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User:Alexander Adam

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I am a physicist looking for a PhD position in theoretical physics in the realm of quantum information or related topics in quantum mechanics concerned with fundamental properties like entanglement or superposition and validity of theoretical assumptions and predictions.

alexander.adam@stud.uni-goettingen.de

To live in an asymmetric world, you should better be symmetric. However if you live in a symmetric world, asymmetry suffices.

I have also a strong interest in number theory because the universe of (natural) numbers is similar to an enigma. It is like the prototyp of all theories of maximal stability and we learn from it fundamental rules how to encode information with minimal relative overhead.

Some results of my personal research (currently most of them is a collection of identities of number theoretic functions I found by myself unless otherwise mentioned which seem important for me, if you find errors feel free to send me a hint - i will apply the correction and a note):


The illusion of the natural number

When we are dealing with a natural number by the fundamental theorem of arithmetic we are dealing obviously with prime numbers. We may decompose . In the linear representation we have which means that we write down only prime numbers. However to quantify the multiplicity of each prime number we resort to natural numbers again. We may overcome this border by recursively decompose each multiplicity and switch to an exponential representation. Again we see only prime numbers, e.g. , but if we want to quantify the exponential levels where the primes appear we use again natural numbers. We might say in the last example that the multiplicity for is and the level is . We can now represent every level again in an exponential representation and also every other quantification of the number like the number of distinct prime numbers in and the number of levels of levels and so on. We say that a representation of a natural number is in prime number language if all of its quantifications are closed under prime numbers meaning that they are expressable only with prime numbers (or - including the - the set of smallest coprime numbers). The general question is now if there is a system that quantifies every natural number in prime number language. If it exists then a natural number would be like an illusion. It seems natural for this question to be answered that square-free numbers play a critical role in this game and also the number of quantifiable properties of a natural number . We now try to make the statement precise which serves us also as a check of the problem if it is well posed: Let be the set of all quantifiable properties of a natural number so that is a quantification of . It is clear that if holds we would be done but this is in general not the case. If we look at . A solution to the illusion problem would be to find such that there is always a finite chain of composition of elements from which evaluates to an element of .

Definition (Basis of a semi-group): Let be a countable semi-group and with . Then is called a basis for iff

.

Example:

Remark: Both examples are commutative and include a neutral element which makes them a commutative monoid. Please note that in the last example the square-free numbers form a semi-group if the multiplicity saturates to under multiplication. Thus we find a morphism between the lhs and the rhs.

Definition (Minimal basis):

Lemma (Semi-group): Let be a countable semi-group such that . Then .

Proof: Let . Then . But then we can decompose recursively and we get a binary tree. If the tree would have a finite level then we could take the leafes as elements of a set . So it follows that there must be at least one path of length but then by assumption.

Symbolic calculus

Definition (Free abelian monoid)

Let be any set. Then with the following properties

is called the free abelian monoid with beeing the concatenation.

Corollary (Free abelian monoid)

Let be a family of monoids and . Then

  • .

Let be any countable set and a bijection. Then

  • where is the occurence number of in .

Proof: First statement: Let . Then . Since is from a monoid we can repeat this step a finite times for and we get a sequence with . Since this implies .

The second statement follows directly from the abelian property of the monoid.

Definition (Primorial)

Let be any countable set and a bijection. Then and .

Definition (Primorial iterates)

Let be any infinite countable set and a bijection. Then we define:

Lemma (Properties of primorial iterates)

  • is an isomorphism.

Proof The first property is directly by definition since : .

Now the second one goes by induction. Because of , is a homomorphism. Since it is also invertible on , is bijective because the image is precisely . Suppose now that is an isomorphism. Now but and are isomorphisms so has to be .

The third property follows again by induction. We have . Now suppose . Then .

The last property can be proven by considering the fixpoints of . Since and we know already two fixpoints. Now consider . Since we will increase the sequence length under every iteration of .

Simple residue rules

  • Let . Then .
    • Let . Then .
    • Let . Then .
    • Let . Then .
  • Let and . Then

Generating function of x mod y

    • (The rhs follows from the special Lambert series.)

Moments of the harmonic summatory function

Proof:

We used for the first step Faulhaber's formula and for the last step the identity: .

  • Let and . Then
    • Let . Then

The sum of the sum of moments of divisors up to a natural number

  • Let where . Then .

A bound on the sum of divisors of a natural number

We have in general the identity: or which is invariant under reflection . First we proof the general property

Proof:

A relative tight upper bound can then be found as where the harmonic number can be coarse bounded as . Using the symmetry of the divisors we get an even better upper bound: .

  • If we restrict ourself only to odd natural numbers we may get an other bound because the spacing of possible divisors is at least . We can express this as:
  • I we consider instead the set we have a minimal periodic spacing of which we can express algebraically as (we have for every periodic sequence algebraically which shows that every such sequence is isomorphic to an alternating sequence) and therefore we get

We might also find always a better upper bound for any multiplicative function under the following assumption: Let and . Then .

On Robin's criterion for RH

  • with

Lemma

Let the strictly increasing sequence of prime numbers and . Then

.

Proof: This follows from the fact that .

Lemma (Exchange of exponents)

Let the strictly increasing sequence of prime numbers and with and . Then

.

Proof: It is enough to consider the derivative of for . We have with because and . Therefore .

Definition (Asymptotic equivalence)

Corollary (Representation of a natural number)

Let . Then there exists with so that . In particular we make the identification .

Proof: Just consider the prime factorization of and regroup the factors according to it's prime multiples in increasing order. The co-prime condition follows immediately.

Remark: Clearly there will some so that .

Definition (Dedekind )

This representation has the advantage that we have .

Corollary (Dedekind ) Let . Then .

Lemma (Dedekind ) Let and .

Definition (Extremal primes)

Let given as above so that we have . We define:

Definition (Primorial)

Corollary (Primorial) Let . Then .

Lemma (Product of primorials)

Let be a product of primorials. Then for some sequences .

Remark: If we understand pointwise then we find a particular short notation of products of primorials: . As an example pick as the colossally abundant number. It is given in this notation as: . However we can get even more structure:

Definition (Primorial products) Let be the set of all finite primorial products. Then we define a completely multiplicative mapping with .

Lemma (Primorial isomorphism) Let be the set of all finite primorial products. Then under the multiplication of we have .

Proof This is because is a bijection and a homomorphism.

Corollary (Properties of primorial isomorphism)

  • is an isomorphism.


Corollary (Product of primorials)

Let be a product of primorials. Then .

Definition (Monotonic bound)

and .

Definition (RH bound)


Lemma (Bound violation)

Let and . Then .

Theorem (Violation of montonic bound) Let and a monotonic bound. Then .

Proof: Let and . Then by our lemma we may assume that . However and . Because .

Lemma: (Violation of bound) Let and a monotonic bound. Then .

Proof: This follows directly by the Theorem and the fact that .

Corollary: (Violation of bound) Let . Then .

Lemma: (On the RH bound for even positive integers)

Let and . Then .

Proof:

We have . Now . Then .

Lemma: (On the RH bound for arbitrary powers)

Let and and such that . Then

.

Proof: Suppose that . Now . Therefore . It follows that .

Application (Square-full numbers):

All proper powers of with fulfill the RH bound. In particular for all such we have: . Because we deduce that all square-full numbers fulfill the RH bound.

Corollary: (On the RH bound for positive integer powers)


Lemma: (On the RH bound)

Let and and . Then .

Proof: . Now and therefore .

Lemma: Let with such that . Further let with . Then .

Proof:

Remark: This shows essentially that we are blind with this technique to numbers with a long square-free tail. Since for we have control over the quadratic component but not over .

Lemma: Let and such that . Then .

Remark: The second assumption is equivalent to: . But this condition is not always fulfilled. A counter example is which is a colossally abundant number. The existence of a counter example leads us to the conclusion that we have to investigate further the numbers which have .

Lemma (Bound on the largest prime factor) Let and may be eventually false. Then with .

Lemma (Necessary condition for the failure of Robin's inequality) Let , where is the exceptionel set where Robin's inequality is false. Then .

Proof: By failure of Robin's inequality we have . But and therefore .

Lemma (Sufficient condition for the truth of Robin's inequality) Let and sucht that . Then .

Proof: By assumption we have . This implies the stronger condition: .

The sum of the count of unordered bipartite factorizations up to a natural number

The modular multiplication table of the finite cyclic group of natural numbers

We may assume here that we have a Hilbert space with dimension and an orthonormal base . The space of -tensors is given by and are the basis coefficients.

Now for some let be the entries of the modular multiplication table . This multiplication table has under the left and right action of the symmetric group symmetries.

The connection with the discrete Fourier transformation

We may define the generalized discrete Fourier transformation (GDFT) as an unitary matrix such that where . The connection follows then by letting . This is also the complex representation of the multiplication of elements of the finite cyclic group .

The connection with the microcanonical ensemble

We establish here a connection to the microcanonical ensemble under the von Neumann measurement. Given some diagonal -tensor and an unitary matrix such that the new measurement base is the von Neumann measurement acts on the diagonal entries as the transition . If we demand now that we have . If we impose and this tells us also that it is always possible to push a quantum system into the state of maximal entropy. It is therefore pretty useful to investigate the space of all GDFTs.

The rank of the modular multiplication table

Proof:

We set . If is given in column notation, we build a new matrix with where we identified . We may now order the according to the divisors of : every divisor of spawns an equivalence class of column vectors of . Elements of different equivalence classes are linearly independent so that our rank is determined by the size of the equivalence classes. The case where will decrease the rank by and will decrease the rank by . In the case where all these classes except have even size and we get:

In the case where also consists of only one element and we get

The result follows then together with the identity .

The binary representation

If we pick as a representative for the smallest natural number possible we have so that counts the number of cycles we passed already. E.g. so . Now we build the column difference matrix with which contains only zeros and ones. This is the binary representation of .

Fig. 1: An example for the binary representations of 1 to 20.

The binary representation has many interesting properties and we will present here some of them. We will show here two derivations but first we introduce as a new -tensor vector space where and . In the first place we are interested in a special element . It is a binary rectangular matrix with all zeros except where the geometric line from to intersects one of the squares. We find now that so we can look at all the elements at once. The identification works because for some and we identify and recall that which is the triangular number and we have .

Fig. 2: The binary matrix for .
Fig. 3: The binary matrix for .
Fig. 4: The composed binary matrix for all .

From figure 2 we can guess that is composed as the direct sum of its irreducible element - that is we may write .

Proof: We draw a line along . To intersect the lattice nodes we demand that which is only possible if . Because this will happen exactly times.

Further from figure 3 we guess that the number of ones for two coprime natural numbers is given as .


Proof: We draw a line along . Because this adds ones. Another one is added if we transit to the next natural number in -direction meaning that we add another ones.

From that it follows that the number of ones in a generic matrix is

The last equation arises by counting the number of zeroes and removing them from the maximal amount of ones which is or by a little bit more complicated transformation of ceiling to floor. It is not new and a reference will be added. It is this trace formula which tells us that the diagonal and the spectrum of carry some important information about the natural numbers. We may also interpret as a bipartite pure quantum state and as the reduced state in subsystem and as the reduced state in subsystem . Further we have the properties:

  • with
    • .
      • Notice that we can generalize it e.g. to so that we get the Euler identity for free: .
      • The last inequality actually tells us that the most simple upper bound which we can choose is pretty worse because we have by far . The equality and upper bound can also easily extended to the slightly more general gcd-sum: . Again we get the Euler identity for free: .

Now we are turning back to the matrix and show how it is related to the binary representation . As a hint we may look at the elements and we observe that the differences of the number of white squares in every row of the half spaces created through the black line of consecutive matrices belong to the binary representation . This is because we have for the number of white squares in every row in the left half space .

Let and . We note that describes the homogeneous commutative two-point nearest neighbor and self interaction of a linear chain[1]. Then the spectrum [2] is given as

with the corresponding normalized eigenbasis [2] . Because we can calculate the entanglement information

which converges to which is less than the expected information of if we would take a pure bipartite quantum state at random under the unitarily invariant Haar measure. Anyway we conjecture here that meaning that it behaves similar like the information content of a random ensemble. Further if then .

Bipartite factorization and the Lorentz scalar

Let such that . If we define and then . But is the squared semi-norm of the -dimensional Minkowski space where . Therefore is just another parametrization of and we have where is the Lorentz boost. It has the nice property that and forms therefore a one-parameter group. This ultimately shows that is really Lorentz scalar under bipartite factorization. Now is invariant under the dilation of with . We find that a boost of with dilates with . This makes it now easy for us to find the boost for certain lattice points.

Example (Boost of integer points): Let and . Then there exists a boost from the point to the point . This is because if we choose then .

Lemma (Boost of integer points): There exists a boost from the point to the point .

Proof: This is because if we choose then .

If we choose then . This shows that if we want to resolve a bipartite factorization of it is sufficient to know the fraction of the factors. This can be seen also by . However without a lot of work there is no chance to infer this fraction with our correspondence principle between dilation and Lorentz boost. But it allows us to state this as an inverse problem by finding so that . We find a bipartite factorization for odd numbers for example if we determine first and then which gives us then the greatest non-trivial factor of . If is not a prime we have . It is unclear how the formalism above would help us here.

Because we may think also of a different parametrization for . We might choose . Then and to make independent of we introduce so that with . Here we cannot revover from (actually we have ) but we may think of recovering from because we have the identity . From the identity it follows that there is not a such that so acts not continuous on the diagonal vector.

Compatibility of bipartitions

We did not talk about compatibility between two vectors and under a boost . We want to calculate the boost parameter such that . Under the boosts we say that is compatible with .

Corollar (Distance of compatible bipartitions): .

Proof: .

But how does the parameter looks like? We solve for in our dilation coordinates and . Remember that , and , . We find that there exists two solutions: which are only real if under the assumption that . If we exclude a complex valued we can say alternatively that lower valued factorizations are only compatible with higher valued ones. However a complex valued boost parameter shifts our points also into the complex plane which means that we can now find a boost which transforms between the complex integers to the real ones! Consider e.g. the factorization . If we take and then and . This means that . Essential we can reach any factorization of some integer under any quadratic field which means that their integer lattices are connected under hyperbolic rotation. Every bipartite factorization of a number in any quadratic field lies on the orbit . In our example we used a decomposition in the Gaussian integers which is the prototype for an imaginary quadratic field. Note that in our setting the physical interpretation of is that of time or space (dependent on the convention) so we would need here imaginary time or space to reach the Gaussian integers (or in general complex numbers).

A connection to Goldbach's conjecture

Our considerations may seem simple but there is already a connection to a really deep problem called the Goldbach conjecture (GC). If we consider the sum of the factors then we see that this yields an even number if or . For the GC to be true we need the latter property (because we have with the only occurence of the first prime ) so we fix with such that . The GC states now that . Please note that the lower bound seems necessary only for the case . The upper and lower bound arises from the fact that the number of divisors of a number composed of two primes is in the case that with and otherwise. The transformations we need to count take to .

Sums of bipartite factorizations

We can easily complicate the things further. Let us generalize our form for as where . For we get back our initial case from above. We can also write by taking elements from the Hilbert space . Here we encounter already the orthogonal symmetry because . But this is just a special kind of dilation because . So our complete symmetry is again defined over the space . We have symmetry of under the transformation with . Now we do the same again and define , and . Then where . We may infer for the finite dimensional case the symmetry group for the elements . It will turn out that they are symmetric under because we have . However if we want to establish such an isomorphism we need a permutation matrix which maps the odd components to the first half and the even components to the second half. E.g. and . Sadly this is not well defined for . Anyway we may conclude that and an additional boost between e.g and . We may write