As indicated below, one could use the Ehrhart polynomials of the simplex in number theory.
Here are the questions without context first:
Questions:
- The sum $$\sum_{k=0}^t (-1)^k ( \operatorname{binomial}(d+k,d)-\operatorname{binomial}(d+k-1,d) )$$
is according to Wolfram Alpha equal to:
$$=((-1)^t \operatorname{binomial}(d+t+1,d) F_{2,1}(1,d+t+2;t+2;-1)+2^{-d-1})-2^{-d-1}(2^{d+1}(-1)^t \operatorname{binomial}(d+t,d)F_{2,1}(1,d+t+1;t+1;-1))$$ where $F_{2,1}(a,b;c;z)$ is the hypergeometric funciton of Gauss.
Can this be proven or how does Wolfram Alpha does it's magic?
Let $G(d,t):=(-1)^t \frac{d}{t+1}F_{2,1}(1,d+t+2;t+2;-1)+(-1)^t F_{2,1}(1,d+t+2;t+2;-1)+\frac{1}{2^{d+1} \operatorname{binomial}(d+t,d)}+(-1)^{t+1}F_{2,1}(1,d+t+1;t+1;-1)$
Is $\lim_{d \rightarrow \infty} G(d,d) = 0$?
Is $\lim_{N \rightarrow \infty} \frac{1}{\operatorname{binomial}(2 \pi(N),\pi(N))}F(N,\pi(N)) = \lim_{N \rightarrow \infty} \frac{1}{|B_{N,\pi(N)}|} \sum_{n \in B_{N,\pi(N)}} \lambda(n) =^? 0$ equivalent to the prime number theorem (in the Landau formulation), where $F(N,t)$ and $B_{N,t}$ are as defined below?
Motivation and context:
Motivation comes from this post at MO, which I have summarized here, for self-consistency:
It is known that for each finite set of primes $p$ we have: $\log(p)$ are linear independent over the rational numbers.
We have $\log(ab) = \log(a)+\log(b)$ and $\log(n) = \sum_{p |n}v_p(n) \log(p)$.
We define $\psi(n) := \sum_{p |n}v_p(n) e_{\pi(p)}$ where $e_p$ is the $p$-th standard basis vector of the Hilbert space of sequences and $\pi$ is the prime counting function.
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 $Q_N:=$ Polytope of $(\{\psi(p)|1\le p \le N, p \text{ prime }\})$
Then the polytope generated by the primes is the simplex, and it is known ( "Computing the continuous discretely" Matthias Beck, Sinai Robins, Second edtion, p.31 ff) that the Ehrhart polynomial is given by:
$$L(Q_N,t) = \operatorname{binomial}(d+t,d)$$
where $d= \pi(N)$ is the dimension of the simplex, and it is equal to the number of primes $\le N$, which is denoted by $\pi(N)$. Let $\hat{\pi}(n):=\{p \text{ prime}: p|n \}$ and let $p_n:=n$-th prime.
Since $L(Q_N,t)$ counts the points $\psi(n)=(x_1,\cdots,x_d)$ in the dilated polytope $t Q_N$ and those points have cooridnates $\ge 0$ with $x_1+\cdots+x_d \le t$, we conclude that, by observing that the sum of the coordinates correspond to $\Omega(n) := \sum_{p|n} v_p(n)$:
$$L(Q_N,t) = |\{\psi(n) : \Omega(n) \le t , \hat{\pi}(n) \subset \hat{\pi}(N!), 1 \le n \le p_{\pi(N)}^t\}|$$
The condition $\hat{\pi}(n) \subset \hat{\pi}(N!)$ is for making sure we look at those numbers which have only prime divisors $\le N$.
We also observe that:
$$L(Q_N,t)-L(Q_N,t-1) = |\{ n : 1 \le n \le p_d^t, \Omega(n)=t, \hat{\pi}(n) \subset \hat{\pi}(N!) \}| = |A_{N,t}|$$
where I have defined:
$$A_{N,t} = \{ n : 1 \le n \le p_d^t, \Omega(n)=t, \hat{\pi}(n) \subset \hat{\pi}(N!) \}$$
We can now define and try to evalute the following sum:
$$F(N,t):= \sum_{k=0}^t \sum_{n \in A_{N,k}} \lambda(n)$$ $$=\sum_{k=0}^t (-1)^k ( L(Q_N,k)-L(Q_N,k-1) )$$ $$=\sum_{k=0}^t (-1)^k ( \operatorname{binomial}(d+k,d)-\operatorname{binomial}(d+k-1,d) )$$
which according to Wolfram Alpha is equal to:
$$=((-1)^t \operatorname{binomial}(d+t+1,d) F_{2,1}(1,d+t+2;t+2;-1)+2^{-d-1})-2^{-d-1}(2^{d+1}(-1)^t \operatorname{binomial}(d+t,d)F_{2,1}(1,d+t+1;t+1;-1))$$ $$=((-1)^t \operatorname{binomial}(\pi(N)+t+1,\pi(N)) F_{2,1}(1,\pi(N)+t+2;t+2;-1)+2^{-\pi(N)-1})-2^{-\pi(N)-1}(2^{\pi(N)+1}(-1)^t \operatorname{binomial}(\pi(N)+t,\pi(N))F_{2,1}(1,\pi(N)+t+1;t+1;-1))$$
where $F_{2,1}(a,b;c;z)$ 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:
$$\sum_{n=1}^N \lambda(n)$$
The naive idea is that in the limit $N\rightarrow \infty, t \rightarrow \infty$, the two sets:
$$\mathbb{N}:= \lim_{N\rightarrow \infty}\{1,\cdots,N\} \text{ and } \lim_{N,t\rightarrow \infty} A_{N,t} $$
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.
def Omega(n):
return sum(valuation(n,p) for p in prime_divisors(n))
def lhsK(N,t,k):
d = prime_pi(N)
return sum((-1)**Omega(n) for n in range(1,nth_prime(d)**t+1) if set(prime_divisors(n)).issubset(set(primes(N+1))) and Omega(n)==k)
def lhsList(N,k):
d = prime_pi(N)
return [n for n in range(1,nth_prime(d)**k+1) if set(prime_divisors(n)).issubset(set(primes(N+1))) and Omega(n)==k]
def rhs0(N,t):
d = prime_pi(N)
return (t-d)*binomial(d-t-1,d)/(d+1)+(t-d-1)*(binomial(d-t,d)+d+1)/(d+1)
def rhsK(N,k):
d = prime_pi(N)
return ((-1)**k*(binomial(d+k,d)-binomial(d+k-1,d)))
#https://doc.sagemath.org/html/en/reference/functions/sage/functions/hypergeometric.html
def bino(x,y):
return gamma(x+1)/(gamma(y+1)*gamma(x-y+1))
def rhsH(N,t):
d = prime_pi(N)
return ((-1)**t*bino(d+t+1,d)*hypergeometric([1,d+t+2],[t+2],-1)+2**(-d-1))-2**(-d-1)*(2**(d+1)*(-1)**t*bino(d+t,d)*hypergeometric([1,d+t+1],[t+1],-1))
def lhs(N,t):
d = prime_pi(N)
return sum((-1)**k*(bino(d+k,d)-bino(d+k-1,d)) for k in range(int(t)+1))
#return sum((-1)**Omega(n) for n in range(1,nth_prime(d)**t+1) if set(prime_divisors(n)).issubset(set(primes(N+1))))
def rhs(N,t):
d = prime_pi(N)
return sum(rhsK(N,k) for k in range(t+1))
for N in range(4,240):
for t in range(1,14):
print(N,t,rhsH(N,t).n(),lhs(N,t).n())
Let $B_{N,t} = \{n : \Omega(n) \le t , \hat{\pi}(n) \subset \hat{\pi}(N!), 1 \le n \le p_{\pi(N)}^t\}$, so that
$$\operatorname{binomial}(d+t,d) = L(Q_N,t) = |B_{N,t}|$$
The arguments above show that:
$$\sum_{n \in B_{N,t}} \lambda(n) = F(N,t) = $$
$$=((-1)^t \operatorname{binomial}(d+t+1,d) F_{2,1}(1,d+t+2;t+2;-1)+2^{-d-1})-2^{-d-1}(2^{d+1}(-1)^t \operatorname{binomial}(d+t,d)F_{2,1}(1,d+t+1;t+1;-1))$$
hence dividing by the number of points under which the sum runs, we get:
$$\frac{1}{|B_{N,t}|} \sum_{n \in B_{N,t}} \lambda(n) = \frac{1}{|B_{N,t}|} F(N,t) = $$ $$=(-1)^t \frac{d}{t+1}F_{2,1}(1,d+t+2;t+2;-1)+(-1)^t F_{2,1}(1,d+t+2;t+2;-1)+\frac{1}{2^{d+1} \operatorname{binomial}(d+t,d)}+(-1)^{t+1}F_{2,1}(1,d+t+1;t+1;-1)$$
where $d:=\pi(N)$.
Notice the similarity to the prime number theorem:
We have in both cases sets $C_N$ such that:
$$\lim_{N\rightarrow \infty} \frac{1}{|C_N|} \sum_{n \in C_N} \lambda(n) = 0$$
The case $C_N := \{1,\cdots,N\}$ corresponds by Landau equivalently to the prime number theorem.
We observe that:
$$\lim_{N \rightarrow \infty} B_{N,t} = \{n \in \mathbb{N} | \Omega(n) \le t \} =: B_{\infty,t}$$
and that
$$\lim_{t \rightarrow \infty}(\lim_{N\rightarrow \infty} B_{N,t} ) = \lim_{t \rightarrow \infty} B_{\infty,t} = \lim_{t \rightarrow \infty} \{n \in \mathbb{N} | \Omega(n) \le t \} = \mathbb{N}$$
Similarily: $$\lim_{N \rightarrow \infty}(\lim_{t\rightarrow \infty} B_{N,t} ) = \lim_{N \rightarrow \infty} B_{N,\infty} = \lim_{N \rightarrow \infty} \{n \in \mathbb{N} | \hat{\pi}(n) \subset \hat{\pi}(N!)\} = \mathbb{N}$$
Hence we have:
$$\lim_{t \rightarrow \infty, N \rightarrow \infty} B_{N,t} = \mathbb{N}$$
(Here we have used that there are infinitely many primes.)
Now let us put $C_N := B_{N,\pi(N)}$.
Then one empirical observation, which I was not able to prove directly, is:
$$\lim_{N \rightarrow \infty} \frac{1}{|C_N|} \sum_{n \in C_N} \lambda(n) =^? 0$$
which reminds a little bit on the prime number theorem where $C_N:=\{1,\cdots,N\}$.
