In an earlier question where I conjectured (and GH from MO confirmed) that the von Mangoldt function is the limit at s=1 of a certain Dirichlet series:
$$\Lambda(m)=\lim_{s\to 1+}\zeta(s)\sum_{d\mid m}\frac{\mu(d)}{d^{s-1}}=\lim_{s\to 1+}\;\sum_{n=1}^\infty\frac{1}{n^s}\left(\sum_{d\mid\gcd(n,m)}d\mu(d)\right),\qquad m>1.\tag{1}$$
joriki at MSE also proved it differently here.
What $(1)$ is saying is that there is a symmetric GCD matrix of numerators for the Dirichlet series expansion of the von Mangoldt function so that:
$$\displaystyle \Lambda(n)= \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}, \qquad n>1 \tag{2}$$ and $$\displaystyle \Lambda(k)= \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{n}, \qquad k>1 \tag{3}$$
By considering only the lower triangular part of a matrix $T$ we have matrix of numerators that has the property that:
$$\displaystyle \Lambda_{H}(n)= \sum\limits_{k=1}^{k=n}\frac{T(n,k)}{k} \tag{4}$$
where
$$\Lambda_{H}(m)=\lim_{s\to 1+} H_m \sum_{d\mid m}\frac{\mu(d)}{d^{s-1}} \tag{5}$$
where $\Lambda_{H}(m)$ stands for the Möbius transform (Möbius inversion) of the $m$-th harmonic number.
The expansion of the partials sums of the Möbius inverse of the Harmonic numbers
$$\sum\limits_{m=1}^{m=n} \Lambda_{H}(m) \tag{6}$$
for $n=1,2,3,4,5,...$ is a lower triangular matrix $M$
$$M=\left(\begin{array}{lllllll} 1 & 0 & 0 & 0 & 0 & 0 & 0&\cdots \\ 2 & -1 & 0 & 0 & 0 & 0 & 0 \\ 3 & 0 & -2 & 0 & 0 & 0 & 0 \\ 4 & -1 & -1 & -1 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & -4 & 0 & 0 \\ 6 & -1 & -2 & -1 & -3 & 2 & 0 \\ 7 & 0 & -1 & 0 & -2 & 3 & -6 \\ \vdots&&&&&&&\ddots \end{array}\right) \tag{7}$$
with the definition:
$$M(m,k)=\sum _{n=1}^m [n\geq k]a(\gcd (n,k)) \tag{8}$$
where $a(n)$ is the Dirichlet inverse of the Euler totient: $$a(n) = \sum_{d|n} d\mu(d) \tag{9}$$ and $[\;]$ is the Iverson bracket.
with the property that the first column equals the row index: $$M(n,1)=n \tag{10.1}$$
and the constraint that the row sum of row $n$ equals $1$: $$\sum\limits_{k=1}^{k=n}M(n,k)=1 \tag{10.2}$$
and the above mentioned expansion that:
$$\sum\limits_{m=1}^{m=n} \Lambda_{H}(m)=\sum\limits_{k=1}^{k=n}\frac{M(n,k)}{k} \tag{10.3}$$
and the two conjectural bounds that I don't know how to prove:
$$M(n,k) \leq +2(k-1) \qquad \tag{10.4}$$ $$M(n,k) \geq -2(k-1) \qquad \tag{10.5}$$
except at prime $k$ where it seems true/obvious.
It then becomes natural to ask what the (negative) maximum (a minimum actually) would be for a linear programming problem with that property, constraint and those bounds. I programmed this in Mathematica with the program:
(*start*)
nn = 42;
TableForm[
L1 = Table[
LinearProgramming[
Table[1/n, {n, 1, k}], {Table[If[n == 1, k, 1], {n, 1, k}]}, {{1,
0}}, Table[
If[n == 1, {-1, 1}, {-2 (n - 1), 2 (n - 1)}], {n, 1, k}]], {k,
1, nn}]];
b=Table[Sum[-Sign[-1 + Sign[L1[[n, k]]]], {k, 1, n}], {n, 1, nn}]
(*end*)
(In the Mathematica program I had to set the property of the first column to be a coefficient in the constraint instead, but the end result is the same.)
This gives a sequence: $$b(n)=\sum _{k=1}^n -\text{sgn}(\text{sgn}(L_{1}(n,k))-1) \tag{11}$$
where $L_{1}(n,k)$ is the $k$-th coefficient in the solution to the $n$-th linear programming problem,
that starts: $$b = 0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14,...$$ and which according to the OEIS appears to be sequence A049472 which is: $$b(n)=\left\lfloor \frac{n}{\sqrt{2}}\right\rfloor \tag{12}$$
So therefore it appears there is a lower bound:
$$\sum\limits_{m=1}^{m=n} \Lambda_{H}(m) > -\left\lfloor \frac{n}{\sqrt{2}}\right\rfloor \tag{13}$$
Sofar this should be provable although I don't know how.
But let me speculate further. What happens if we would be allowed to set the first bound on the variables to zero:
$$M(n,k) \leq 0 \qquad \tag{14.4}$$ $$M(n,k) \geq -2(k-1) \qquad \tag{14.5}$$ and with $(10.1)$, $(10.2)$ and $(10.3)$ unchanged?
This as a Mathematica program is:
(*start*)
nn = 42;
TableForm[
L2 = Table[
LinearProgramming[
Table[1/n, {n, 1, k}], {Table[If[n == 1, k, 1], {n, 1, k}]}, {{1,
0}}, Table[
If[n == 1, {-1, 1}, {-2 (n - 1), 0}], {n, 1, k}]], {k, 1, nn}]];
c = Table[1 - Sum[Sign[L2[[n, k]]], {k, 1, n}], {n, 1, nn}]
(*end*)
This gives a sequence: $$c(n)=1-\sum_{k=1}^n \text{sgn}(L_{2}(n,k)) \tag{15}$$
where $L_{2}(n,k)$ is the $k$-th coefficient in the solution to the $n$-th linear programming problem,
that starts:
$$c = 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5,...$$
and which according to the OEIS appears to be sequence A000194, the nearest integer to square root of n.
Since the bound (13.4) is less than equal to zero we don't get any bounds on $(6)$, the partial sums of the Möbius inverse of the Harmonic numbers - which is the goal if one is pursuing the Riemann hypothesis.
Question:
In the Linear Programming problem:
Minimize: $$\sum\limits_{k=1}^{k=n}\frac{x_{n,k}}{k} \tag{16.1}$$
subject to constraints: $$x_{n,1}=n \tag{16.2}$$
$$\sum\limits_{k=1}^{k=n} x_{n,k}=1 \tag{16.3}$$
$$-f(n,k) \leq x_{n,k} \leq g(n,k) \tag{16.4}$$
What is the minimum of the objective function in terms of functions $f(n,k)$ and $g(n,k)$ for different $n$?
One remark. In the bounds from sequences $b(n)$ and $c(n)$ I ignored the positive parts from lower triangular matrices $L_{1}$ and $L_{2}$ of linear programming solutions. So the (conjectured) bounds arrived at are more complicated. I remembered this now, and I will try to edit and give the exact versions tomorrow. And as always with my questions, if the latex is wrong trust the Mathematica programs provided you are able to interpret them.
