Are my calculations of a recursive prime-generating function based on logarithms correct?

Question

I am trying to devise a recursive prime-generating function following an intuition of a possible analogy to Mills and Wright prime-representing functions, in the present case based on logarithms. The proposed prime generating function $f(n)$ will provide not a subset but the complete set of primes being $f(1)=2$, $f(2)=3$, $f(3)=5$... and the prime-generating constant will be called $L$. As in the standard results from Mills and Wright the decimal precision of the constant is important in order to recover the embedded primes and it is not known if $\lim_{n \to \infty} L_n$ is rational or not.

This is how it works and the questions are at the end:

1. Start with $n=1$, current prime $p_1=2$, previous accumulated value of the constant will be defined as $L_0=0$ (starting value). Calculate the value for the constant for $n=1$. It will be $$L_1=\frac{Ln(2+L_0)}{Ln(1+1)}$$

(Where $Ln$ is the natural logarithm).

3. Calculate the value for the constant for $n=2$, $p_2=3$, $$L_2=\frac{Ln(3+L_1)}{Ln(2+1)}$$

So if we apply recursively the formula, for $n$:

$$L_n=\frac{Ln(p_n+L_{n-1})}{Ln(n+1)}$$

For instance, the following PARI_GP code calculates $L_{500}$

\p2000;
testlimit=500;current_pr=2;L=0;for(n=1,testlimit,L=log(current_pr+L)/log(n+1);current_pr=nextprime(current_pr+1););print("n is ",testlimit," and L is ",L);

$$L_{500} = 1.3159864456...$$

Reviewing the results of the tests, it seems that $\lim_{n \to \infty} L_n$ oscillates (due to the gaps between primes applied in the formula) but in the long term it is stable and tends to decrease and goes down to a value closer to $1$ and lower than $2$.

For example:

\p3;
testlimit=500000;current_pr=2;L=0;for(n=2,testlimit,L=log(current_pr+L)/log(n);current_pr=nextprime(current_pr+1););print("n is ",testlimit," and L is ",L);

$L_{5000}$ is around $1.23...$ and the above code shows that $L_{500000}$ is around $1.21$. Other similar tests show that it tends to go down to some specific limit near the lower bound of $[1,2]$.

As the proccess of recovering the primes is recursive, the way of using the constant is as follows:

For instance, assuming that we have $L_{500}$, we need to start obtaining the last prime back:

$$f(500)=p_{500}=\lfloor (500+1)^{L_{500}} \rfloor - 1 = 3571$$

Then recover $L_{499}$ and recover $f(499)=p_{499}$

$$L_{499}=((500+1)^{L_{500}})-p_{500}=1.31586811...$$

$$f(499)=p_{499}=\lfloor (499+1)^{L_{499}} \rfloor - 1 = 3559$$

$$...$$

So in general the recursive process to recover $L_{n}$ and $f(n)=p_n$ from $L_{n+1}$ and $f(n+1)=p_{n+1}$ is:

$$L_{n}=((n+2)^{L_{n+1}})-p_{n+1}$$

$$f(n)=p_{n}=\lfloor (n+1)^{L_{n}} \rfloor - 1$$

The following code, having $L_{500}$ calculates backwards the complete set of primes $\{p_{500}..p_{1}\}$

curr_L=L;for(n=1,testlimit,curr_n=testlimit-n+2;curr_p=(floor(curr_n^curr_L))-1;print("n is ",testlimit-n+1," ; Current prime is ",curr_p," and is_prime check = ",isprime(curr_p));curr_L=(curr_n^curr_L)-curr_p;);

There is a little correction required for the calculation of $p_1$. Depending on the $L_n$ calculated, sometimes the recovered value $\lfloor (2)^{L_{1}} \rfloor = 2$ instead of $3$. And for that reason, when the value is $2$, $f(1)=p_{1}=\lfloor (2)^{L_{1}} \rfloor - 1 = 2-1 = 1$ instead of the expected $p_1=2$. To handle this special case, we can express the prime-generating function as:

$$f(n)=p_{n}=\lfloor (n+1)^{L_{n}} \rfloor - 1 + \delta_{f(n),2}$$

Where $\delta_{f(n),2}$ is the Kronecker delta function (kindly provided by @MitchellSpector in this question). Basically it is a little trick that will assure that always $f(1)=2$ independently of the value of $\lfloor (n+1)^{L_{n}} \rfloor$ ($2$ or $3$). It would be possible a definitions-by-case of $f(n)$ as well instead of using one single expression.

This is a graph of the evolution of $L_n$ (only four decimals of precision):

I would like to ask the following questions:

1. Are the calculations correct or is there a mistake in the assumptions?

2. How could I prove that $L_n$ is decreasing in the long term and there is indeed a limit? The tests show that but I am a little bit lost about how to demonstrate that it really is decreasing (because each step depends on the gaps between primes). A hint about the starting step would be great!

3. Initially I think this kind of recursive prime-generating function is a little bit different from the original results of Mills and towers of powers of Wright, but it might be possible that a similar idea had been explored before in recent literature. Initially I did not find such references. Are there similar solutions as the one I am devising? Any references to papers would be very appreciated. Thank you!

Answer 1

So say that for some $C\in\Bbb{R}$, we have: $$p_n\sim C^{(\log2)(n+3)}$$ Now: $$L_n\sim\dfrac{\log(C^{(\log 2)(n+3)}+L_{n-1})}{\log 2^{n+3}}$$ We already know that $L_{n-1} < 1$, while $C^{(\log 2)(n+3)}$ tends to infinity: $$L_n\sim \dfrac{\log(C^{(\log 2)(n+3)})}{\log 2^{n+3}} = \dfrac{1}{\log 2^{n+3}}\cdot \log(C^{(\log 2)(n+3)})=\\ \log(C^{\dfrac{(\log 2)(n+3)}{\log 2^{n+3}}})=\log(C^{\dfrac{(\log 2)(n+3)}{(\log 2)(n+3)}})= \log C$$ We can now say: $$L_n \sim C \Longleftrightarrow p_n \sim e^{C^{(\log 2)(n+3)}}$$ So you'll still have some Mills-like constant involved.

Answer 2

Thanks to @barakmanos I have been able to understand that due to the oscillating gaps between primes, the proposed constant $L_n$ will not be able to have a proper limit. To avoid that issue, I have restricted the conditions to force $L_n$ to be strictly decreasing, so a prime-generating function of this kind requires some more restrictions to have a constant with a limit when $n \to \infty$. Based on these two points:

1. The gap between the embedded primes will be strictly increasing. The first prime selected will be $11$ then $19$ (gap 8), then 29 (gap 10), then 41 (gap 12), etc.

2. The formula will be modified so each $n$ forces the value of $L$ to decrease on each iteration by using powers of $2$ associated to each $n$.

$$L_n=\frac{Ln(p_n+L_{n-1})}{Ln(2^{n+3})}$$

The initial status is $n=1$, $p_1=11$, $L_0=0$

So this looks like this ($16$ initial values):

Now $L_n$ is strictly decreasing, and a limit (rational or not) is assured.

The PARI/GP code to generate the constant:

\p20000;
testlimit=500;prev_gap=6;current_pr=11;L=0;for(n=1,testlimit,print("added prime p_",n,"=",current_pr,"; gap=",prev_gap);L=log(current_pr+L)/log(2^(n+3));prev_pr=current_pr;current_pr=nextprime(current_pr+prev_gap+1);prev_gap=current_pr-prev_pr;);print("n is ",testlimit," and L is ",L);

The constant now has this value for $n=500$:

$$L_{500}=0.0396890...$$

So in general the recursive process to recover $L_{n}$ and $f(n)=p_n$ from $L_{n+1}$ and $f(n+1)=p_{n+1}$ now will be:

$$L_{n}=(2^{n+1})^{L_{n+1}}-p_{n+1}$$

$$f(n)=p_{n}=\lfloor (2^{n})^{L_{n}} \rfloor$$

The following code, having $L_{500}$ calculates backwards the complete set of primes $\{p_{500}..p_{1}\}$

curr_L=L;for(n=1,testlimit,curr_n=testlimit-n+1;curr_p=floor((2^(curr_n+3))^curr_L);print("n = ",testlimit-n+1,"; current prime is ",curr_p," and is prime check = ",isprime(curr_p));curr_L=((2^(curr_n+3))^curr_L)-curr_p;);

Now it looks much more like a Mills-like constant but recursive!