# Primes

Consider the infinite table $\begin{array}{c | c c c c c c c c} k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \dots \\ \hline P_k & 2 & 3 & 5 & 7 & 11 & 13 & 17 & \dots\\ S^1_k & 2 & 1 & 2 & 2 & 4 & 2 & 4 & \dots\\ S^2_k & 2 & -1 & 1 & 0 & 2 & -2 & 2 & \dots\\ S^3_k & 2 & -3 & 2 & -1 & 2 & -4 & 4 & \dots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{array}$

Where $$P_k$$ is the sequence of prime numbers and $$S^1_k$$ is the difference of terms in $$P_k$$, $$S^2_k$$ is the difference of terms in $$S^1_k$$ etc. One can define a set of generating functions for each column such that for a given generating function $$G_n(a_n)(z)$$, when $$z=0$$, $$G_n=P_n$$, the, $$n^{th}$$ coeficcient of the $$n^{th}$$ power of $$z$$ is the $$n^{th}$$ prime.

The functions are then $G_1=\frac{-2}{z-1} = 2+2z+2z^2+2z^3 \dots \\ G_2=\frac{3-5z}{(z-1)^2} = 3+1z-1z^2-3z^3 \dots\\ G_3=\frac{-10z^2+13z-5}{(z-1)^3} = 5+2z+1z^2+2z^4 \dots\\ G_4=\frac{-17z^3+34z^2-26z+7}{(z-1)^4} \\ G_5=\frac{-28z^4+78z^3-92z^2+51z-11}{(z-1)^5} \\ \vdots$

Here one can see that the coefficients of the resulting polynomial match the table columns. It seems a better idea to tabulate the coefficients of this new set of functions.

$\begin{array}{c | c c c c c c c} & c & z & z^2 & z^3 & z^4 & z^5 & z^6 & z^7 & z^8 & z^9\\ \hline G_1(z-1)^1 & -2 \\ G_2(z-1)^2 & 3 & -5 \\ G_3(z-1)^3 & -5 & 13 & -10 \\ G_4(z-1)^4 & 7 & -26 & 34 & -17 \\ G_5(z-1)^5 & -11 & 51 & -92 & 78 & -28 \\ G_6(z-1)^6 & 13 & -76 & 181 & -222 & 143 & -41\\ G_7(z-1)^7 & -17 & 115 & -331 & 521 & -477 & 245 & -58\\ G_8(z-1)^8 & 19 & -150 & 514 & -996 & 1186 & -876 & 378 & -77 \\ G_9(z-1)^9 & -23 & 203 & -794 & 1802 & -2606 & 2474 & -1520 & 562 & -100 \\ G_{10}(z-1)^{10} & 29 & -284 & 1247 & -3230 & 5456 & -6260 & 4910 & -2564 & 823 & -129 \\ \end{array}$

By construction the first column must be the alternating sign prime number sequence such that when $$z=0$$, the equation reduces to the primes. We can attempt to find the sequences however for the diagonals of this table! The main diagonal appears to be negative the sequence OEIS A007504, which is the sum of the first few primes. This is achieved also by extending $$table 1$$ back by one extra row.

Looking a the next diagonal down, this appears to be the partial sum of the sequence OEIS A117495, which is the product of a prime with the number of primes less than itself. That is $$p_n\pi(p_n-1)$$ when $$\pi(x)$$ is defined as the number of primes smaller or equal to $$x$$.

The next diagonal was found to be another sum over a prime product with a different element in the pi function.

To re create the table above we can add piecewise terms. $M_{ij}=-\delta^i_j\sum_{k=1}^jP_k+\delta^i_{j+1}\sum_{k=1}^jP_{k+1}\pi(P_{k+1}-1) -\delta^i_{j+2}\sum_{k=1}^j P_{k+2}\pi(P_{k(k+1)/2})+\dots$

If we let the $$M_{ij} \to M_{i,1}$$ then this is just the alternating sequence of primes!, The sums disappear.

$M_{i1}=(-1)^iP_i=-\delta^i_1P_1+\delta^i_{2}P_{2}\pi(P_{2}-1) -\delta^i_{3}P_{3}\pi(P_{2}-1)+\dots$

However this is a statement of the obvious! It implies consitency with what was done so far.

Further diagonal were found, the key being the polynomial in $$k$$ in the index of the prime number inside the $$\pi$$ function the next terms in the series were found to be $+\delta^i_{j+3}\sum_{k=1}^j P_{k+3}\pi(P_{k(k+1)(k+2)/6})$

Can assume a form of polynomial (which one is better ?/same thing) $Q(k)=\frac{k(k+1)(k+2)...(k+m)}{(m+1)!} \\ Q_n(k)=\frac{1}{n!}\prod_{i=1}^n(k+i-1)$

Also, removing the unecesarry pi function, we can simply state the number of primes below or equal to the $$k^{th}$$ prime is $$k$$. I.e $$\pi(P_k)=k$$. Then we have $M_{ij}=\sum_{k=1}^j -\delta^i_jP_{k+0}Q_0+\delta^i_{j+1}P_{k+1}Q_1-\delta^i_{j+2}P_{k+2}Q_2+ \dots$

Which can be written as an ’infinite’ sum $M_{ij}=-\sum_{l=0}^{\infty}\sum_{k=1}^j (-1)^{l}\delta^i_{j+l}P_{k+l}Q_l(k) \\ M_{ij}=-\sum_{l=0}^{\infty}\sum_{k=1}^j \bigg[ \frac{(-1)^{l}}{l!}\delta^i_{j+l}P_{k+l}\prod_{m=1}^l(k+m-1) \bigg ]$

But it is not really infinite, as the kronecker deltas will all be zero at some $$l$$ and greater, so this can be expressed as $M_{ij}=-\sum_{k=1}^j \bigg[ \frac{(-1)^{i-j}}{(i-j)!}P_{k+i-j}\prod_{m=1}^{i-j}(k+m-1) \bigg ]$

We can now express the coeficcients of a generating function’s polynomial directly for example $G_4=\frac{-17z^3+34z^2-26z+7}{(z-1)^4}=\frac{M_{44}z^3+M_{43}z^2+M_{42}z+M_{41}}{(z-1)^4}$

For any general column generating function of the original table we than have $G_q=\frac{\sum_{r=1}^q M_{qr}z^{r-1}}{(z-1)^q}$

which is $G_q=\frac{-1}{(z-1)^q}\sum_{r=1}^q \sum_{k=1}^r \bigg[ \frac{(-1)^{q-r}}{(q-r)!}P_{k+q-r}\prod_{m=1}^{q-r}(k+m-1) \bigg ]z^{r-1}$

We should then reclaim the fact that when $$z=0$$ the series $$G_q$$ will be that of the prime numbers. The only non zero term will be when $$r=1$$ $P_q=\frac{-1}{(-1)^q}\bigg[ \frac{(-1)^{q-1}}{(q-1)!}P_{q}\prod_{m=1}^{q-1}(m) \bigg ]$

Which is indeed reclaimed!

# Abstract

Rough. List of curiosities that have caught my attention over the last period of time...

# 1

$\frac{2!!}{3!!}=0.\dot{6}...(period \; 1) \\ \frac{2!!}{3!!}\frac{5!!}{7!!}=0.0\dot{9}5238\dot{0}... (period \; 6) \\ \frac{2!!}{3!!}\frac{5!!}{7!!}\frac{11!!}{13!!}= \frac{1}{137} + \frac{1}{37401} = 0.00\dot{7}3260\dot{0}... (period \; 6)\\ \frac{2!!}{3!!}\frac{5!!}{7!!}\frac{11!!}{13!!}\frac{17!!}{19!!}=0.000\dot{3}8557933294775400\dot{0}... (period \; 18) \\ \frac{2!!}{3!!}\frac{5!!}{7!!}\frac{11!!}{13!!}\frac{17!!}{19!!}\frac{23!!}{29!!}=0.0000000\dot{1}9697539358761379330042366944968275881766504876 \\ 00275864038719453117517243349064280703724139900788418634 \\ 75862265940910828993103645251255656579310541802979794510 \\ 34502456160048416551743835470393244137950732022117382068 \\ 985214780738071724157628573841520000\dot{0}...(period \; 252) \\ \frac{2!!}{3!!}\frac{5!!}{7!!}\frac{11!!}{13!!}\frac{17!!}{19!!}\frac{23!!}{29!!}\frac{31!!}{37!!}=(period \; 756) \\ \frac{2!!}{3!!}\frac{5!!}{7!!}\frac{11!!}{13!!}\frac{17!!}{19!!}\frac{23!!}{29!!}\frac{31!!}{37!!}\frac{41!!}{43!!}=(period \; 756)$

Period series is then, $$1,6,6,18,252,756,756,...$$. Could not go any further with the analysis system being used at the moment in terms of repeating digits.

$$\alpha$$   1/137... Perhaps... 1*6/1 is 6, then 1*6*6/2 is 18 then 1*6*6*18/

756 is 252*3... 252 is 18*14... 18 is 6*3... 6 is 1*6...

Continued fractions for this series goes [0; 1, 2],[0; 10, 2],[0; 136, 2],[0; 2593, 2],[0; 50767762, 2],[0; 2169560330437, 2],[0; 93291094208812, 2],[0; 12356125554674589187, 2],[0; 753723658835149940437, 2],[0; 3692492204633399558203312, 2],...

The significant series being $$1,10,136,2593,50767762,2169560330437,93291094208812,12356125554674589187,753723658835149940437,3692492204633399558203312...$$ as the alternating component is always 2.

Each appears to be an egyptian fraction of two things...? (verify in a table)

The periods of the same sequence without the double factorials is 1,6,6,18,252,252,252,3276...

#include <stdio.h> #include <math.h> #include <stdlib.h>

int main()

//RAND_MAX is: 2147483647 //Draw 4 means 536870911.75 period. srand(time(NULL));

int i,j; long int n,col; double a1,a2,t1,t2; double R,r;

R=1.0; r=(8.0/103.0)*M_LN2; col=0; n=0;

for(j=0; j<50; j++)

while(n<536870911)

for(i=0; i<100000; i++)

a1=((double)rand()/RAND_MAX)*(R-(r/2.0)); a2=((double)rand()/RAND_MAX)*(R-(r/2.0)); t1=((double)rand()/RAND_MAX)*M_PI*2.0; t2=((double)rand()/RAND_MAX)*M_PI*2.0;

if( sqrt( pow(a1*sin(t1)-a2*sin(t2),2) + pow(a1*cos(t1)-a2*cos(t2),2) ) < r ) col++;

n++;

printf(" n=0; col=0;

return 0;