AN APPLIANCE APPLIANCE OF COMBINA COMBINATORIA TORIAL L NUMBER NUMBER THEORY THEORY M. N. G. EINSTEIN Based on the constructed combinatorial number theory by M. N. G. Einstein (work still in revision) we want to find a different application than the combinatorial − combinatorial − and Dirichlet Dirichlet series. series. While While the roots of the construction construction may lie in the combinatorial number theory, it is completely separated from − and can be derived without it. Therefore we will show this latter case. Abstract.
Contents
1. Canonical combinatorial Matrix 2. Bertrand’s postulate 3. Goldstream 3.1. Miscellanea 4. Claim 5. Prime Canonical Combinatorial Matrix 6. Density References
1. Canonical
1 4 7 7 10 15 17 18
combinatorial combinatorial Matrix Matrix
Before we can start any analysis we need a (triangular) matrix with special conditions. Definition 1.1. Let T = T ij ij be the canonical combinatorial matrix:
T ij ij =
with the following properties:
a00 a10 a20
a01 . . . a11 . . . a21 . . .
.. .
.. .
an 0
an1 . . .
..
.
a0n a1n a2n
...
ann
∀i = 0 : a 0 := 1 ∧ ∀ j ≥ 1 : a0 i
j
:= 1
i = 0 : a i0 := 0
and: i T
= κ i with the column κ and row vectors κi (whereby the last condition depends on the length of the taken chain if the matrix T ij ij is defined differently). For our purpose i
−
∀i ≥ 1 ∧ ∀ j ≥ 1 : (κ )
1
2
M. N. G. EINSTEIN
the values aij are:
T i≥1 j≥1 =
κ1j = κ2j = κ3j =
∀ a
ii = T ii ii : a ii is
2 3 4 .. .
.. . κnj =
3 4 5 .. .
4 5 6 .. .
n + 1
n
... ... ...
n n + 1 n + 2
..
... n + n
.
n + 2 . . .
With that definition the canonical combinatorial matrix T ij ij has some amazing conseq consequen uences ces.. The most importan importantt one is that that the elements elements of the trace T ii ii are even and are constructed by the indices i: even
∧a
ii = i + i
The proof of that is trivial, since a11 = 2 urther, ai+1 i+1 = i + 1 + i + 1. Further, the diagonales diag(aij ) constructed by (i 1, j + 1) from from a i≥2 j ≥1 to a i=1 j with j i (= A 3, whereby A is the value of the cell), have = max to the starting point of i always the same value. Moreover they are constructed by the indices as well! The proof of that is again trivial. trivial. It follows follows in its simplest simplest form from the third condition condition i T 1 1 : (κ ) = κi . That That the valu valuee of aij in those diagonals diag(aij ) i j ai−1 i+1 = are constructed by the indices can be shown easily with aii = i + i i + j = diag(aij ).
−
⇒ −
∀ ≥ ∧ ∀ ≥
⇒
We want to add here, that in combinatorial number theory the matrix above follows from: Definition 1.2. Let
n , p ∈ n
N
form:
i
i
and κ i represent a multiplication chain in the
k
= k
= some some cons consta tan nt p
κ1
= κ1
= n3 , n4 , n5 , . . . , nn
κ2
k1
κ3
=
=
k1
.. .
·
k2
k2
· ·
=
k3
=
k1 k2
= n3 , n3 , n4, . . . , n(n−1) = n4 , n5 , n6, . . . , nn
k1 k2 k3
= n3 , n4 , n5 , . . . , n(n−2) = n4 , n5 , n6 , . . . , n(n−1) = n5 , n6 , n7 , . . . , nn
| | k ∩ k 1 ∩ k 2 ∩ k3 ∩ · · · ∩ k
when i increases, the length of nn decreases, with the condition:
We call then:
i
=
ki =
i
∅
i C ij ij (κ )) i a combinatorial function, where C ij ij (κ ) means the j th combination of the chain i (κ ) with: C 1j (κ1 )
−
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
3
simply representing the numbers of κ1 = n 3 , n 4 , n5 , . . . , n n . Therefore C 1j (κ1 ) returns successively n3j=1 , n4j=2 , n5j =3 , . . . , n nj =|n|−1 , while C i0 (κi ) delivers a one: i
and:
i
∀i = 0 : C 0(κ ) := 1 i = 0 : C i0 (κi ) := 0
i We see, that C ij ij (κ ) delivers exactly the values T ij ij .
4
M. N. G. EINSTEIN
2.
Bertrand’s Bertrand’s postulate postulate
between every every 1 1 Theorem 2.1 (theorem of Bertrand) . In between there lies at least one prime number N:
≤ n ∈ N and its double
∀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 < 2 pi−1 . Hence every every interv interval al pi : n < pi n 4000 contains one of these pi prime numbers.
≤
With the identity:
≤ 2n}, with
x ≥ 2 : p ≤ 4 −1 ≤ If one notes now that for largest P p ≤ x , then it is: p = p an and 4 −1 ≤ 4 −1 R
x
p x
P
j
≤
pj
x
≤
p x
p pj
Therefore it is enough to check for the proposal the case x = p j P. By induction we 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: 2n + 1 p = p p 4n 4 n 22n = 42n
∈ {
∈
p 2n+1
≤
p n+1
≤
n+1
≤
}
≤
≤
n
This follows from three things. First due to the holdings of:
≤ 4
p
p n+1
≤
n
by induction; secondly from the observation that: 2n + 1 (2n + 1)! = n n!(n + 1)! is an integer, where the primes we consider are all factors of the numerator (2 n +1)! and only the numerator. Thus follows the inequality: 2n + 1
≤ ≤ p
n+1
≤
and finally the inequalit inequality: y:
2n + 1 n
holds due to:
2n + 1 n
and
n
2 2n
2n + 1 n + 1
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
5
are two equal summands that appear in the series: 2n+1
2n + 1
= 22n+1
k
k=0
For the next step we need Legendre’s theorem. Theorem 2.2 (theorem of Legendre) . The factorial n! contains the prime factor p exactly: n pk
k 1
≥
times.
Proof. Exactly pn of the factors of n! are divisible by p, which accounts for pn p factors. factors. Next, p n2 of the factors of n! are even divisible by p2 which accounts for the next p n2 prime factors p of n!. By continuing this process follows the sum n k≥1 pk .
−
With Legendre’s theorem we get that the binomial coefficient tains the prime factor p exactly:
− − − − 2n
2
pk
k 1
≥
2n n
=
(2n)! conn!n!
n pk
times. Each summand in the series is 1 due to the satisfaction of: n n 2n 2n 2 k < k 2 k 1 =2 k p
p
p
p
Furthermore the summands vanish whenever pk > 2 n. Thus the binomial coefficient 2n contains p exactly: n
− ≤ 2n
k 1
≥
2
pk
n pk
max v : p v
{
≤ 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:
√
≤ ≤ √ √ ≤ ≤ n
2n
>
n
n
2
for n > 3, we get: 4n 2n
2n n
√ p≤ 2n
2n
and thus, since there are not more than 4n
2 n 3
2n,
p
√ n
2 n 3
≤ ≤
p
n
2n primes p
√ 1+ 2n
(2 n)
p
√ 2n
p
n
≤
for n > 3. If there there is no prime p with n < p 2 n the second product is equal 1. Now we substitute: p 4 x−1 2 : R x
≥
P
p≤x
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M. N. G. EINSTEIN
into: 4n
√ 2n
≤ (2n)1+
p
√ 2n
so we get:
2 n 3
√ n
≤ (2n)1+ 2 4 √ 4 ≤ (2 n)1+ 2
4n or:
1 n 3
p
n
≤
2 n 3
n
which becomes false for large enough n! Using a + 1 < 2 a (which holds a > 2, by induction) we get:
√
∀
√ √ √ 2n√ + 1)6 < 26 2 ≤ 26 2 and thus for n > 50 (and hence 18 < 2 2n) we obtain from both previous inequal2n = ( 2n)6 < ( 6
ities: 22n
≤ (2n)3(1+
√ 2n)
< 2
6
√ 2n(18+18√ 2n)
6
1
This implies (2n) 3 < 20, and thus n < 4000. An accurate proof can be found here [1].
6
< 2 20
n
√ 2n√ 2n
6
6
n
2
= 220(2n) 3
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
7
3. Goldstream For the entire following discussion the rows ( i0), (i1) and the columns (0 j ), (1 j ) are completely cut out, since they are obsolete. We keep them in the matrix T = T ij ij as a silence witness, although they are not necessary to secure the results imposed by the definition. definition. 3.1. Miscellanea. We want first to capture some facts about the canonical combinatorial matrix. One that is more obvious is the: Corollary Corollary 3.1. The cardinality p pAij of the prime lattice points pij with pij := pi pj P in the triangle:
∧ ∈
|
|
[2 2]
[2 j ]
[ j j j ] A 22 ) among the even numbers above the trace (including the cells A ii , but excluding A Aij for j is given by: n n2 + n p pAij = k = 2
|
|
whereby p1 = 3.
k=1
Proof. The proof is more than trivia trivial. l. With With j there are (n 1) prime numbers pn−1 j , each of which has a lattice point at pi j an addition Sincee i P. Sinc to j adds, with each new prime number pn 1) points are P, n points ((n delivered by the pn−1 prime numbers and one by pn ) to it, this leads to the sum p pAij = p p1 + p p2 + pn = 1 + 2 + + p + n, which can be computed by the above given formula.
≤
|
∈ ∈ − ···
| | | | | ··· | |
−
≤ − −
This follows a more complex one:
|
|
pAji of the prime lattice points pij with pij := Corollary Corollary 3.2. The cardinality p pi pj P in the triangle:
∧ ∈
[2 2]
[ j j 2] j j ] [ j under the trace (excluding the cells Aii ) for j among the even numbers is given by: n
p pAji =
|
|
whereby p1 = 5.
k =
n2 + n
k=1
2
Proof. The simplest way to see it, is with the construction law (κi )T = κi , after which it follows, that the series is one index behind the sum from corollary 3.1. Otherwise it follows from the exclusion of the cells of the trace, which starts therefore p j : j = 3 p1 = 5. the counting by p 1 > pi : i = 3
⇐⇒ ⇐ ⇒
⇒
8
M. N. G. EINSTEIN
Both results can be expressed with the aid of the prime number counting function π (n), as follows: π(j ) 1
p pAij =
|
|
and:
−
k =
|
|
− 1)2 + (π( j ) − 1) 2
k=1
π(j ) 2
p pAji =
(π( j )
−
k =
(π( j )
− 2)2 + (π( j ) − 2) 2
k=1
Interesti Interestingly ngly enough both are triangular numbers numbers (which (which comes as no surprise). surprise). Remark Remark 3.3. We want to point out, that the row i = 2 has not to be excluded
here in any of our analysis, simply because it is not possible that it harbours any prime lattice points among the even numbers, which is a singularity behaviour in the matrix since two is a prime number and p i P. But j pushes the odd prime to the next next odd value. value. The same is true for the column column j = 2 (obviously). (obviously). Thus, Thus, all values j = 2 p i P and i = 2 pj P are equally odd.
∈
∧ ∈
−
∧ ∈
≥ 2 in the triangle:
Corollary 3.4. The total amount of lattice points q ij ij with j
[2 2]
[2 j ]
j j ] [ j above the trace (including the cells Aii ) among the even numbers are given by, if j is even: j 2 q ij ij = 2 or if j is odd then by: j 1 j + 1 q ij ij = 2 2 Proof. We have have the sequence: sequence: 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + . . . which leads to the sum 1, 2, 4, 6, 9, . . . . Those values 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 3 20 , + 1 1 1
and on the other hand by: 16 12 4 25 = + , 1 j =8 1 1 1
=
j =9
=
j =10
16 4 30 , + 1 1 1
20 5 36 + , 1 1 1
=
j =11
=
j =12
which leads to:
⇒ 4 · 5 j = 11 ⇒ 5 · 6 j = 13 ⇒ 6 · 7 j = 9
.. .
.. .
⇒ 5 · 5 j = 12 ⇒ 6 · 6 j = 14 ⇒ 7 · 7 j = 10
.. .
.. .
25 5 ,... + 1 1 30 6 + ,... 1 1
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
9
These two sequences, constructed by j when j is either even or odd, are represented resented exactly by the formulas formulas given given at the beginning. beginning. Thus Thus the proof is completed. Remark 3.5. The total amount of prime lattice points p ij with p ij := p i
in the square: [2 2]
[2 j ]
j 2] [ j
∧ p ∈ P j
j j ] [ j
among the even numbers is: π (j ) 1
|
| |
|
p pAij + p pAji =
−
π (j ) 2
−
k1 =1
k1 +
k2 =
k2 =1
(π( j )
− 1)2 + (π( j ) − 1) + (π( j) − 2)2 + (π( j ) − 2) 2
2
(whereby p 1 = 3 as before) while the total amount of lattice points is given by, if j is even: j q ji 1 +1 ji + q ij ij = j 2 or if j is odd: j 1 q ji j 1 ji + q ij ij = 2 since the sequence is 1, 1, 3, 3, 5, 5, . . . . If the the cel celll a22 should be included in p pAij + p pAji , it has only to be added to the formula (+1).
| | | |
| | | |
|
| |
− − −
|
Corollary 3.6. Up to the value A the triangle constructed by those diagonals have (A 1)2 lattice points in total. The chain with A 4 has the length (A 3). 3). 2
− −
≥
−
1 ) Remark Remark 3.7. Astonishing about the results is the fact, that the factors ( j − 2 j +1 j ( 2 ) in the series, as well as ( 2 1) ( j 1) from the second lattice, appear
∧
− ∧ − here, here, because because we recogn recognise ise the same same factor factorss that that we have have had in the distribu distribu-tion of the values of the combinatorial (cosine − and sine) − and therefore in the Dirichlet−series (especially the combinatorial η −function).
10
M. N. G. EINSTEIN
4. Claim In this section we want to apply the coloring method, while making use of the canonical combinatorial matrix to crack the Goldbach hypothesis. Theorem 4.1. The Goldbach hypothesis is true.
Proof. After the presumption we have i + j = A with A being the value of the cell aij . Let N A and be even. Then A lies in the diagonal:
∈ − ∈
i
A
j
2,
A
2,
− 3,
...,
3,
...,
A
2
+ 1,
A
− 1, 2
A
A
A
2
2
2
,
− 1,
A
A
2
2
,
+ 1,
− 2,
A
2
...,
3,
+ 2, . . . , A
− 3,
− 2
A
2
since all values are equal besides the last one (the value for T i= A − A j = A + A is 2 2 2 2 by definition definition equal 1), which which is uninteresting uninteresting to us anyway anyway. Now we concentrate concentrate our focus on the indices. indices. It is crucial crucial to recognise recognise immediately immediately,, that they come in pairs of numbers, either even or odd!
∧
First, we notice that i j contain all the prime numbers up to: pi pj p n A 2, leading to:
∧ ≤ ≤ −
∈
∈
i
j
p1 ,
p2 ,
...,
pn−2 ,
pn−1 ,
pn
p1 ,
p2 ,
...,
pn−2 ,
pn−1 ,
pn
≤ p ≤ A − 2 n
Due to the fact that there is no way to compute the gap between two prime numbe numbers, rs, we are forced forced to use a colori coloring ng method. method. Theref Therefore ore,, we start start with with the prime lattice points pij := p i pj with pi P and pj P. Beginning with ( pi pi ) we get the chains (diagonals):
∧
∈
− 1) (6 p2 − 1) (8 p3 − 1)
(4 p1 (8 p 2 + 1)
...
(12 p2 + 1)
...
.. . ( pi + pi
.. .
− 2 p + p + 2) i
i
...
.. . ( pi + 1 pi
− 1)
∈
( p1 p1 )
(2 p1 + 1)
( p2 p2 )
(4 p2 + 1)
...
(2 p2 + 3)
( p3 p3 )
(6 p3 + 1)
...
(2 p3 + 5)
.. .
.. .
.. . (2 pi + pi
.. . ( pi pi )
( pi
− 1 p + 1) i
...
whereby p1 = 3. In each row row (diagonal (diagonal)) the value value is the same. Then, Then, we have have the chains for ( pi pj ): (4 p2 (8 p 3 + 1)
...
(12 p4 + 1)
...
− 1) (6 p3 − 1) (8 p4 − 1)
( p1 p2 )
(2 p2 + 1)
( p1 p3 )
(4 p3 + 1)
...
(2 p3 + 3)
( p1 p4 )
(6 p4 + 1)
...
(2 p4 + 5)
− 2)
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
.. . (2 p 1 + pj
.. .
− 2)
.. . ( p1 + 1 pj
.. .
− 1)
.. . ( p1
.. .
.. . (2 p1 + pj
− 1 p + 1) . . . − 2) p ) with p ≤ p ≤ A − 2 ∧ p ≤ p ≤ A − 2 where
...
( p1 pj )
11
j
and every combination of ( pi j i n j n pi and p j is equal to 3, 5, 7, . . . , p n , excluding the case p i > T ii . We do not need to ii include every combination of ( pj pi ), because those p i ’s under the trace are already integrated in the chains of ( pi pj ) due to the constru constructi ction on law: (κi )T = κi . No Now w we use Bertrand’s postulate in two ways:
∀ p ≤ A − 2 : ∀ p ≤ A − 2 : t
k
<
pt
≤ 2k
t
pt
<
pt+1
< 2 pt
Both may not give us an inside about the distribution of prime numbers, but it sets a maximum distance between two primes. Either condition limits the distance of primes in j and i, while the second one sets the maximal distance of the points within the prime lattice. Since pi p n i = j = A 2 p j p n j = i = A 2 pt+1 < 2 pt , we have and pt have created a web which which includes every diagonal. diagonal. This follows from the coloring method, that specifies the distance of the chains simply by pj pj −1 with every combination of it and corollary 3.1. Thus, independently every possible distance is constructed for the chains and includes every line due to the combinations of ( pi pj ), because after corollary 3.1 each prime number p ν p n in j includes not only ν prime lattice points, but due to corollary 3.6 it covers moreover ν diagonals.
≤ ≤
≤
− ∧ ≤ ≤
−
−
≤
To recognise recognise that all diagonals are covered, covered, we have have two two possibilities possibilities.. First, First, with j we have also ( j 1) diagonals. diagonals. This is the same same as the amount of different numbers A within the square: [2 j ] [2 2]
| |
− −
−
[ j j 2] j j ] [ j pn prime numbers, which cover: But at the same time we have also p
| |
π (j ) 1
σ =
−
k =
(π ( j )
2
k=1
diagonals. We have: A
− 1)2 + (π( j) − 1)
| | σ , since after the prime number theorem it is: π (j ) 1
σ
=
−
k
k=1
=
(π ( j )
− 1)2 + (π( j) − 1) 2
12
M. N. G. EINSTEIN
∼
2
j
log( j )
− 1
+
j
log( j )
2
− 1
Second, we can construct a method so each number that is even is paired with two two prime numbers. numbers. We say Aij = 2k = n, which is even, is our starting point, whereby it consists already of a prime lattice point n p ij . An addition from:
∈ − 2,
∈
i i
n
3, 2, n
with two, whereby pj = n
∈
∈
i
j
where n + 2
p
n
...,
− 3,
2 n
...,
2
n
,
− 1, 2
n
,
2
− 1 ∈ P, leads to:
2,
3,
n,
n
n
...,
− 1,
2 n
...,
ij automatically.
2
,
n
2 n
,
2
...,
n
+ 1,
...,
2
+ 1,
...,
n
+ 1,
...,
2
−2
It follows that the numbers in the sequence:
n + 4, n + 6, n + 10, n + 12, . . .
which is represented by:
− 1)
n + ( pi
consist all a prime lattice point, leading to the sequence:
≥ n −1 + ( p −1 − 1), n −1 + ( p − +2 − 1), . . . , n −1 + ( p1 − 1)
n = n k
k
j
k
j i
k
all of which are actually different numbers (at least in the purely greater case) and therefore therefore points. This evokes evokes the formula:
− ( pj + 1) + ( pi even
which yields:
1) = n
even
even
pj + pi = n
even if it were:
− 1) + ( p − 1) = n
( pj or:
i
( pj + 1) + ( pi + 1) = n it would still be correct as n is even. Thus, the proof is completed.
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
13
Remark 4.2. The use of Bertrand’s postulate in the two ways:
∀ p ≤ A − 2 : ∀ p ≤ A − 2 : t
k
<
pt
≤ 2k
t
pt
<
pt+1
< 2 pt
has not been utilized simply as a diameter between two primes, but as a sequence of diameters diameters between the primes primes pt p n as indicated by the indice t and since the prime numbers never end, so does not the sequence for n . Especia Especially lly the the second second condition condition is important important as it guarantee guaranteess that the Goldbach Goldbach hypothesis hypothesis holds due to the inequality pt < pt+1 < 2 pt . This This shows shows that in the matrix matrix a diagon diagonal al cannot exist which is not covered by one prime lattice point as a result of the length of the diagonal, which is always longer than the maximal distance of twice the length of a prime.
≤
→ ∞
Remark 4.3. It is false to assume that there are gaps, due to that the first covering sequence is given by n + 4, n + 6, n +10, n +12, . . . . The so called holes are closed by p j −1 p j −2 . . . , because their distance (with the next prime in j ) is definitely n n, , . . . , meaning that even if pj −1 does not cover n + 8, n + 20, . . . , one 2 of pj −t , with t = 2, 3, . . . , i, will. Further, urther, after n steps there is a new prime number included in j and a new sequence starts starts from the beginning, beginning, while lastly up to n , from p2 = 5, all the even numbers start an equal sequence of primes, making an overlapping unavoidable in which gapes cannot accumulate in an isolation.
≤ ≤
∨
∨
≤ −
result is not remark remarkabl ablee due to that it was was expected. expected. What What Remark Remark 4.4. The result though is astonishing, is the clarity and the simplicity of the proof. Moreover, it is surprising, that the result does not depend on any distance of the prime numbers and their appearance, otherwise some diameter would have appeared in the proof at some point. Remark 4.5. It is important to mention, that with the aid of Bertrand’s postulate, it appears to follow, that with p 5 the prime numbers pj pi in pj + pi = n are actually different due to for ( ij ) ( ii) a prime number in j appears that is not to be found in the sequence i.
≥ ≤
∧
Remark 4.6. To be particular thorough we give here another possibility for a proof. The diagonals with the value A has the length (A 3) (from the trace it is ( A 1)). 2
−
−
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 < j < A 3, which, as we can see, cannot hold, if A 2 n due to that the total length is ( A 3) 2 n A and in the interval 3 < j is still one prime to be found. Thus, the increase of 2 the length of the chain secures the proof and is only an upper border, which makes the distance between prime numbers obsolete.
≤
≈
− − ≈
14
M. N. G. EINSTEIN
· ·
0,535
< pk+1 for the gap of primes Remark 4.7. If we use the relation pk < C pk with a constant C , the proof becomes much more easy and clear, especially if we
use the prime canonical canonical combinator combinatorial ial matrix of the next paragraph. paragraph. We want to show it by contradic contradiction. tion. Hence, Hence, we can assume without loss of generalit generality y that the gap appears between pk and pk+1 . Sinc Sincee pk evoks k and pk+1 (k + 1) prime lattice points, this covers k, respectively ( k + 1) diagonals in the columns ( pi pj ), where pj p k pk+1 . Thus, Thus, a gap can only arise under the diagonal diagonal diag( pk pk ), which is 2 pk long; long; and over over the diagonal diagonal diag( pk+1 pk+1), which is 2 pk+1 long. Further, between diag( pk pk ) and diag( p1 pk+1 ) a prime number cannot appear either after the assumption. assumption. This cannot hold as the distance between between the columns columns ( p1 pk ) and ( p1 pk+1) is bounded by p1 + pk , respectively p1 + p k+1 and neither by pn < pn+1 < 2 pn nor by pk < C p0k,535 < pk+1 . Either Either ineq inequal ualit ity y only determines the distance between pk and pk+1 . Since, Since, only even even numbe numbers rs appear in the prime canonical combinatorial matrix, the maximal distance between ( p1 pk ) and ( p1 pk+1 ) is < pk . This This stands stands in contrad contradict iction ion to the needed needed length length of the diagonal, which should be > 2 pn long. After the inequality pk < C p0k,535 < pk+1 and the construction of the prime canonical combinatorial matrix follows therefore, that diag( pk pk ) lies before the column of ( p1 pk+1 ). Hence, Hence, a gap cannot cannot appear. appear. Thus, the proof is complete.
−
∈ ∨ ≥ −
−
−
≥
· ·
· ·
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
5. Prime
15
Canonical Combinatorial Combinatorial Matrix Matrix
Definition 5.1. Let P = P p pi pj be the prime canonical combinatorial matrix:
P pi pj =
with the following properties: a pi pj =
and:
a p1 p1
a p1 p2 . . . a p2 p2 . . .
..
a p1 pn a p2 pn
...
.
a pn pn
2n if
pi + pj = 2n
∈ N
0 otherwise i T
∀ p ≥ 3 = p1 ∧∀ ∧ ∀ p ≥ 3 = p1 : (κ )
= κ i with the column κ and row vectors κi (whereby the last condition depends on the length of the taken chain if the matrix P ij ij is defined differently). For our purpose the values aij are: i
i
−
j
− 6
P pi pj =
− 14 − 16 − 18 − − − −
8 10
10 12 14
... ... ... ...
2n 2(n + 1) 2(n + 2)
..
... 2(n + n)
.
−
With the above definition the prime canonical combinatorial matrix P pi pj has some amazing consequences. The most important one is that all elements and not only the one of the trace P pi pi are even and are constructed by the indices p i pj :
∀a
pi pj
= P p pi pj : a pi pj is even and a pi pj = p i + pj
The proof of that is trivial, since a p1 p1 = 6 a pi+1 pi+1 = p i+1 + pi+1. Further, the values of the diagonales diag(aij ) constructed by ( pi−1 , pi+1) from a i≥2 j ≥1 to a pi pi are always always the same. Moreov Moreover, er, they are constructed constructed by the indices indices as well! well! The proof of that is again trivial. It follows in its simplest form from the definition of the matrix. That the value of a a pi pj in those diagonals diag(aij ) are constructed by the indices can be shown easily with a pi pi = p i + pi a pi−1 pi+1 = p i + pj = diag(aij ). It is new that the matrix has empty rows and columns, which we have chosen, thus all values in each diagonal match each other.
⇒
⇒
Another characteristics of the prime canonical combinatorial matrix, is that, all appearing a pi pj are prime lattice points, since pi , pj P. This This makes makes the matrix matrix very uniqe.
∈
One very big difference to the canonical combinatorial matrix, is that, the values of the prime canonical combinatorial matrix are all even numbers as a result of a pi pj = p i + pj . Further, prime twins, prime triplets and so forth appear next to each other in the matrix matrix,, which which is very very handy handy.. Thus, Thus, when when two, two, three, three, four, . . . , prime numbers numbers emerge in neighbouring rows pj , pj +1 , then we know immediately they are prime twins, triplets etc.
16
M. N. G. EINSTEIN
Theorem 5.2. The Goldbach hypothesis is true.
Proof. We will show show it by contradi contradicti ction. on. Assumi Assuming ng the Goldba Goldbach ch hypothesi hypothesiss is false, it follows that there exists at least one diagonal diag( ij ) in the prime canonical combinatorial matrix:
P pi pj =
a p1 p1
a p1 p2 . . . a p2 p2 . . .
..
.
a p1 pn a p2 pn
...
a pn pn
that is completely completely empty empty. Without Without loss of generality generality we can presume that the gap appears between p n and p n+1. Since p n evoks n and pn+1 ( n + 1) prime lattice points, this covers n, respectively (n + 1) diagonals in the columns ( p1 pj ), where pj pn pn+1 . Thus, Thus, the gap can therefore therefore only arise under under the diagonal diag( pn pn ) and above the diagonal diag( p1 pn+1 ). The value of the cells of all three diagonals are given by the indices, which are:
−
∈ ∨
−
−
A pn pn
= p n + pn
Aij
= p n + pn + 2 = 2 pn + 2 (the next even number)
A p1 pn+1
= p 1 + pn+1
−
= 2 pn ,
= 3 + pn+1
The assumption that Aij = pn + p n + 2 > A p1 pn+1 = pn+1 + p 1 cannot hold as the distance between ( p1 pn ) and ( p1 pn+1) is bounded by p1 + p n , respectively p1 + pn+1 and not by pn < pn+1 < 2 pn nor by n < p < 2 n. Either inequality only determines the distance between pn and pn+1 . However, the inequality: pn + 2 < pn+1 + 2 < 2 pn + 2
≤ is odd
yields:
is odd
pn+1 + 3 = p n+1 + p1 is even
is even
2 pn + 2 is even
since pn+1 + 2 is odd, while 2 pn + 2 is even even.. This This stan stands ds in cont contra radi dict ctio ion n to the needed value value of the diagonal diagonal in order to be empty. empty. After the construction construction of the prime canonical canonical com combinato binatorial rial matrix follows follows therefore, therefore, that diag(ij ) lies in diag( p1 pn+1 ). Hence, a gap cannot appear. Thus, the proof is complete.
≤
AN APPLIANCE OF COMBINATORIAL NUMBER THEORY
17
6. Density The relation between the total amount and the prime lattice points p ij on the one hand, and the actual numbers in each square on the other hand, seems very interesting. We had with pi pj P in the square: [2 j ] [2 2]
−
∧ ∈
j 2] [ j
j j ] [ j
among the even numbers: π (j ) 1
P =
−
π (j ) 2
k1 +
k1 =1
−
k2 =
(π ( j )
− 1)2 + (π( j ) − 1) + (π( j ) − 2)2 + (π( j) − 2) 2
k2 =1
2
(whereby p1 = 3 as before) as the total amount of prime lattice points, while the total amount of lattice points is given by, if j is even:
− − −
Q = q ji ji + q ij ij = j
| | | |
or j is odd:
j
2
1 +1
| | | | j 2 1 j 1 although we have only |A| = j − 1 different numbers. So despite the fact, that we have: Q P we have also: |A| P Q = q ji ji + q ij ij =
which which is astonishing. astonishing. But this is what secures the proof, since the density density of the prime lattice points are incredible more dense, than the actual amount of even numbers presented in the same interval in the canonical combinatorial matrix.
Example 6.1. For j = 106 we have only π ( j ) = 78498 prime numbers, following:
− 784972 + 78497 (78497 − 1)2 + (78497 − 1) P = + = 6 .161.779.009
2 2 guarantee guaranteed d prime lattice p oints, oints, but only: A = 106
| |
− 1 = 999.999
different even numbers, while the entire lattice has:
106 1 + 1 = 499.999.000.001 2 points. At the same time we have: Q = 106
D =
diagonals covered.
−
784972 + 78497 = 3 .080.928.753 2
18
M. N. G. EINSTEIN
As an overview, we have for j = x and π ( j ) = π (x) the following table: j π ( j ) A D ( j )
| |
| |
A D(x)
10 4 9 6 1, 5
100 25 99 300 0, 33
1.000 168 999 14.028 0, 071214 . . .
10.000 1.229 9.999 754.606 0, 013250 . . .
100.000 9.592 99.999 45.998.436 0, 002173 . . .
1.000.000 78498 999.999 3.080.928.753 0, 000324 . . .
10.000.000 664579 9.999.999 220.832.291.331 0, 000045 . . .
| |
where A are the different even numbers and D is the amount of the covered diagonals. This yields:
| |
A D (x)
∼
0
We can see how the amount of the prime lattice points is exploding, although after the prime number theorem we have for the amount of prime numbers only the growth: π (x)
∼
x
log(x)
References
[1] Martin Aigner Aigner and G¨ unter M. Ziegler. Proofs from THE BOOK.
