A Novel Generalization of the Liouville Function $\lambda(n)$ and a Convergence Result for the Associated Dirichlet Series

Properties 1 and 2 follow directly from the arithmetic of fractions with coprime denominators. We therefore begin by proving Property 3.

Observe that if the prime factorization of $n$ is given by $n=p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{J}^{k_{J}}$, then $w(n)=e^{i\theta(n)}$ where

From the definition of $w(n)$ in (1) it is clear how the sum in (3) results from each prime factor $p_{j}^{k_{j}}$ of $n$ contributing a factor of $\exp\bigl{(}i\psi\bigl{(}p_{j}^{k_{j}}\bigr{)}\bigr{)}$ to $w(n)$, and likewise the term $\frac{\pi}{2}\left(1-\lambda(n)\right)$ results from the alternating signs in the series in (1), whereby each prime factor of $n$ contributes a factor of $-1=e^{i\pi}$ to $w(n)$, thereby shifting $\theta(n)$ by $\pi$. Next, observe that $\psi(p^{k})$ is the $k^{\mathrm{th}}$ partial sum of the geometric series with $a=\frac{\pi}{p^{3}G}$ and $r=\frac{p-1}{p}$. Therefore, $\lim_{k\to\infty}\psi(p^{k})=\frac{\pi}{p^{2}G}$, and so $\sum_{p}\left(\lim_{k\to\infty}\psi(p^{k})\right)=\pi$, since we defined $G$ to be the sum of the reciprocals of the primes squared.

Imagine a hypothetical number $n$ whose prime factorization contains all prime numbers in infinite multiplicity. Then for all $p_{j}^{k_{j}}$ in its prime factorization, $k_{j}\to\infty$ for each $j\in\mathbb{N}$. So the value of $\sum_{j=1}^{\infty}\psi\bigl{(}p_{j}^{k_{j}}\bigr{)}$ in the argument of $w(n)$ for such a theoretical $n$ would be $\pi$, and for any actual $n\in\mathbb{N}$ the sum would be less than $\pi$. Therefore, if $n$ has an even number of prime factors, the argument of $w(n)$ would be in $[0,\pi)$ (zero being the argument of $w(1)$), and since if $n$ has an odd number of prime factors we introduce an extra factor $-1$, having the effect of adding $\pi$ to $\arg(w(n))$, in that case $\arg(w(n))\in(\pi,2\pi)$. This proves Property 3. ∎

We saw above that $w(n)=e^{i\theta(n)}$ where

with $G=\sum_{p}\frac{1}{p^{2}}$ and where $k_{p}$ is the multiplicity of the prime factor $p|n$. Thus, to show $\{w(n)\}$ dense in $C$, we will show $\{\theta(n)\}$ dense in $(0,2\pi)$.

To begin, let $x\in(0,\pi)$ and choose $\epsilon>0$. We will show there exists an infinite sequence $\{n_{k}\}$ such that $\theta(n_{k})$ is within an $\epsilon$-radius of $x$ for all sufficiently large $k$. We will construct $\{n_{k}\}$ by identifying prime factors to use to build $\theta(n_{k})$ to be in the proper interval when $k$ is sufficiently large. We will accomplish this essentially by identifying terms of $\frac{\pi}{G}\sum_{p}^{\infty}\frac{1}{p^{2}}$ which will yield a sum in the $\epsilon$-neighborhood of $x$.

First, let $x_{0}=x$ and choose $b_{0}$ as large as possible such that

Next, if $B_{0}>x_{0}+\epsilon$, choose $t_{0}$ as large as possible such that

(If we have that $B_{0}\leq x_{0}+\epsilon$, instead choose $t_{0}$ sufficiently large so that $T_{0}$ is small enough for $\left|x_{0}-\left(B_{0}-T_{0}\right)\right|<\epsilon$). Clearly $t_{0}+1\geq b_{0}$, since $B_{0}>B_{0}-(x_{0}+\epsilon)$ and $t_{0}$ is chosen to be the largest such that (6) holds. If $t_{0}+1>b_{0}$ then let $S_{0}=B_{0}-T_{0}=\frac{\pi}{G}\sum_{j=b_{0}}^{t_{0}}\frac{1}{p_{j}^{2}}$. Otherwise, if $t_{0}+1=b_{0}$, then by the choice of $t_{0}$ it follows that

and so since $t_{0}+1=b_{0}$ we have

and therefore $x_{0}+\epsilon\leq\frac{\pi}{G}\frac{1}{p_{j}^{2}}$ for $j=t_{0}+1=b_{0}$. In this case, let $p=p_{j}$ and choose $k_{0}$ the largest such that $\psi(p^{k_{0}})\leq x_{0}$, and let $S_{0}=\psi(p^{k_{0}})$. Note that this choice of $k_{0}$ is always possible: if $K$ is the set of all $k\in\mathbb{N}$ such that $\psi(p^{k})\leq x_{0}$, then $K$ is finite, since otherwise $\lim_{k\to\infty}\psi(p^{k})=\frac{\pi}{G}\frac{1}{p_{j}^{2}}\leq x_{0}$, but this is a contradiction since we had that $x_{0}+\epsilon\leq\frac{\pi}{G}\frac{1}{p_{j}^{2}}$. It remains to be seen that $K$ is nonempty, so assume to get a contradiction that

but since $b_{0}$ was chosen to be the largest such that $\frac{\pi}{G}\sum_{j=b_{0}}^{\infty}\frac{1}{p_{j}^{2}}>x_{0}-\epsilon$, therefore

and if we let $q$ be the next prime after $p$, so $q=p_{b_{0}+1}$, then we have in particular that $\frac{\pi}{G}\frac{1}{q^{2}}<x_{0}-\epsilon$. However, for any successive primes $p,q$, from Bertrand’s Postulate we have that $q<2p$, or equivalently, $\frac{1}{p}<\frac{2}{q}$, and so it follows that

which is a contradiction unless $q<8$. However, if $q<8$ then $p\in\{2,3,5\}$, in which case we know numerically that

and the sum on the right is less than or equal to $x_{0}-\epsilon$ by our choice of $b_{0}$. This contradicts our assumption that $x_{0}<\psi(p^{1})$. Hence that assumption is false; $K$ is nonempty, and there exists some maximum $k_{0}$ such that $\psi(p^{k_{0}})\leq x_{0}$.

We have now that $S_{0}=B_{0}-T_{0}$ or $S_{0}=\psi(p^{k_{0}})$, and in either case $S_{0}$ has been constructed such that $S_{0}<x+\epsilon$. Finally, let $\epsilon_{0}=\left|x-S_{0}\right|$, and if $\epsilon_{0}<\epsilon$, then we have obtained a sum which is within an $\epsilon$-radius of $x$, but if $\epsilon_{0}\geq\epsilon$, it follows that $S_{0}<x-\epsilon$. In that case repeat the process above, this time letting $x_{1}=\epsilon_{0}$, choosing the maximum possible $b_{1}$ and $t_{1}$ to form $B_{1}>x_{1}-\epsilon$ and $T_{1}>B_{1}-(x_{1}+\epsilon)$ as in (5) and (6), and letting $S_{1}=S_{0}+(B_{1}-T_{1})$, or if $B_{1}=T_{1}$ letting $S_{1}=S_{0}+\psi(p^{k_{1}})$ for $p=p_{b_{1}}$ and $k_{1}$ maximum such that $\psi(p^{k_{1}})\leq x_{1}$. Let $\epsilon_{1}=\left|x-S_{1}\right|$, and note that we have either $T_{1}>B_{1}-(x_{1}+\epsilon)$, from which it follows that $\left(B_{1}-T_{1}\right)<x_{1}+\epsilon$, or else we have $\psi(p^{k_{1}})\leq x_{1}$, and in either case we have $S_{1}<x+\epsilon$; furthermore, note that $\epsilon_{1}<\epsilon_{0}$.

Continue until we have $\epsilon_{k}<\epsilon$ for some $k$. This is possible since given a sum $S_{m}$ and $\epsilon_{m}=\left|x-S_{m}\right|$ with $\epsilon_{m}\geq\epsilon$, we can always find a tail of $\frac{\pi}{G}\sum_{p}\frac{1}{p^{2}}$ using only primes larger than were used to form $S_{m}$ and which when added to $S_{m}$ takes us greater than $x-\epsilon$, and we have already shown a method for cutting off the tail in the event that it takes us greater than $x+\epsilon$; this process reduces $\epsilon_{m}$ and the process terminates when we have $\epsilon_{k}<\epsilon$ for some $k$, yielding $S_{k}$, a sum in the range $\left(x-\epsilon,x+\epsilon\right)$.

Recall that we are in the process of identifying primes to use to form $\{n_{k}\}$, so at each iteration of this process we must necessarily identify primes which have not yet been used. To see that we can always find a sufficiently large tail of the series which does not use any primes already used, assume we have obtained a sum $S_{m}$ and $\epsilon_{m}=x-S_{m}$ as described above, with $\epsilon_{m}\geq\epsilon$, and for notational convenience let $n=m+1$. Let $b_{n}$ be maximum such that

(following the same method as before, having at this stage replaced $x$ with $x_{n}=\epsilon_{m}$, the distance from $S_{m}$ to $x$). We must consider two cases: the first case in which the last term added to form $S_{m}$ was $(B_{m}-T_{m})$, in which we need to show that $b_{n}>t_{m}$, and the second case in which the last term added was $\psi(p^{k_{m}})$ for $p=p_{b_{m}}$ and appropriate choice of $k_{m}$

In the first case, note that if the term containing $p_{t_{m}+1}$ were included in $B_{m}-T_{m}$, then we would have $B_{m}-T_{m}>x_{m}+\epsilon$, by choice of $t_{m}$ as in (6). Therefore, if we let $b_{n}=t_{m}+1$, then we would have already $B_{n}>x_{n}-\epsilon$. Hence, since $b_{n}$ is the maximum which allows $B_{n}$ to fulfill this condition, we have that $b_{n}>t_{m}$.

Now in the second case, in which $S_{m}=S_{m-1}+\psi(p^{k_{m}})$ where $p=p_{b_{m}}$, since $k_{m}$ was chosen as the maximum such that $\psi(p^{k_{m}})\leq x_{m}$ (or equivalently, such that $S_{m}=S_{m-1}+\psi(p^{k_{m}})\leq x$), and since we must have $S_{m}\leq x-\epsilon$ or else we would not need to continue, then the difference $\psi(p^{k_{m}+1})-\psi(p^{k_{m}})>x_{n}-\epsilon$ (since the distance from our current $S_{m}$ to $x$ is exactly $\epsilon_{m}=x_{n}$). Hence it suffices to show there exists a $B_{n}$ greater than this difference, but note that $\psi(p^{k_{m}+1})-\psi(p^{k_{m}})$ is the difference between successive partial sums of a convergent geometric series, and so we have that $\psi(p^{k_{m}+1})-\psi(p^{k_{m}})\leq\psi(p^{2})-\psi(p)<\psi(p^{2})$. Finally, since

we see that if we can find a $B_{n}$ larger than $\frac{2\pi}{p^{3}G}$ we are done. In fact, if we let $q$ be the next prime after $p$, so $q=p_{b_{m}+1}$, then since $q<2p$ implies $2q^{2}<8p^{2}$, which implies

we have that if $p>8$ it follows that

and so $\psi(p^{2})<\frac{\pi}{q^{2}G}<B_{n}$, letting

On the other hand if $p<8$ it is easily verifiable numerically that $\psi(p^{2})<B_{n}$ with $B_{n}$ as defined in (17).

We have that, in either case, the algorithm producing a value $S_{m}$ with $\epsilon_{m}\geq\epsilon$ will always continue with a $B_{n}>x_{n}-\epsilon$, with appropriate choice of $T_{n}$ or $\psi(p^{k_{n}})$ (with none of the terms of $B_{n}$ using primes used to form $S_{m}$), providing $S_{n}<x+\epsilon$, with $S_{n}>S_{m}$. Note the $S_{k}$ are increasing and bounded above by $x+\epsilon$, and the algorithm continues until we have that $\epsilon_{k}=\left|x-S_{k}\right|<\epsilon$ for some $k$.

Suppose however that the algorithm fails to terminate because the above criterion is never met. In that case, suppose to get a contradiction that there exists some $\delta>0$ so that $\epsilon_{k}\geq\epsilon+\delta$ for all $k$; that is, the algorithm continues, with each successive $S_{k}$ increasing, but never getting closer than $\delta$ to the interval $\left(x-\epsilon,x+\epsilon\right)$. However, note that for any given $S_{n}=S_{m}+(B_{n}-T_{n})$, the term $B_{n}>\delta$ by the argument above. Meanwhile, recall that if the term containing $p_{t_{n}+1}$ were included in $B_{n}-T_{n}$, then we would have $B_{n}-T_{n}>x_{n}+\epsilon$, by choice of $t_{n}$ as in (6). Hence, if we let $p=p_{t_{n}+1}$ we must have that the term $\frac{\pi}{p^{2}G}>\delta$, and that inequality must hold at every iteration of the algorithm. This is a contradiction since $\frac{\pi}{p^{2}G}<\delta$ for sufficiently large $p$. A similar argument shows that as the algorithm continues, eventually the first term of $B_{n}$ will itself be less than $\delta$, so that the choice of $t_{n}$ will no longer require the special case involving $\psi(p^{k_{n}})$. Hence, no such $\delta$ exists, and the algorithm will eventually terminate with some $S_{k}\in\left(x-\epsilon,x+\epsilon\right)$, as desired.

Now, let $I$ index the primes used in all the sums of the form $\frac{\pi}{G}\sum_{j=b}^{t}\frac{1}{p_{j}^{2}}$ and $J$ index the primes and multiplicities used in the form $\psi(p^{k})$. Let $n=\prod_{j\in I}p_{j}$ and let $m=\prod_{j\in J}p_{j}^{k_{j}}$. (If $I$ is empty, instead let $I$ index a sufficiently large prime $p$ so that $\frac{\pi}{p^{2}G}+S_{k}$ is still within $\epsilon$ of $x$, and if $J$ is empty, instead let $m=1$). Now if $\lambda(m)=-1$ ($\dagger$) replace $m$ by $pm$ for some $p$ indexed by $I$. Then for all $k\in\mathbb{N}$, let $n_{k}=mn^{2k}$, noting that $\lambda(n_{k})=1$, and as $k\to\infty$ the contribution to $\theta(n_{k})$ by the primes indexed by $I$ will converge on $\frac{\pi}{G}\sum_{j\in I}\frac{1}{p_{j}^{2}}$, and so $\{\theta(n_{k})\}$ will have infinitely many terms within the interval $\left(x-\epsilon,x+\epsilon\right)$, as desired. Hence, $\{\theta(n)\}$ is dense in $(0,\pi)$.

A similar argument, adjusting line $(\dagger)$ appropriately, shows there exists an infinite sequence of integers $\{n_{k}\}$, with each $n_{k}$ having $\lambda(n_{k})=-1$, which sequence $\{\theta(n_{k})\}$ converges to any $x\in(\pi,2\pi)$. Hence we have finally that $\{w(n)\}$, $n\in\mathbb{N}$, is dense in the unit circle in $\mathbb{C}$, as desired. ∎

Let $x\in(0,\pi)$, and define

Now let $X=A_{x}\cup B_{x}$. $X$ is a subset of $\mathbb{N}$, and since the Density Lemma implies that $X$ is nonempty, it has a least element; call it $d_{0}$, and let $X_{0}=d_{0}\mathbb{N}=\left\{d_{0}n\ \big{|}\ n\in\mathbb{N}\right\}$. Note that $\delta(X_{0})=\frac{1}{d_{0}}$. The Prime Number Theorem implies that the natural density of $n\in\mathbb{N}$ for which $\lambda(n)=1$ equals the natural density of $m\in\mathbb{N}$ for which $\lambda(m)=-1$, and it follows that $\delta(X_{0}\cap A_{x})=\delta(X_{0}\cap B_{x})$.

We have, again from the Density Lemma, that $X-X_{0}$ is not empty, so let $d_{1}$ be its smallest element and define $X_{1}=d_{1}\mathbb{N}$ similarly to above, and note that PNT again implies $\delta(X_{1}\cap A_{x})=\delta(X_{1}\cap B_{x})$. Let $m_{1}=\mathrm{lcm}(d_{0},d_{1})$ be the least common multiple, and note that if we similarly define $M_{1}=m_{1}\mathbb{N}$ then we have that $M_{1}=X_{0}\cap X_{1}$ and that $\delta(M_{1}\cap A_{x})=\delta(M_{1}\cap B_{x})$. Since natural density is additive, we have that

and likewise if we replace $A_{x}$ with $B_{x}$. It follows that

Continue in this way to let $d_{2}$ be the least element of $X-(X_{0}\cup X_{1})$, letting $X_{2}=d_{2}\mathbb{N}$ and $m_{2}=\mathrm{lcm}(m_{1},d_{2})$, concluding ultimately that for all $k$

implying that $\delta(A_{x})=\delta(B_{x})$. Now let $y\in(0,\pi)$ with $x<y$ and define $A_{y},B_{y},Y$ similarly to above. Then if we define $W=X-Y$, $A=A_{x}-A_{y}$, and $B=B_{x}-B_{y}$, we have that $\delta\left(W\cap A\right)=\delta\left(W\cap B\right)$ and so $\delta(A)=\delta(B)$ as desired. ∎