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Samstag, 2. März 2024

Application of Ehrhart polynomials for counting of numbers and lattices for natural numbers.

This is a transcription of the MO question written here: It is known that log(p) is independent as a vector over the rational numbers where p is a prime. We have log(ab)=log(a)+log(b) and log(n)=p|nvp(n)log(p). We define ψ(n):=p|nvp(n)ep where ep is the p-th standard basis vector of the Hilbert space of sequences. for q=a/bQ>0 we define: ψ(a/b):=ψ(a/gcd(a,b))ψ(b/gcd(a,b)). We have for nN: ψ(n)2=p|nvp(n)2 is a natural number Furthermore the function K(a,b)=ψ(a),ψ(b)=p|gcd(a,b)vp(a)vp(b) is a positive definite kernel on the natural numbers. Let us define the -dimensional lattice Γ as: Γ:={ψ(q)|qQ>0,ψ(q)20mod(2)} and let us also define: η(n):=(1)ψ(n)2=p|n(1)vp(n)2 For ψ(a),ψ(b)Γ we have: ψ(ab)2=ψ(a)+ψ(b)2=ψ(a)2+ψ(b)2+2K(a,b)0+0+00mod(2) Therefore ψ(a)+ψ(b)Γ and ψ(1)Γ. The lattice shares some properties with the Leech-lattice and the E8 lattice: 1. It is unimodular in the sense that: For a finite set of primes, the gram matrix (K(p,q))p,q is prime has det=1. 2. It is even: xΓ:|x|20mod(2) 3. For all nonzero vectors in the lattice the squared norm is at least 2. My questions are these: 1) Probably an easy question: η(mn)=η(m)η(n)m,nN 2) Does it satisfy a Riemann-Hypothesis type of random variable equality: ϵ>0:limN1N1/2+ϵNn=1η(n)=0 3) Is there a relationship to the Riemann Hypothesis and this lattice? Here is some Sagemath code for experiments. And here is some picture concerning the 2. question:
**Edit**: Let λ(n)=(1)Ω(n) be the Liouville function. Then because vp(n)2vp(n)mod(2), we have indeed: η(n)=λ(n) This answers the first question since it is known that λ is multiplicative. This also answers the second and third question. Modified question: Is this lattice known in literature? Thanks for your help. ------------------------------------------------------------------------------------------------- A maybe useful point of view, is to use the Ehrhart polynomial of a polytope, namely the simplex, to count the numbers as points in a simplex: Let QN:= Polytope of ({ψ(p)|1pN,p prime }) Then the polytope generated by the primes is the simplex, and it is known ( "Computing the continuous discretely" Matthias Beck, Sinai Robins, Second edition, p.31 ff) that the Ehrhart polynomial is given by: L(QN,t)=binomial(d+t,d) where d=π(N) is the dimension of the simplex, and it is equal to the number of primes N, which is denoted by π(N). Let ˆπ(n):={p prime:p|n} and let pn:=n-th prime. Since L(QN,t) counts the points ψ(n)=(x1,,xd) in the dilated polytope tQN and those points have coordinates 0 with x1++xdt, we conclude that, by observing that the sum of the coordinates correspond to Ω(n):=p|nvp(n): L(QN,t)=|{ψ(n):Ω(n)t,ˆπ(n)ˆπ(N!),1nptπ(N)}| The condition ˆπ(n)ˆπ(N!) is for making sure we look at those numbers which have only prime divisors N. We also observe that: L(QN,t)L(QN,t1)=|{n:1nptd,Ω(n)=t,ˆπ(n)ˆπ(N!)}|=|AN,t| where I have defined: AN,t={n:1nptd,Ω(n)=t,ˆπ(n)ˆπ(N!)} We can now define and try to evaluate the following sum: F(N,t):=tk=0nAN,kλ(n) =tk=0(1)k(L(QN,k)L(QN,k1)) =tk=0(1)k(binomial(d+k,d)binomial(d+k1,d)) which according to Wolfram Alpha is equal to: =((1)tbinomial(d+t+1,d)F2,1(1,d+t+2;t+2;1)+2d1)2d1(2d+1(1)tbinomial(d+t,d)F2,1(1,d+t+1;t+1;1)) =((1)tbinomial(π(N)+t+1,π(N))F2,1(1,π(N)+t+2;t+2;1)+2π(N)1)2π(N)1(2π(N)+1(1)tbinomial(π(N)+t,π(N))F2,1(1,π(N)+t+1;t+1;1)) where F2,1(a,b;c;z) is the hypergeometric function. I must admit that the formula is not really "nice" but I think it should be useful, because using the RHS of the last equality with d could maybe be extended to other values for d,t other than the natural numbers. I can also not see, how to relate the sum F(N,t) to the Liouville sum: Nn=1λ(n) The naive idea is that in the limit N,t, the two sets: N:=limN{1,,N} and limN,tAN,t should be intuitively equal. There is also an equivalent formulation of the Riemann Hypothesis with the points on a different polytope, but while it uses the Ehrhardt polynomials, which in this case are very complicated at t=1 I can not see how to derive an analytic formula in this case. Here you can find some sanity check in SageMath. Let BN,t={n:Ω(n)t,ˆπ(n)ˆπ(N!),1nptπ(N)}, so that binomial(d+t,d)=L(QN,t)=|BN,t| The arguments above show that: nBN,tλ(n)=F(N,t)= =((1)tbinomial(d+t+1,d)F2,1(1,d+t+2;t+2;1)+2d1)2d1(2d+1(1)tbinomial(d+t,d)F2,1(1,d+t+1;t+1;1)) hence dividing by the number of points under which the sum runs, we get: 1|BN,t|nBN,tλ(n)=1|BN,t|F(N,t)= =(1)tdt+1F2,1(1,d+t+2;t+2;1)+(1)tF2,1(1,d+t+2;t+2;1)+12d+1binomial(d+t,d)+(1)t+1F2,1(1,d+t+1;t+1;1) where d:=π(N). Notice the similarity to the prime number theorem: We have in both cases sets CN such that: limN1|CN|nCNλ(n)=0 The case CN:={1,,N} corresponds by Landau equivalently to the prime number theorem. We observe that: limNBN,t={nN|Ω(n)t}=:B,t and that limt(limNBN,t)=limtB,t=limt{nN|Ω(n)t}=N Similarly: limN(limtBN,t)=limNBN,=limN{nN|ˆπ(n)ˆπ(N!)}=N Hence we have: limt,NBN,t=N (Here we have used that there are infinitely many primes.) Now let us put CN:=BN,π(N). Then one empirical observation, which I was not able to prove directly, is: limN1|CN|nCNλ(n)=?0 which reminds a little bit on the prime number theorem where CN:={1,,N}.

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