On a stricter Twin Primes Conjecture and on the Polignacs Conjecture in general Giulio Morpurgo

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On a stricter Twin Primes Conjecture, and on

the Polignac’s Conjecture in general

Giulio Morpurgo

March 13, 2023

Abstract

The Polignac’s Conjecture, ﬁrst formulated by Alphonse de Polignac

in 1849, asserts that, for any even number M, there exist inﬁnitely many

couples of prime numbers P,P+M. When M= 2, this reduces to the

Twin Primes Conjecture. Despite numerical evidence, and many theoret-

ical progresses, the conjecture has resisted a formal proof since.

In the ﬁrst part of this paper, we investigate a stricter version of the

conjecture, expressed as follows: ”Let pnbe the n-th prime. Then, there

always exist twin primes between (pn

−2)2and p2

n”. To justify this con-

jecture, we formulate a prediction (based on a double-sieve method) for

the number of twin prime pairs in this range, and compare the prediction

with the real results for values of pnup to 6500000. We also analyse what

should happen for higher values of pn.

In the second part, we investigate the validity of the general Polignac’s

Conjecture. We predict the ratio of the number of solutions for any value

of M divided by the number of solutions for M= 2, and explain how this

ratio depends on the factorization of M. We compare the predictions with

the real values for Mup to 3000 (and for the special case 30030) in the

range of from the 1000000-th prime to the 21000000-th prime.

Part 1: The Twin Prime Conjecture

1Introduction

The Twin Prime Conjecture, which is a particular case of the more general

Polignac’s Conjecture [1]) hypothesizes that there exist an inﬁnite number of

pairs k,k+ 2 where both kand k+ 2 are prime (”twin” primes). Despite

numerical evidence (higher and higher twin prime pairs have been found), and

many theoretical progresses [2], [3], the conjecture has resisted a formal proof

since.

In this paper we introduce a stricter version of the conjecture, expressed

as follows: ”Let pnbe the n-th prime. Then, there always exist twin primes

between (pn−2)2and p2

n”. We will also formulate a prediction for the number

of twin prime pairs in this range.

From now on, although many arguments we present apply to the general case,

we will focus on this stricter conjecture, i.e., that there exists at least one pair

of twin primes between (pn−2)2and p2

n. One of the reasons to choose such a

restricted range is that, as the range starts with a value higher than p2

n−1, all

arXiv:2210.15487v2 [math.GM] 10 Mar 2023

the prime numbers up to pn−1 can eliminate ”candidates” (deﬁned in the next

section) in the entire range. This would not be true, for instance, if we used a

range from 1 to p2

n; in such a case, piwould only be signiﬁcant in the portion

of the range above p2

i(introducing complexity in a prediction’s formulation).

We proceed along the following path:

•count the number of potential candidates for twin prime pairs between

(pn−2)2and p2

•use a double-sieve to estimate the probability for a candidate to actually

be a twin prime pair.

•derive a prediction for the number of twin primes in the selected range.

•compare the prediction with the real result (obtained from a computer

program).

•investigate what should happen for larger pnvalues, to show that when

pngrows, also the predicted number of twin pairs grows.

Finally, note that for pn= 3 and pn−2 = 1 the solutions 3,5 and 5,7, and for

pn= 5 and pn−2 = 3, the solutions 11,13 and 17,19 satisfy the conjecture. In

what follows we will therefore assume pn≥7.

2Looking for twin primes between (pn−2)2and p2

2.1 Counting the candidates

Any pair of twin primes can always be expressed as N−1, N+ 1, where Nis a

multiple of 6 (3 must divide N, otherwise it would divide one of N−1, N+ 1).

To search for twin pairs satisfying our version of the conjecture for a given pn,

we have to examine all the integer numbers Nm= 6m, contained in the range

from (pn−2)2to p2

n. These numbers will be our ”candidates”, and for each of

them we will check if both N−1 and N+ 1 are prime. We list these candidates

in increasing order, from Nm1(the smallest value above (pn−2)2to be divisible

by 6), to Nm last (which is equal to p2

n−7, as explained in the next paragraph).

For all pn>3, pnmodulo 3 is either 1 or 2, hence p2

nmodulo 3 is 1. N=p2

n−1

is an even number, and its modulo 3 is 0, hence it is divisible by 6. But we

can always exclude this candidate, because N+1 is p2

n, which is not a prime.

Therefore the largest candidate is p2

n- 7.

For a given pn, the size of the range within which we look for twin primes is

n−(pn−2)2, i.e. 4(pn−1). If pn−2 is also prime, or if 3 does not di-

vide it, Nm1is equal to (pn−2)2+ 5, because also (pn−2)2modulo 6 is 1.

So, between (pn−2)2and p2

nthere is an ”excluded” region of size 12 (7+5)

that cannot contain useful candidates. The number of candidates will then be

(4(pn−1) −12)/6 + 1.

If pn−2 is divisible by 3, then (pn−2)2modulo 6 = 3; this makes Nm1=

(pn−2)2+ 3, and reduces the size of the excluded region to 10. In this case the

number of candidates is (4(pn−1) −10)/6 + 1.

In both cases, when pnis large, a good approximation for the number of candi-

dates is

n of candidates ∼

3pn(2.1)

2.2 How the values of Nmodulo pidetermine if N−1,N+ 1 is a

twin prime pair

For any candidate Nm,Nm–1 and Nm+1 are always smaller than p2

n. Therefore,

for any value of m, we know that Nm–1 and Nm+ 1 are prime if they are not

divisible by any prime from 5 to pn−1(we already excluded 2 and 3, by imposing

the condition that Nmis a multiple of 6).

Without loss of generality, let’s call None of these candidates.

We look at the residuals of Nwith respects to each prime from 5 to pn−1.

•For 5, if Nmodulo 5 is 1, 5 divides N−1. If Nmodulo 5 is 4, 5 divides

N+ 1. If Nmodulo 5 is 0, 2, or 3, N−1 and N+ 1 are not divisible by

5. So we discard all the Nmvalues for which Nmmodulo 5 is 1 or 4 (two

cases), and we keep the others (three cases).

•For 7, we discard all the Nmvalues for which Nmmodulo 7 is 1 or 6, (two

cases) and we keep those for which modulo 7 is 0, 2,3,4, or 5 (ﬁve cases).

•...

•For pn−1, we discard all the Nmvalues for which Nmmodulo pn−1is 1 or

pn−1–1 (two cases), and we keep all the others (pn−1–2 cases).

In fact, we are applying a double-sieve to the set of Nmvalues (which form a

ﬁnite arithmetic progression).

For pn, the number of primes for which we have to check the set of Nmvalues

is n-3. For example, for p5(= 11), we have to check p4(=7) and p3(=5). For

each of these primes pj, with jgoing from 3 to n−1, we discard Nmif the

value of Nmmodulo pjis either 1 or pj−1.

2.3 Predicting the number of twin primes

To predict the number of prime twins for a given pn, we assess the eﬀect of

applying our double-sieve on the set of Nmvalues. In particular, we try to

estimate the probability that a candidate is spared by the double sieve.

We start by evaluating the product of terms (pi−2)/pi, where the index i

goes from 3 (p3being 5) to n−1. We recall that, by choosing only Nmvalues

divisible by 6, we already took into account the eﬀect of p1(=2) and p2(=3).

For a given pn, this product, which we call product pminus2 over p, consists

of n-3 terms.

product pminus2over p =

pn−1

primes≥p3

p−2

p=3

13 ··· pn−1−2

pn−1

(2.2)

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OnastricterTwinPrimesConjecture,andonthePolignac'sConjectureingeneralGiulioMorpurgoMarch13,2023AbstractThePolignac'sConjecture,rstformulatedbyAlphonsedePolignacin1849,assertsthat,foranyevennumberM,thereexistinnitelymanycouplesofprimenumbersP,P+M.WhenM=2,thisreducestotheTwinPrimesConjecture.Despite...

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On a stricter Twin Primes Conjecture and on the Polignacs Conjecture in general Giulio Morpurgo

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