I've applied the same concept in creating a prime sequence. This method is straightforward because it involves keeping track of the prime count. The principle used here is p ∣ (n, k) for all k except 1 and n. However, finding the n-th prime using this approach requires the prime counting function π(x), which provides an approximation of the number of primes less than or equal to x using x/ln(x), rather than the exact value. Consequently, we cannot determine with certainty the exact value of n for which a specific prime is found.
I also tried using the method p | (m, floor(m/2) menus 1), but it may not work for all cases. I'm currently working on an alternative method to find all prime
Thank fo you work, ihave been working on this concept a half year ago. I did not want to put it into world. If you put into world already, lets do it complex as it is without virtual numbers but with redefine "e" as it is in reality. You can mail mail me.
There is already a symbol to represent the remainder operator, instead of writing R(a, b) just simply write a%b. Also using sieves is faster than using this formula and if you just want to know if a number is prime, there are plenty of quick methods of doing so that only depend on the logarithm of that prime instead of depending oon the size of the prime like the BPSW primality check.
Looks good and interesting. I wonder how it works. Guess I'll just prove it on my own then. Unfortunately, your formula is not really useful because you use the prime counting function in. This is just an equivalent of it. It is not that hard to find the n-th prime with the pi function, the hard thing is to compute it with a nice formula.
This video doesn’t use virtual numbers. The counting prime function pi(x) is approximately equal to x/ln(x); if x=-1, pi(-1) is approximately equal to -1/j. Since Ln(-1)=j. Missed opportunity for virtual number inclusion.
A general formula for P would take a number n and find m such that P(n)=m. What your formula does is take a number m and find n such that P(n)=m, and your formula does this by defining n=π(m). All this video shows is that P is the inverse function of π if and only if its argument is prime :/ That's not a technique for computing P for any input
We can also find the nth prime. For example, if π(m) = 100, this represents the 100th prime. By using the π formula, we can determine the value of m and then proceed with the calculation.
@ That's not at all what it means to have a general formula!! If you want to compute P(100000), you need to first know the smallest m such that π(m)=100000, and that m is the output of P. Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.
@@log_minus_1 Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.
This video is incomplete. The general formula presented (a) is not proven, and (b) if true, then still needs (at worst) floor(m/2) divisions against the set to verify if m is prime. [Remember that formation of set needs lots of computation!] If true, then this can beat the simple division test of m that always needs floor(m/2) divisions.
"The Hindu religion is the only one of the world's great faiths dedicated to the idea that the Cosmos itself undergoes an immense, indeed an infinite, number of deaths and rebirths. It is the only religion in which the time scales correspond to those of modern scientific cosmology. Its cycles run from our ordinary day and night to a day and night of Brahma, 8.64 billion years long. Longer than the age of the Earth or the Sun and about half the time since the Big Bang." “Most cultures imagine the world to be a few hundred human generations old. Hardly anyone guessed that the cosmos might be far older but the ancient Hindus did,” "It is the only religion in which the time scales correspond, to those of modern scientific cosmology" ~ CARL SAGAN (famous astronomer, cosmologist)
It is not first ever formula for prime numbers. There exists others, but they are so inefficient that it is faster to compute them directly.
Cool it's like CP Willians sequential prime number formula except with virtual numbers.
Today I learnt that X/log(X) gives the number of primes below X. Crazy!!!!
Dude This Channel Is A Fking Joke @@mrbutish
I've applied the same concept in creating a prime sequence. This method is straightforward because it involves keeping track of the prime count. The principle used here is p ∣ (n, k) for all k except 1 and n. However, finding the n-th prime using this approach requires the prime counting function π(x), which provides an approximation of the number of primes less than or equal to x using x/ln(x), rather than the exact value. Consequently, we cannot determine with certainty the exact value of n for which a specific prime is found.
I also tried using the method p | (m, floor(m/2) menus 1), but it may not work for all cases. I'm currently working on an alternative method to find all prime
At 6:15 it is said none of m=9 elements are divisible by 9, even though first two elements are.
Must divide set S9 , all elements
Then a language barrier? Correct way is to say the whole set is not divisible by 9.
yes , 9 must divide all elements of S9 as R(S9,9) must be 0
Thank fo you work, ihave been working on this concept a half year ago. I did not want to put it into world. If you put into world already, lets do it complex as it is without virtual numbers but with redefine "e" as it is in reality. You can mail mail me.
Are u talking about primes formula or virtual numbers?
There is already a symbol to represent the remainder operator, instead of writing R(a, b) just simply write a%b. Also using sieves is faster than using this formula and if you just want to know if a number is prime, there are plenty of quick methods of doing so that only depend on the logarithm of that prime instead of depending oon the size of the prime like the BPSW primality check.
Looks good and interesting. I wonder how it works. Guess I'll just prove it on my own then. Unfortunately, your formula is not really useful because you use the prime counting function in. This is just an equivalent of it. It is not that hard to find the n-th prime with the pi function, the hard thing is to compute it with a nice formula.
It is interesting, but I don't think that it's very efficient for large numbers. Computing the Newton's symbol for large values...
This video doesn’t use virtual numbers. The counting prime function pi(x) is approximately equal to x/ln(x); if x=-1, pi(-1) is approximately equal to -1/j. Since Ln(-1)=j. Missed opportunity for virtual number inclusion.
I didn't use virtual numbers in this video,
This was purely on primes which are real
A general formula for P would take a number n and find m such that P(n)=m. What your formula does is take a number m and find n such that P(n)=m, and your formula does this by defining n=π(m). All this video shows is that P is the inverse function of π if and only if its argument is prime :/ That's not a technique for computing P for any input
We can also find the nth prime. For example, if π(m) = 100, this represents the 100th prime. By using the π formula, we can determine the value of m and then proceed with the calculation.
@ That's not at all what it means to have a general formula!! If you want to compute P(100000), you need to first know the smallest m such that π(m)=100000, and that m is the output of P. Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.
We can find m using pi function
@@log_minus_1 Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.
We can find all the primes using formula.. I will make it better in future videos
This is totally misleading: there is no "general formula for primes"....
This formula works to find nth prime
This video is incomplete. The general formula presented (a) is not proven, and (b) if true, then still needs (at worst) floor(m/2) divisions against the set to verify if m is prime. [Remember that formation of set needs lots of computation!] If true, then this can beat the simple division test of m that always needs floor(m/2) divisions.
Yes, it's not perfect, but it works well. In my next video, I will create a formula for primes using a different method.
"The Hindu religion is the only one of
the world's great faiths dedicated to
the idea that the Cosmos itself
undergoes an immense, indeed an
infinite, number of deaths and
rebirths. It is the only religion in which
the time scales correspond to those of
modern scientific cosmology. Its
cycles run from our ordinary day and
night to a day and night of Brahma,
8.64 billion years long. Longer than
the age of the Earth or the Sun and
about half the time since the Big
Bang."
“Most cultures imagine the world to be a few hundred human generations old. Hardly anyone guessed that the cosmos might be far older but the ancient Hindus did,”
"It is the only religion in which the time scales correspond, to those of modern scientific cosmology"
~ CARL SAGAN (famous astronomer, cosmologist)
💕
Can you solve 1/x = 0 Please make a video if it is possible.
Take ln on both sides ,
-lnx = ln0
-Lnx = e^j * zeta(1)
Lnx = zeta(1)
X = e^zeta(1)
@@log_minus_1 Thank you ❤️
@@log_minus_1 zeta(1) is undefined because its a harmonic series which is undefined. I can leave a proof if you want