Combinatorial number theory

An appliance of combinatorial number theorywith an inclusion of linear algebra

Preprint

vorgelegt von

M. N. G. Einstein

aus Bonn

Contents

Preface i

1 Introduction 1

1.1 Canonical combinatorial Matrix . . . . . . . . . . . . . . . . . . . . 1

1.2 Bertrand’s postulate . . . . . . . . . . . . . . . . . . . . . . . . . . 3

I Applied combinatorial number theory 7

2 Goldstream 9

2.1 Miscellanea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Claim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Bibliography 17

i

Chapter 1

Introduction

Based on the constructed combinatorial number theory by M. N. G. Einstein (workstill in revision) we want to find a different application than the combinatorial −and Dirichlet series. While the roots of the construction may lie in the combinato-rial number theory, it is completely separated from − and can be derived withoutit. Therefore we will show this latter case.

1.1 Canonical combinatorial Matrix

Before we can start any analysis we need a (triangular) matrix with special con-ditions.

Definition 1.1. Let T = Tij be the canonical combinatorial matrix:

Tij =

a00 a01 . . . a0na10 a11 . . . a1na20 a21 . . . a2n...

.... . .

...an0 an1 . . . ann

with the following properties:

∀i 6= 0 : ai0 := 1

i = 0 : ai0 := 0

and:∀i ≥ 1 ∧ ∀j ≥ 1 : (κi)T = κi

1

2 § 1 Introduction

with the column κi − and row vectors κi (whereby the last condition depends on thelength of the taken chain if the matrix Tij is defined differently). For our purposethe values aij are:

Ti≥1j≥1=

κ1j = 2 3 4 . . . nκ2j = 3 4 5 . . . n+ 1κ3j = 4 5 6 . . . n+ 2...

......

.... . .

...κnj = n n+ 1 n+ 2 . . . n+ n

With that definition the canonical combinatorial matrix Tij has some amazingconsequences. The most important one is that the elements of the trace Tii areeven and are constructed by the indices i:

∀ aii = Tii : aii = even ∧ aii = i+ i

The proof of that is trivial, since a11 = 2 ⇒ ai+1 i+1 = i + 1 + i + 1. Furtherthe diagonales dlr(aij) constructed by (i − 1, j + 1) from ai≥2j≥1

to ai=1j with j= max to the starting point of i (= A − 3, whereby A is the value of the cell),have always the same value. Moreover they are constructed by the indices aswell! The proof of that is again trivial. It follows in its simplest form fromthe third condition ∀i ≥ 1 ∧ ∀j ≥ 1 : (κi)T = κi. That the value of aij inthose diagonals dlr(aij) are constructed by the indices can be shown easily withaii = i+ i⇒ ai−1 i+1 = i+ i = dlr(aij).

We want to add here, that in combinatorial number theory the matrix abovefollows from:

Definition 1.2. Let N 3 ni, p ∈ ni and κi represent a multiplication chain in theform:

k = k = some constant p

κ1 = κ1 = n3, n4, n5, . . . , nn

κ2 = k1 · k2 =

{k1 = n3, n3, n4, . . . , n(n−1)k2 = n4, n5, n6, . . . , nn

κ3 = k1 · k2 · k3 =

k1 = n3, n4, n5, . . . , n(n−2)k2 = n4, n5, n6, . . . , n(n−1)k3 = n5, n6, n7, . . . , nn

...

1.2 Bertrand’s postulate 3

when i increases, the length of |nn| decreases, with the condition:

k ∩ k1 ∩ k2 ∩ k3 ∩ · · · ∩ ki =⋂i

ki = ∅

We call then:Cij(κ

i))

a combinatorial function, where Cij(κi) means the j−th combination of the chain

(κi) with:C1j(κ

1)

simply representing the numbers of κ1 = n3, n4, n5, . . . , nn. Therefore C1j(κ1)

returns successively n3j=1, n4j=2

, n5j=3, . . . , nnj=|n|−1

, while Ci0(κi) delivers a one:

∀i 6= 0 : Ci0(κi) := 1

and:i = 0 : Ci0(κ

i) := 0

We see, that Cij(κi) delivers exactly the values Tij.

1.2 Bertrand’s postulate

Theorem 1.3 (theorem of Bertrand). In between every 1 ≤ n ∈ N and its doublethere lies at least one prime number N:

∀n ∃p ∈ P : n < p ≤ 2n

Proof. We start with the sequence of prime numbers:

p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 13, p6 = 23, p7 = 43,

p8 = 83, p9 = 163, p10 = 317, p11 = 631, p12 = 1259, p13 = 2503, p14 = 4001

for which we have pi < 2pi−1. Hence every interval {pi : n < pi ≥ 2n}, withn ≤ 4000 contains one of these |pi| prime numbers.

4 § 1 Introduction

With the identity:

R 3 x ≥ 2 :∏

P 3 p≤x

p ≤ 4x−1

If one notes now that for largest P 3 pj ≤ x, then it is:∏p≤x

p =∏p≤pj

p and 4pj−1 ≤ 4x−1

Therefore it is enough to check for the proposal the case x = pj ∈ P. By inductionwe assume then that it is valid for all x ∈ N in the continuous set {2, 3, 4, . . . , 2n},while we consider odd primes p = 2n+ 1, for which we split the product to:∏

p≤2n+1

p =∏

p≤n+1

p∏

n+1<p≤2n+1

p ≤ 4n(

2n+ 1

n

)≤ 4n22n = 42n

This follows from three things. First due to the holdings of:∏p≤n+1

p ≤ 4n

by induction; secondly from the observation that:(2n+ 1

n

)=

(2n+ 1)!

n!(n+ 1)!

is an integer, where the primes we consider are all factors of the numerator (2n+1)!and only the numerator. Thus follows the inequality:∏

n+1<p≤2n+1

p ≤(

2n+ 1

n

)and finally the inequality: (

2n+ 1

n

)≤ 22n

holds due to: (2n+ 1

n

)and

(2n+ 1

n+ 1

)are two equal summands that appear in the series:

2n+1∑k=0

(2n+ 1

k

)= 22n+1

For the next step we need Legendre’s theorem.

1.2 Bertrand’s postulate 5

Theorem 1.4 (theorem of Legendre). The factorial n! contains the prime factorp exactly: ∑

k≥1

⌊n

pk

⌋times.

Proof. Exactly bnpc of the factors of n! are divisible by p, which accounts for bn

pc

p−factors. Next, b np2c of the factors of n! are even divisible by p2 which accounts

for the next b np2c prime factors p of n!. By continuing this process follows the sum∑

k≥1bnpkc.

With Legendre’s theorem we get that the binomial coefficient(2nn

)= (2n)!

n!n!contains

the prime factor p exactly:

∑k≥1

(⌊2n

pk

⌋− 2

⌊n

pk

⌋)

times. Each summand in the series is 1 due to the satisfaction of:⌊2n

pk

⌋− 2

⌊n

pk

⌋<

2n

pk− 2( npk− 1)

= 2

Furthermore the summands vanish whenever pk > 2n. Thus the binomial coeffi-cient

(2nn

)contains p exactly:

∑k≥1

(⌊2n

pk

⌋− 2

⌊n

pk

⌋)≤ max{v : pv ≤ 2n}

times. Hence the largest power of p that divides the binom(2nn

)is not larger than

2n. In particular, primes p >√

2n appear at most once in(2nn

). With:(

nn2

)>

2n

n

for n > 3, we get:

4n

2n≤(

2n

n

)≤∏

p≤√2n

2n∏

√2n<p≤ 2

3n

p∏

n<p≤2n

p

6 § 1 Introduction

and thus, since there are not more than√

2n primes p ≤√

2n,

4n ≤ (2n)1+√2n

∏√2n<p≤ 2

3n

p∏

n<p≤2n

p

for n > 3. If there is no prime p with n < p ≤ 2n the second product is equal 1.Now we substitute:

R 3 x ≥ 2 :∏

P 3 p≤x

p ≤ 4x−1

into:4n ≤ (2n)1+

√2n

∏√2n<p≤ 2

3n

p∏

n<p≤2n

p

so we get:

4n ≤ (2n)1+√2n4

23n

or:4

13n ≤ (2n)1+

√2n

which becomes false for large enough n! Using a+ 1 < 2a (which holds ∀a > 2, byinduction) we get:

2n = (6√

2n)6 < (b 6√

2nc+ 1)6 < 26b 6√2nc ≤ 26 6√2n

and thus for n > 50 (and hence 18 < 2√

2n) we obtain from both previous in-equalities:

22n ≤ (2n)3(1+√2n) < 2

6√2n(18+18√2n) < 220 6√2n

√2n = 220(2n)

23

This implies (2n)13 < 20, and thus n < 4000.

An accurate proof can be found here [1].

Part I

Applied combinatorial numbertheory

7

Chapter 2

Goldstream

For the entire following discussion the rows (i0), (i1) and the columns (0j), (1j)are completely cut out, since they are obsolete. We keep them in the matrix Tijas a silence witness, although they are not necessary to secure the results imposedby the definition.

2.1 Miscellanea

We want first to capture some facts about the canonical combinatorial matrix.One that is more obvious is the:

Corollary 2.1. The cardinality |pAij| of the prime lattice points pij with i∧ j ∈ P

in the triangle above the trace (including the cell Aii, but excluding A22) amongthe even numbers Aij for j are given by:

|pAij| =

n∑k=1

k =n2 + n

2

whereby p1 = 3.

Proof. The proof is more than trivial. With j there are n prime numbers pn ≤ j,each of which has a lattice point at i ∈ P. Since i ≤ j an addition to j adds, witheach pn, n−points ((n−1)−points are delivered by the pn−1 prime numbers and oneby n) to it, this leads to the sum |pAij

| = |p1|+ |p2|+ · · ·+ |pn| = 1+2+ · · ·+n.

9

10 § 2 Goldstream

This follows a more complex one:

Corollary 2.2. The cardinality |pAji| of the prime lattice points pij with i∧j ∈ P in

the triangle under the trace (excluding the cells Aii) for j among the even numbersare given by:

|pAji| =

n∑k=1

k =n2 + n

2

whereby p1 = 5.

Proof. The simplest way to see it, is with the construction law (κi)T = κi, afterwhich it follows, that the series is one index behind the sum from corollary 2.1.Otherwise it follows from the exclusion of the cells of the trace, which startstherefore the counting by p1 > pi : i = 3.

Both results can be expressed with the aid of the prime number counting functionπ(n), as follows:

|pAij| =

π(j)−1∑k=1

k =(π(j)− 1)2 + (π(j)− 1)

2

and:

|pAji| =

π(j)−2∑k=1

k =(π(j)− 2)2 + (π(j)− 2)

2

Interestingly enough both are triangular numbers (which comes as no surprise).

Remark 2.3. We want to point out, that the row j = 2 has not to be excludedhere in any of our analysis, simply because it is not possible that it harbours anyprime lattice points among the even numbers, which is a singularity behaviour inthe matrix since 2 is a prime number and pi ∈ P. But j pushes the odd prime −to the next odd value. The same is true for the line i = 2 (obviously). Thus allvalues j = 2 ∧ i ∈ P and i = 2 ∧ j ∈ P are equally odd.

Corollary 2.4. The total amount of lattice points qij with j ≥ 2 in the triangleabove the trace (including the cell Aii) among the even numbers are given by, ifj = even:

|qij| =(j

2

)2or if n = odd by:

|qij| =(j − 1

2

)(j + 1

2

)

2.1 Miscellanea 11

Proof. We have the sequence: 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + . . . which leads to thesum 1, 2, 4, 6, 9, . . . . Those can be constructed on the one hand by the binoms(we start with greater numbers to make the patterns obvious):(

12

1

)j=7

=

(9

1

)+

(3

1

),

(20

1

)j=9

=

(16

1

)+

(4

1

),

(30

1

)j=11

=

(25

1

)+

(5

1

), . . .

and on the other hand by:(16

1

)j=8

=

(12

1

)+

(4

1

),

(25

1

)j=10

=

(20

1

)+

(5

1

),

(36

1

)j=12

=

(30

1

)+

(6

1

), . . .

which leads to:

j = 9⇒ 4 · 5 j = 10⇒ 5 · 5

j = 11⇒ 5 · 6 j = 12⇒ 6 · 6

j = 13⇒ 6 · 7 j = 14⇒ 7 · 7

......

......

These two sequences, constructed by j = even ∨ odd, are represented exactly bythe formulas given at the beginning. Thus the proof is finished.

Remark 2.5. The total amount of prime lattice points pij with i ∧ j ∈ P in thesquare [j 2]− [2 j] among the even numbers is:

|pAij|+|pAji

| =π(j)−1∑k1=1

k1+

π(j)−2∑k2=1

k2 =(π(j)− 1)2 + (π(j)− 1)

2+

(π(j)− 2)2 + (π(j)− 2)

2

(whereby p1 = 3 as before) while the total amount of lattice points is given by, ifj = even:

|qji|+ |qij| = j(j

2− 1)

+ 1

or j = odd:

|qji|+ |qij| =(j − 1

2

)(j − 1

)since the sequence is 1, 1, 3, 3, 5, 5, . . . . If the cell a22 should be included in|pAij|+ |pAji

|, it has only to be added to the formula (+1).

12 § 2 Goldstream

Corollary 2.6. Up to the value A the triangle constructed by those diagonals have(A2− 1)2−lattice points in total. The chain with A ≥ 4 has the length (A− 3).

Remark 2.7. Astonishing about the results is the fact, that the factors ( j−12

) ∧( j+1

2) in the series, as well as ( j

2− 1) ∧ (j − 1) from the second lattice, appear

here, because we recognise the same factors that we have had in the distributionof the values of the combinatorial (cosine − and sine) − and therefore in theDirichlet−series (especially the combinatorial η−function).

2.2 Claim

In this section we want to apply the coloring method, while making use of thecanonical combinatorial matrix, to crack the Goldbach hypothesis.

Theorem 2.8. The Goldbach hypothesis is true.

Proof. After the presumption we have i+ j = A with A being the value of the cellaij. Let N 3 A = even be given. Then A lies in the diagonal:

i ∈{A− 2, A− 3, . . . ,

A

2+ 1,

A

2,

A

2− 1,

A

2− 2, . . . , 3, 2

}j ∈

{2, 3, . . . ,

A

2− 1,

A

2,

A

2+ 1,

A

2+ 2, . . . , A− 3, A− 2

}because all values are equal besides the last one (the value for Ti

=A2 −

A2j=A

2 +A2

is by

definition equal 1), which is uninteresting to us anyway. Now we concentrate ourfocus on the indices. It is crucial to recognise immediately, that they come in pairsof numbers, either even or odd for an even or an odd number!

First we notice that i ∧ j contain all the prime numbers up to: pi ≤ A − 2 ∧pj ≤ A− 2, leading to:

i ∈{pn y1 , pn−1 y2 , . . . , p3 yt p2 yt+1 p1 yt+2

}j ∈

{p1 x1 , pn−1 x2 , . . . , p3 xt p2 xt+1 p1 xt+2

}with the position xl ∧ yl in i ∧ j. Due to the fact that there is no way to computethe gap between two prime numbers, we are forced to use a coloring method.

2.2 Claim 13

Therefore we start with the prime lattice points pij = pi ∧ pj with pi = i ∈ P ∧pj = j ∈ P. Starting with (pi pi) we get the chains:

(4 p1 − 1) (p1 p1) (2 p1 + 1)

(8 p2 + 1) . . . (6 p2 − 1) (p2 p2) (4 p2 + 1) . . . (2 p2 + 3)

(12 p2 + 1) . . . (8 p3 − 1) (p3 p3) (6 p3 + 1) . . . (2 p3 + 5)

......

......

......

...(pi + pi − 2 pi + pi + 2) . . . (pi + 1 pi − 1) (pi pi) (pi − 1 pi + 1) . . . (2 pi + pi − 2)

whereby p1 = 3! Then we have the chains for (pi pj):

(4 p2 − 1) (p1 p2) (2 p2 + 1)

(8 p3 + 1) . . . (6 p3 − 1) (p1 p3) (4 p3 + 1) . . . (2 p3 + 3)

(12 p4 + 1) . . . (8 p4 − 1) (p1 p4) (6 p4 + 1) . . . (2 p4 + 5)

......

......

......

...(2 p1 + pj − 2) . . . (p1 + 1 pj − 1) (p1 pj) (p1 − 1 pj + 1) . . . (2 p1 + pj − 2)

and every combination of (pi pj) with pi ≤ A − 2 ∧ pj ≤ A − 2 where pi ∧ pjis = 3, 5, 7, . . . , p ≤ j, excluding the case pi = pj. We do not need to includeevery combination of (pj pi), because those are already integrated in the chains of(pi pj) due to the construction law (κi)T = κi. Now we use Bertrand’s postulatein two ways: n < pt ≤ 2n ∧ pt < pt+1 < 2pt. Both may not give us an insideabout the distribution of prime numbers, but it sets a maximum distance betweentwo primes. Since pi ≤ j = A− 2 ∧ pj ≤ j = A− 2, we have created a web, whichincludes every diagonal. This follows from the coloring method, that specifies thedistance of the chains simply by pj−pj−1 with every combination of it and corollary2.1. Thus independently every possible distance is constructed for the chains andincludes every line due to the combinations of (pi pj), because after corollary 2.1each prime number pn in j includes not only n prime lattice points, but due tocorollary 2.6 it covers moreover n diagonals.

To see the covering of all diagonals, we have two possibilities. First, with j we havealso (j − 1)−diagonals. This is the same − as the amount of different numbers|A| inside the square [j 2] − [2 j]. But at the same time we have also pn ≤ j

14 § 2 Goldstream

prime numbers, which cover σ =∑π(j)−1

k=1 k = (π(j)−1)2+(π(j)−1)2

diagonals. We have:|A| � σ.

Second we can construct a method so each number = even is paired with twoprime numbers. We say Aij = 2k = n = even is our starting point, whereby itconsists already of a prime lattice point n 3 pij. An addition from:

i ∈{

2, 3, . . . ,n

2,

n

2− 1, . . . , n− 2

}i ∈{n− 2, n− 3, . . . ,

n

2,

n

2+ 1, . . . , 2

}with two, whereby pj = n− 1 ∈ P leads to:

i ∈{

2, 3, . . . ,n

2,

n

2+ 1, . . . , n

}j ∈

{n, n− 1, . . . ,

n

2,

n

2+ 1, . . . , 2

}where n+2 3 pij automatically. It follows that the numbers in the sequence n+4,n+ 6, n+ 10, n+ 12, . . . , which is represented by n+ (pi − 1) consist all a primelattice point, leading to the sequence n = nk ≥ nk−1+(pj−1−1), nk−1+(pj−i+2−1),. . . , nk−1 + (p1 − 1), all of which are actually different numbers (at least in thepurely greater case) and therefore points. This evokes the formula:

(pj + 1)︸ ︷︷ ︸even

+ (pi − 1)︸ ︷︷ ︸even

= n︸︷︷︸even

which leads to:pj + pi = n

even if it were:(pj − 1) + (pi − 1) = n

Thus the proof is finished.

Remark 2.9. It is false to assume that there are holes due to that the first coveringsequence is given by n+4, n+6, n+10, n+12, . . . . The so called holes are closedby pj−1 ∨ pj−2 ∨ . . . , because their distance is definitely ≤ n, ≤ n

2, . . . , meaning

that even if pj−1 does not cover n+ 8, n+ 20, . . . , one of pj−t, with t = 2, 3, . . . ,i, will. Further after ≤ n−steps there is a new prime number included in j anda new sequence starts from the beginning, while lastly up to n, from p2 = 5, allthe even numbers started an equal sequence, making an overlapping unavoidable inwhich holes can not accumulate in an isolation.

2.3 Density 15

Remark 2.10. The result is not remarkable due to that it was expected. Whatthough is astonishing, is the clarity and the simplicity of the proof. Moreover, it issurprising, that the result does not depend on any distance of the prime numbersand their appearance, otherwise xl ∧ yl would have to appear in the proof at adifferent point than that one time.

Remark 2.11. It is important to mention that with the aid of Bertrand’s postulate,it follows actually that with p ≥ 5 the prime numbers pj ∧ pi in pj + pi = n areactually different due to for (ij) ≤ (ii) a prime number in j appears that is not tobe found in the sequence i.

Remark 2.12. Another way of understanding the proof is, that the coloring methodallows us to specify the distance of any chain, which is pj − pj−t, with t = 2, 3,. . . , i. Thus, we are able to create every possible distance that we want for thechain and include every line due to the combinations of pi pj. Further in eachrow pj, the distance of pi to pi+1 is equal and constant, but in pj−1 this primenumber gap (n < p ≤ 2n = k) moved down ≤ (k + 1), while in pj+1 it moved up≤ (k + 1); with it also the interval n < p ≤ 2n moves ≤ ±(k + 1). The primenumber gap ≤ 2n between each pj and pj+1 conducts the movement and secures anoverlaying, because in each row pj, pi starts from the beginning with p1. ThereforeApi pj stands in ai+(k+1) pj(k+1)

and the other way around or in other words, aij isoverlayed by Apipj whereby pi ≤ i ± (k + 1) ∧ pj ≤ j ± (k + 1) is creating a web,since for j → A − 3 the distance for the next prime number pj is ≤ (k + 1), but

for pi it is less than (k+1)2

< pi ≤ (k + 1).

Remark 2.13. To be particular thorough we give here another possibility for aproof. The diagonals with the value A has the length (A− 3) (from the trace it is(A2− 1)). So, in order to secure that there is no prime lattice point in between, a

prime number should not appear at least in the interval above the trace: A2− 2 <

j < A − 3, which, as we can see, can not hold, if A ≈ 2n due to that the totallength is (A − 3) ≈ 2n and in the interval 3 < j ≤ A

2− 2 is still one prime to be

found. Thus the increase of the length of the chain secures the proof and is onlyan upper border, which makes the distance between prime numbers obsolete.

Remark 2.14. If we take the relation pk < C · p0,535k < pk+2 for the gap of primeswith a constant C, the proof becomes much more easy and clear.

2.3 Density

The relation between the total amount − and the prime lattice points pij on theone hand and the actual numbers in each square on the other hand seems very

16 § 2 Goldstream

interesting. We had with i ∧ j ∈ P in the square [j 2] − [2 j] among the evennumbers:

P =

π(j)−1∑k1=1

k1 +

π(j)−2∑k2=1

k2 =(π(j)− 1)2 + (π(j)− 1)

2+

(π(j)− 2)2 + (π(j)− 2)

2

(whereby p1 = 3 as before) as the total amount of prime lattice points, while thetotal amount of lattice points is given by, if j = even:

Q = |qji|+ |qij| = j(j

2− 1)

+ 1

or j = odd:

Q = |qji|+ |qij| =(j − 1

2

)(j − 1

)although we have only |A| = j − 1 different numbers. So despite the fact, that wehave:

Q� P

we have also:P � |A|

which is astonishing. But this is what secures the proof, since the density of theprime lattice points are incredible more dense, than the actual amount of evennumbers presented in the same interval in the canonical combinatorial matrix.

Example 2.15. For j = 106 we have only π(j) = 78498−prime numbers, follow-ing:

P =784972 + 78497

2+

(78497− 1)2 + (78497− 1)

2= 6.161.779.009

guaranteed prime lattice points, but only:

|A| = 106 − 1 = 999.999

different even numbers, while the entire lattice has:

Q = 106(106

2− 1)

+ 1 = 499.999.000.001

points. At the same time we have:

D =784972 + 78497

2= 3.080.928.753

diagonals covered.

Bibliography

[1] Martin Aigner and Gunter M. Ziegler, Proofs from THE BOOK, vol. 1, 1990.

17