Theorem 1: Given a primorial [itex]p_n^{\sharp}>2[/itex], [itex]p_{n+1}[/itex] is the second largest element in the reduced residue system modulo [itex]p_n^{\sharp}[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Proof

Clearly [itex]p_{n+1}[/itex] is an element of the r.r.s. modulo [itex]p_n^{\sharp}[/itex] since every prime less than [itex]p_{n+1}[/itex] divides [itex]p_n^{\sharp}[/itex]. Next notice that the smallest element of the r.r.s. modulo [itex]p_n^{\sharp}[/itex] is invariably one.

So it remains to show that the second largest element in r.r.s. modulo [itex]p_n^{\sharp}[/itex] is in fact prime. But wait this is already done since every prime less than this second element is a factor of [itex]p_n^{\sharp}[/itex] and the two are relatively prime, the second element must be prime and it must be the next prime.

q.e.d.

Lemma 1: It follows immediately that the third element in the r.r.s. modulo [itex]p_n^{\sharp}[/itex] is either prime or divisible by the second element in the r.r.s. modulo [itex]p_n^{\sharp}[/itex]. Further extension along these lines is possible.

Theorem 2: We can construct a sequence of prime numbers using the sequence of primorials [itex]p^{\sharp}[/itex] greater than two and the reduced residue systems of this sequence.

Proof

We are given [itex]p_2^{\sharp}=6[/itex] and r.r.s. modulo 6 = {1,5}

By Theorem1 [itex]5=p_3[/itex]. We can write the r.r.s. modulo [itex]p_3^{\sharp}=30[/itex] using the reduced residue system modulo 6 as follows

[tex]r.r.s. mod 30 = \{6x+r_{i}|r_{i} \in r.r.s. mod 6, 0<x<4\}-\{5,25\}[/tex]

In general we can write

Eq. 1 [tex]r.r.s. modulo p_{n+1}^{\sharp} = \{p_n^{\sharp}x+r_{i}|r_{i} \in r.r.s. mod p_n^{\sharp}, 0<x<p_{n+1}-1\}-\{p_{n+1}r_{j}r_{k}|r_{j}r_{k}= 1 mod p_n^{\sharp}\} [/tex]

Using this construction we generate the sequence of primes like this

2 -given

3 - given

5 - Theorem 1 using base 6

7 - Theorem 1 using base 30, also given as 6*1+1, since the only 6*0 elments were 6*0+1, which is excluded and 6*0+5 which of course goes into the base 30.

11 - Theorem 1 using base 210, also given as 210*0+11

13 - Theorem 1 using base 2310, also given as 2310*0+13

You see it happens in an odd way. For each step the second lowest number is prime by Theorem 1. Reduced residue systems of primorials are very rich in primes in the smaller values.

So then the next prime 17 was added to the reduced residue system at base 30 and it just sat there as we climbed the ladder to [itex]p_7^{\sharp}=30030[/itex]. So there you have the power of [itex]p_n^{\sharp}[/itex] to reveal primes greater than [itex]p_n[/itex].

The draw back from a computing standpoint is that the size of the reduced residue system for primorials grows relatively fast compared to the guaranteed number of primes you get back, so it is not a computational theory unless we can find a way to calculate recursively and in batches. Equation 1 might be worth looking at more closely along those lines.

Define the following operations on sets.

Given two sets A and B let [itex]A\otimesB = \{(a_{i},b_{j})\}[/itex] where the indices cover all elements of the two sets. Example [itex]\{1,3\}\otimes\{2,4\}=\{(1,2),(1,4),(3,2),(3,4)\}[/itex]. Given a set of ordered pairs [itex]A\otimesA[/itex] and a scalar s define [itex]sA\otimesA= \{(sa_{i}, a_{j})\}[/itex] where the indices cover all elements of A. Example let A = {4,5}, [itex]A\otimesA=\{(4,4),(4,5),(5,4),(5,5)\}[/itex] and let s = 3 under standard multiplication then [itex]sA\otimesA=\{(12,4),(12,5),(15,4),(15,5)\}[/itex] Now we need an operation to evaluate the ordered pairs, this can be done using a prefix as follows. Let [itex]\mu_{}A\otimesA=\{a_{i}a_{j}\}[/itex] where the indices cover all elements of A, andis a base for modular multiplication. This notation avoids confusion when using subscripts to denote the element of a sequence so for instance [itex]a_{n}a_m[/tex] is well understood as is [itex]a_{n}+_{}a_m[/itex] this an alternative to mod b notation. Whenis omitted it means standard multiplication over the natural numbers.

Now we can write Equation 1 in a simpler form.

Equation 2: [tex]r.r.s._{[p_{n+1}^{\sharp}]} = p_n^{\sharp}\{0,...,p_{n+1}-1\}\oplus r.r.s._{[p_n^{\sharp}]}- p_{n+1} Ker \mu r.r.s._{[p_{n}^{\sharp}]}\otimes r.r.s._{[p_{n}^{\sharp}]}[/tex]

It follows that

[tex]r.r.s._{[p_{n+1}^{\sharp}]} = \{0,p_n^{\sharp}, 2p_n^{\sharp},...,p_n^{\sharp}(p_{n+1}-1)\}\oplus r.r.s._{[p_n^{\sharp}]}- p_{n+1}Ker \mu r.r.s._{[p_{n}^{\sharp}]}\otimes r.r.s._{[p_{n}^{\sharp}]}[/tex]

Ker their is the kernel of the map [itex]\mu_{[p_n^{\sharp}]} : r.r.s._{[p_n^{\sharp}]} \otimes r.r.s._{[p_n^{\sharp}]} \rightarrow r.r.s._{[p_n^{\sharp}]}[/itex]. But the notation is still clumsy. Something has to be gotten that does not interfere with existing notation but allows easy use of multiple bases in modular equations, we could use the term muddled equations to describe euqations like [itex]x \equiv y mod a mod b[/itex]. In the notation given here so far that might be written [itex]x =_{[a]}y[/itex] or perhaps [itex]x =_{[a,b]}y[/itex] not necessarily the same as [itex]x =_{[b,a]}y[/itex] .

Still something better will have to be gotten. I'll try to work that out more later, Equation 2 is really a prescription for an algorithm it could encompass more than one algorithm. I'll have to look into that.

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# Primorials and the Next Prime

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