Is $small sumlimits_{pleq x}pileft(frac{x^2}{p}right)-sumlimits_{pgt x}pileft(frac{x^2}{p}right)=pi^2(x)$ of...
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I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.
$$sumlimits_{pleq x}pileft(frac{x^2}{p}right)-sumlimits_{pgt x}pileft(frac{x^2}{p}right)=pi^2(x)$$
There is a similar formula I already explored in an other context (Goldbach).
$$sumlimits_{pleq x}pi(small 2x-p normalsize)-sumlimits_{pgt x}pi(small 2x-p normalsize)=pi^2(x)$$
Any paper covering one of them? Were they used anywhere?
Edit: this seems to work also for
$$sumlimits_{pleq x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)-sumlimits_{pgt x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$ with $kge 1$
and for some $kgt gamma_x$ we even have
$$sumlimits_{p}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$
To illustrate the influence of $k$, here is a chart with $k=1$ in blue, $k=2$ in purple and $k=16$ in yellow. $x=9$ in this example:
wrapping arround $pi^2(x)$
I guess there are a lot of other working formulas...
prime-numbers
asked Jan 13 at 20:34
Collag3nCollag3n
769211
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add a comment |
$begingroup$
I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.
$$sumlimits_{pleq x}pileft(frac{x^2}{p}right)-sumlimits_{pgt x}pileft(frac{x^2}{p}right)=pi^2(x)$$
There is a similar formula I already explored in an other context (Goldbach).
$$sumlimits_{pleq x}pi(small 2x-p normalsize)-sumlimits_{pgt x}pi(small 2x-p normalsize)=pi^2(x)$$
Any paper covering one of them? Were they used anywhere?
Edit: this seems to work also for
$$sumlimits_{pleq x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)-sumlimits_{pgt x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$ with $kge 1$
and for some $kgt gamma_x$ we even have
$$sumlimits_{p}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$
To illustrate the influence of $k$, here is a chart with $k=1$ in blue, $k=2$ in purple and $k=16$ in yellow. $x=9$ in this example:
wrapping arround $pi^2(x)$
I guess there are a lot of other working formulas...
prime-numbers
asked Jan 13 at 20:34
Collag3nCollag3n
769211
$endgroup$
1
$begingroup$
No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
$endgroup$
– reuns
Jan 14 at 16:22
$begingroup$
That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04
add a comment |
$begingroup$
I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.
$$sumlimits_{pleq x}pileft(frac{x^2}{p}right)-sumlimits_{pgt x}pileft(frac{x^2}{p}right)=pi^2(x)$$
There is a similar formula I already explored in an other context (Goldbach).
$$sumlimits_{pleq x}pi(small 2x-p normalsize)-sumlimits_{pgt x}pi(small 2x-p normalsize)=pi^2(x)$$
Any paper covering one of them? Were they used anywhere?
Edit: this seems to work also for
$$sumlimits_{pleq x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)-sumlimits_{pgt x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$ with $kge 1$
and for some $kgt gamma_x$ we even have
$$sumlimits_{p}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$
To illustrate the influence of $k$, here is a chart with $k=1$ in blue, $k=2$ in purple and $k=16$ in yellow. $x=9$ in this example:
wrapping arround $pi^2(x)$
I guess there are a lot of other working formulas...
prime-numbers
asked Jan 13 at 20:34
Collag3nCollag3n
769211
$endgroup$
I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.
$$sumlimits_{pleq x}pileft(frac{x^2}{p}right)-sumlimits_{pgt x}pileft(frac{x^2}{p}right)=pi^2(x)$$
There is a similar formula I already explored in an other context (Goldbach).
$$sumlimits_{pleq x}pi(small 2x-p normalsize)-sumlimits_{pgt x}pi(small 2x-p normalsize)=pi^2(x)$$
Any paper covering one of them? Were they used anywhere?
Edit: this seems to work also for
$$sumlimits_{pleq x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)-sumlimits_{pgt x}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$ with $kge 1$
and for some $kgt gamma_x$ we even have
$$sumlimits_{p}pileft(smallsqrt[k]{2 x^k-p^k}normalsizeright)=pi^2(x)$$
To illustrate the influence of $k$, here is a chart with $k=1$ in blue, $k=2$ in purple and $k=16$ in yellow. $x=9$ in this example:
wrapping arround $pi^2(x)$
I guess there are a lot of other working formulas...
prime-numbers
prime-numbers
asked Jan 13 at 20:34
Collag3nCollag3n
769211
asked Jan 13 at 20:34
Collag3nCollag3n
769211
edited Jan 14 at 14:32
Collag3n
asked Jan 13 at 20:34
Collag3nCollag3n
769211
asked Jan 13 at 20:34
Collag3nCollag3n
769211
asked Jan 13 at 20:34
Collag3nCollag3n
769211
769211
1
$begingroup$
No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
$endgroup$
– reuns
Jan 14 at 16:22
$begingroup$
That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04
add a comment |
1
$begingroup$
No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
$endgroup$
– reuns
Jan 14 at 16:22
$begingroup$
That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04
1
1
$begingroup$
No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
$endgroup$
– reuns
Jan 14 at 16:22
$begingroup$
No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
$endgroup$
– reuns
Jan 14 at 16:22
$begingroup$
That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04
$begingroup$
That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04
add a comment |
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No because $pi(x)^2 = sum_{p le x, q le x} 1 = sum_{pq le x^2} 1 - sum_{pq le x^2, p > x} 1-sum_{pq le x^2, q > x} 1=sum_{pq le x^2} 1 -2sum_{pq le x^2, p > x} 1 $ $=sum_{pq le x^2,p le x} 1 -sum_{pq le x^2, p > x} 1 =sum_{ p le x}pi(x^2/p)-sum_{ p>x} pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $pi(x) = sum_{n le x, n in A} 1$
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– reuns
Jan 14 at 16:22
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That was my guess. It was not very usefull on Goldbach either.
$endgroup$
– Collag3n
Jan 14 at 18:04