Formula for nearest prime $p_x$ less than a number $k$
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In my attempts to get around Brocard's problem, I've come up with the following formula for the prime factorization of $k!$:
$$prod_{n=1}^xleft(prod_{s=1}^{lfloorlog_{p_n}krfloor}left(p_n^{lfloorfrac{k}{p_n^s}rfloor}right)right).$$
My problem: $x$ needs to be the greatest value for which $frac{k}{p_x}$ is greater than or equal to $1$. I'm currently too busy to think about it, so: does anyone know of a formula to find $x$?
prime-numbers diophantine-equations
edited Nov 21 at 13:01
Klangen
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
add a comment |
up vote
1
down vote
favorite
In my attempts to get around Brocard's problem, I've come up with the following formula for the prime factorization of $k!$:
$$prod_{n=1}^xleft(prod_{s=1}^{lfloorlog_{p_n}krfloor}left(p_n^{lfloorfrac{k}{p_n^s}rfloor}right)right).$$
My problem: $x$ needs to be the greatest value for which $frac{k}{p_x}$ is greater than or equal to $1$. I'm currently too busy to think about it, so: does anyone know of a formula to find $x$?
prime-numbers diophantine-equations
edited Nov 21 at 13:01
Klangen
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
1
Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my attempts to get around Brocard's problem, I've come up with the following formula for the prime factorization of $k!$:
$$prod_{n=1}^xleft(prod_{s=1}^{lfloorlog_{p_n}krfloor}left(p_n^{lfloorfrac{k}{p_n^s}rfloor}right)right).$$
My problem: $x$ needs to be the greatest value for which $frac{k}{p_x}$ is greater than or equal to $1$. I'm currently too busy to think about it, so: does anyone know of a formula to find $x$?
prime-numbers diophantine-equations
edited Nov 21 at 13:01
Klangen
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
In my attempts to get around Brocard's problem, I've come up with the following formula for the prime factorization of $k!$:
$$prod_{n=1}^xleft(prod_{s=1}^{lfloorlog_{p_n}krfloor}left(p_n^{lfloorfrac{k}{p_n^s}rfloor}right)right).$$
My problem: $x$ needs to be the greatest value for which $frac{k}{p_x}$ is greater than or equal to $1$. I'm currently too busy to think about it, so: does anyone know of a formula to find $x$?
prime-numbers diophantine-equations
prime-numbers diophantine-equations
edited Nov 21 at 13:01
Klangen
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
edited Nov 21 at 13:01
Klangen
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
edited Nov 21 at 13:01
Klangen
1,36511131
edited Nov 21 at 13:01
Klangen
1,36511131
edited Nov 21 at 13:01
Klangen
1,36511131
1,36511131
asked Jan 28 at 16:55
weatherman115
1447
asked Jan 28 at 16:55
weatherman115
1447
asked Jan 28 at 16:55
weatherman115
1447
1447
1
Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14
add a comment |
1
Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14
1
1
Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14
add a comment |
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Yes $p_{pi(k)}$, since $p_{pi(k)} leq k<p_{pi(k)+1}$ ... if you find a better one, let us now :)
– rtybase
Jan 28 at 17:03
I found $sum_{n=2}^klfloorfrac{2}{sigma_0left(nright)}rfloor$ for $kgeq2$, but it feels impractical for this situation.
– weatherman115
Jan 28 at 19:14