - 1]$ is given by $pi(2^n)-pi(2^{n-1})$.
We have:
$pi(2^{n-1}) <frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) implies$
$-frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < -pi(2^{n-1}) $
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) < pi(2^n)$
adding up these two:
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) - frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < pi(2^n) - pi(2^{n-1})$
At this point I could do various simplifications but I cannot find the right approximation. Any ideas?
abstract-algebra number-theory cryptography
asked Jan 9 at 13:38
JavierJavier
2,01821133
$endgroup$
$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35
add a comment |
$begingroup$
From the prime number theorem, one can deduce the following inequality:
For $x ge 355991$, if $pi(x) = |{p le x:p text{ is prime}}|$
then we have
$frac{x}{ln(x)}Big(1+frac{1}{ln(x)}Big) < pi(x) <
frac{x}{ln(x)}Big(1+frac{1}{ln(x)}+frac{2.51}{(ln(x))^2}Big)$
Form here I have to deduce that for $n ge 20$ one has that there are at least $0.6 cdot frac{2^n}{n}$ primes in $[2^{n-1},2^n - 1]$.
My try
Observation: $pi(2^n -1) = pi(2^n)$ since $2^n$ is not prime.
So the number of primes in $[2^{n-1},2^n - 1]$ is given by $pi(2^n)-pi(2^{n-1})$.
We have:
$pi(2^{n-1}) <frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) implies$
$-frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < -pi(2^{n-1}) $
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) < pi(2^n)$
adding up these two:
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) - frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < pi(2^n) - pi(2^{n-1})$
At this point I could do various simplifications but I cannot find the right approximation. Any ideas?
abstract-algebra number-theory cryptography
asked Jan 9 at 13:38
JavierJavier
2,01821133
$endgroup$
From the prime number theorem, one can deduce the following inequality:
For $x ge 355991$, if $pi(x) = |{p le x:p text{ is prime}}|$
then we have
$frac{x}{ln(x)}Big(1+frac{1}{ln(x)}Big) < pi(x) <
frac{x}{ln(x)}Big(1+frac{1}{ln(x)}+frac{2.51}{(ln(x))^2}Big)$
Form here I have to deduce that for $n ge 20$ one has that there are at least $0.6 cdot frac{2^n}{n}$ primes in $[2^{n-1},2^n - 1]$.
My try
Observation: $pi(2^n -1) = pi(2^n)$ since $2^n$ is not prime.
So the number of primes in $[2^{n-1},2^n - 1]$ is given by $pi(2^n)-pi(2^{n-1})$.
We have:
$pi(2^{n-1}) <frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) implies$
$-frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < -pi(2^{n-1}) $
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) < pi(2^n)$
adding up these two:
$frac{2^n}{ln(2^n)}Big(1+frac{1}{ln(2^n)}Big) - frac{2^{n-1}}{ln(2^{n-1})}Big(1+frac{1}{ln(2^{n-1})}+frac{2.51}{(ln(2^{n-1}))^2}Big) < pi(2^n) - pi(2^{n-1})$
At this point I could do various simplifications but I cannot find the right approximation. Any ideas?
abstract-algebra number-theory cryptography
abstract-algebra number-theory cryptography
asked Jan 9 at 13:38
JavierJavier
2,01821133
asked Jan 9 at 13:38
JavierJavier
2,01821133
edited Jan 9 at 13:46
Javier
asked Jan 9 at 13:38
JavierJavier
2,01821133
asked Jan 9 at 13:38
JavierJavier
2,01821133
asked Jan 9 at 13:38
JavierJavier
2,01821133
2,01821133
$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35
add a comment |
$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35
$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35
add a comment |
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$begingroup$
In the estimation in the yellow box you have x/lnx in the upper bound, but when you plug in $2^{n-1}$, you write 1/ln($2^{n-1}$).
$endgroup$
– Student7
Jan 9 at 13:43
$begingroup$
@Student7 is fixed now, thank you
$endgroup$
– Javier
Jan 9 at 13:47
$begingroup$
Using $ln(2^n) = n ln(2)$ etc. and factoring out $frac{2^n}{n}$ does not help? Then we have a sum which we have to bound by $0.6$.
$endgroup$
– Stockfish
Jan 9 at 14:15
$begingroup$
@Stockfish not really, this is the kind of simplification i tried
$endgroup$
– Javier
Jan 9 at 21:35