# Prove that the lower bound for the partial sums of the Möbius inverse of the Harmonic numbers minus $n$ is greater than $1$ minus a Harmonic number

Let $$\varphi^{-1}(n)$$ be the Dirichlet inverse of the Euler totient function:

$$\varphi^{-1}(n) = \sum\limits_{d\mid n} d \cdot \mu(d) \tag{1}$$

and let the matrix $$T$$ be:

$$T(n,k)=\varphi^{-1}(\gcd(n,k)) \tag{2}$$

$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}, \tag{3}$$

therefore the Möbius inverse of the Harmonic numbers $$H_{n}^{-1}$$ is:

$$H_{n}^{-1}=\sum _{k=1}^{n} \frac{\varphi^{-1}(\gcd (n,k))}{k} \tag{4}$$

The partial column sums of the lower triangular part of $$T(n,k)$$ is: $$M(n,k)=\underset{m\geq k}{\sum _{m=1}^n} \varphi^{-1}(\gcd (m,k)) \tag{5}$$

The partial sums of the Möbius inverse of the Harmonic minus $$n$$ is therefore:

$$-n +\sum_{n=1}^{N} H_{n}^{-1} = \sum_{k=2}^{n} \frac{M(n,k)}{k} \tag{6}$$

Conjecture and question:
Show that:
$$|M(p_i-1,k)|>0 \tag{7}$$ where $$p_i$$ is the $$i$$-th prime number, $$i=1,2,3,4,5,...$$.
for all $$1 \leq k \leq p_i-1$$.

Consider the linear programming problems:

$$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{k=1}^{n} \frac{x_{k}}{k} \\ \text{subject to constraints:} & n + \displaystyle\sum_{k=2}^{n}x_{k}=1 \\ & x_1 \geq 1 \\ k>1: & -1 \leq x_k \leq 1 & \tag{8} \end{array}$$

where the objective function above to be minimized is what we are seeking to (lower) bound in RHS of $$(6)$$ for $$k=1,2,3,4,5,...,n$$:

$$\sum_{k=1}^{n} \frac{x_{k}}{k} \iff \sum_{k=2}^{n} \frac{M(n,k)}{k} \tag{9}$$

and where the constraint is the property:

$$n + \sum_{k=1}^{n} \frac{x_{k}}{k} = 1 \iff \sum_{k=1}^{k=n} M(n,k) = 1 \tag{10}$$ proven here by Max Alekseyev.

and where the variable bounds are: $$k=1: \; x_1 \geq 1 \iff M(n,1)=n \tag{11}$$

$$k>1: \; -1 \leq x_k \leq 1 \iff |M(p_i-1,k)|>0 \tag{12}$$

The solutions to linear programming problem $$(8)$$ is:

$$1+\sum _{k=1}^n -\frac{1}{k} \tag{13}$$ Which depends on the conjecture in $$(7)$$ and which means that lower bound of RHS of $$(6)$$ is:

$$\text{lower bound}\left(\sum_{k=2}^{n} \frac{M(n,k)}{k}\right)<1+\sum _{k=1}^n -\frac{1}{k} \tag{14}$$ which in turn is: $$\text{lower bound}\left(-n +\sum_{n=1}^{N} H_{n}^{-1}\right)<1+\sum _{k=1}^N -\frac{1}{k} \tag{15}$$ which is the question in the title:

"Prove that the lower bound for the partial sums of the Möbius inverse of the Harmonic numbers minus $$n$$ must be more than $$1$$ minus a Harmonic number, when $$n=p-1$$ for $$p$$ a prime number."

• Is there a more elementary way to prove the question in the title? Commented Mar 28, 2020 at 11:38
• Asked on mathoverflow: mathoverflow.net/q/355991/25104 Commented Mar 28, 2020 at 19:49

This answer is not complete yet.

It is known that the $$k$$-th columns in: $$\gcd(1..k+1,k) \tag{1}$$ and: $$\gcd(k..2k+1,k) \tag{2}$$ are palindromic for all $$k$$ and therefore the $$k$$-th columns in: $$T(1..k+1,k) \tag{3}$$ and: $$T(k..2k+1,k) \tag{4}$$

are also palindromic for all $$k$$.

$$T(n,k)=\varphi^{-1}(\gcd(n,k)) \tag{5}$$ is as stated in the question.

Since the sequences in the columns are periodic and periods sum to zero, the palindromicity means that for $$n=1..k$$ the $$k$$-th columns of the sum:

$$\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k)) \tag{6}$$

for $$n=(k-1)..2(k-1)$$ are the reversed sequences of the $$k$$-th columns of the negated sum:

$$-\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k)) \tag{7}$$

for all $$k>1$$.

Mathematica program to demonstrate that palindromicity gives negated reversals of columns.

This is as far as I got. The next step must have something to with the cases when:

$$T(n,k)=\varphi^{-1}(\gcd(n,k))=1 \tag{8}$$

and how such cases can't cause a zero in:

$$\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k))$$ when $$n=1,2,3,4,5,...$$ is a prime number, and how the reversal is shifted by one step.

But I don't know how to prove it.

Edit 29.3.2020:

Let $$T(n,k)=\varphi^{-1}(\gcd(n,k)) \tag{9}$$

The following holds:

$$\varphi^{-1}(1)=1 \tag{10}$$

$$T(1,k)=\varphi^{-1}(\gcd(1,k))=1 \tag{11}$$ for all $$k \geq 2$$.

$$\sum\limits_{m=1}^{k} T(m,k) = \sum\limits_{m=1}^{k} \varphi^{-1}(\gcd(m,k)) = 0 \tag{12}$$ for all $$k \geq 2$$.

Let:

$$W(n,k)=\sum\limits_{m=1}^{n} \varphi^{-1}(\gcd(m,k)) \tag{13}$$

Since prime numbers $$p_i$$ are not multiples of any $$k$$ and

$$\varphi^{-1}(n) \neq 0 \tag{14}$$ for any $$n$$ and in particular $$n=1$$,

and since it is possible for some $$k$$ that: $$\sum\limits_{m=1}^{m_1} \varphi^{-1}(\gcd(m,k))=0 \tag{15}$$ only for some $$1,

and since the columns in $$W(n,k)$$ are periodic for $$k>1$$: $$\sum\limits_{m=1}^{n} \varphi^{-1}(\gcd(m,k))=\sum\limits_{m=1}^{n+k} \varphi^{-1}(\gcd(m,k)) \tag{16}$$ which is the same as: $$W(n,k)=W(n+k,k) \tag{17}$$

Therefore

$$\sum\limits_{m=1}^{p_i} \varphi^{-1}(\gcd(m,k)) \neq 0 \tag{18}$$

for $$k=1..(p_i-1) \tag{19}$$

I am unsatisfied with having $$k$$ both in the upper summation limit in (12) and in the argument in (19), because it is confusing. This is the best I can do at the moment.