Return to Answer

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

(Using exact values of $\zera(2k)$ up to $k=6$ and Microsoft Excel for the calculations I extended the list above to include $13$ and $17$.)

Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerical computation.

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

(Using exact values of $\zeta(2k)$ up to $k=6$ and Microsoft Excel for the calculations I extended the list above to include $13$ and $17$.)

Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerical computation.

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerucal computation.

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

(Using exact values of $\zera(2k)$ up to $k=6$ and Microsoft Excel for the calculations I extended the list above to include $13$ and $17$.)

Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerical computation.

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes.

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerucal computation.