On the Smallest Counterexample to the
Log-Concavity of the D’Arcais Polynomials
January 01, 1970
Recently, Starr used asymptotic methods to disprove a conjecture by Heim–Neuhauser and Abdesselam about the log-concavity of the D’Arcais polynomials, without giving an explicit counterexample. We refine the asymptotics, to give the necessary estimates on convolutions of \(\sigma_{-1}\), and identify the first counterexample at \(\lambda = 65\,214\,507\,758\,400\). We also consider the asymptotic density of such counterexamples.
The D’Arcais polynomials [1] (or Nekrasov-Okounkov polynomials [2]–[4] in combinatorics) are a sequence of polynomials given by \[\label{eqn:darcais:exp} \sum_{n=0}^\infty P_n^\sigma(X) q^n \mathrel{\vcenter{:}}= \prod_{m=1}^\infty (1 - q^m)^{-X} = \exp\Bigg( X \sum_{j=1}^\infty \sigma(j) \frac{q^j}{j} \Bigg) \,,\tag{1}\] where \(\sigma(n) = \sigma_1(n)\), with \(\sigma_a(n) \mathrel{\vcenter{:}}= \sum_{d \mid n} d^a\) the generalised sum-of-divisors function. These polynomials are important in understanding the Fourier coefficients of \(\eta(q)^{-X}\); the Lehmer conjecture [5] on the non-vanishing of the Ramanujan \(\tau\)-function is equivalent to the conjecture that \(P_n^\sigma(X)\) has no root at \(X = -24\). The polynomials have been generalised to replace \(\sigma\) with an arbitrary arithmetic function \(g \colon \mathbb{N} \longrightarrow \mathbb{Z}\), with \(g(1) = 1\), connecting to areas such the Weil conjectures [6] and the study of subgroups of a given index in \(\mathbb{Z}^\ell\) [7].
We are interested in the log-concavity of the polynomial \(P_n^\sigma(X)\). Recall that a sequence \((a_n)_{n=1}^\infty\) of real numbers is said to be log-concave at \(k\) if \[a_k^2 \geq a_{k + 1}a_{k-1} \,.\] For any polynomial \(H(X) = \sum_{i=0}^d h_i X^i\), the finite sequence \((h_i)_{i=0}^d\) of coefficients can be extended by 0 to obtain an infinite sequence. We say \(H(X)\) is log-concave at \(k\) if this extended sequence is log-concave at \(k\). In their study of the D’Arcais polynomials, Heim and Neuhauser [8] conjectured that they are always log-concave. Abdesselam further generalised this [7] to a certain family of polynomials \(P_n^{g_\ell}(X)\) related to the number of subgroups of a given index in \(\mathbb{Z}^\ell\) (here \(g_2 = \sigma\), and \(g_\ell\) is some recursively defined generalisation). Starr examined this case, and disproved the conjecture for \(\ell = k = 2\).
Extracting the coefficient of \(X^k\) in Eqn. 1 , via the Taylor series of \(\exp(X)\), shows \[\sum_{n=0}^\infty p_n^\sigma(k) q^n = \frac{1}{k!} \Bigg( \sum_{j=1}^\infty \frac{\sigma(j)}{j} q^j \Bigg)^k = \frac{1}{k!} \Bigg( \sum_{j=1}^\infty \sigma_{-1}(j) q^j \Bigg)^k \,,\] Hence \[p_n^\sigma(k) = \frac{1}{k!} \sum_{\substack{j_1 + \cdots + j_k = n \\ j_k \geq 1}} \sigma_{-1}(j_1) \sigma_{-1}(j_2) \cdots \sigma_{-1}(j_k) = \frac{1}{k!} (\underbrace{\sigma_{-1} \ast \cdots \ast \sigma_{-1}}_{k})(n) \,,\] where \((a \ast b)(n) \mathrel{\vcenter{:}}= \sum_{i=1}^{n-1} a(i) b(n-i)\) is the Cauchy (or additive) convolution of two arithmetic functions \(a,b \colon \mathbb{N} \longrightarrow \mathbb{C}\). Starr used asymptotic estimates on \(\sigma_{-1}\) convolutions investigate the log-concavity of \(P_n^\sigma(X)\) at \(k = 2, 3, 4, \ldots\). Starr gives the following asymptotic.
Lemma 1 (Starr [9]). Let \(a, b \geq 0\) and \(r, s \geq 1\), then we have \[\begin{gather} \sum_{k = 1}^{n - 1} k^a \sigma_{-r}(k) (n - k)^b \sigma_{-s}(n - k) \sim \\[-1ex] \frac{\Gamma(a + 1)\Gamma(b + 1)}{\Gamma(a + b + 2)} \cdot \frac{\zeta(r + 1)\zeta(s + 1)}{\zeta(r + s + 2)} \cdot n^{a + b + 1}\sum_{d \mid n} d^{-r-s-1}, \qquad \text{as } n \to \infty. \end{gather}\]
This is connected to sums \(\sum_{k=1}^{n-1} \sigma_r(k) \sigma_s(n-k)\) considered by Ramanujan [10], involving sum-of-divisor functions with \(r,s\) positive odd integers. In Starr’s case, more precise asymptotics, with the leading order remainder terms, are given by [11]–[13]. From this asymptotic, Starr calculates the limiting behaviour of the ratio \(\tfrac{p_n^\sigma(k)^2}{p_n^\sigma(k-1) p_n^\sigma(k+1)}\), which controls the log-concavity of \(P_n^\sigma(X)\) at \(k\).
Theorem 1 (Starr [9]). The following limit behaviour holds \[\begin{align} \liminf_{n \to \infty} \frac{p_n^{\sigma}(2)^2}{p_n^{\sigma}(3)p_n^{\sigma}(1)} &= 0 \,, \\ \intertext{and for 3 \leq k \leq n - 1,} \liminf_{n \to \infty} \frac{p_n^{\sigma}(k)^2}{p_n^{\sigma}(k+1)p_n^{\sigma}(k-1)} &= \frac{k}{k-1} \cdot \frac{\zeta(2k+2)\zeta(2k-2)}{\zeta(2k)^2} \cdot \frac{\zeta(2k-1)^2}{\zeta(2k-3)\zeta(2k+1)} > 1 \,. \end{align}\]
In particular \(P_n^\sigma(X)\) is eventually not log-concave at \(k=2\) for some infinite list of \(n\) but is log-concave at \(k \geq 3\), for all but finitely many \(n\). We establish that the first counterexample to the log-concavity at \(k=2\) occurs at \(\lambda = 65\,214\,507\,758\,400 = 29\sharp \cdot 2^5 \, 3^2 \, 5^1 \, 7^1\) (the 80th superabundant number), where \(m\sharp\) is the primorial of \(m\) (the product of primes \(\leq m\)).
Theorem 2. Let \(\lambda = 65\,214\,507\,758\,400\). Then \(n = \lambda\) is first counterexample to the log-concavity of \(P_n^\sigma(X)\) at \(k=2\), meaning
the polynomial \(P_\lambda^\sigma(X)\) is not log-concave at \(k=2\), i.e., \[\label{eqn:p22p1p3} \frac{(p_\lambda^{\sigma}(2))^2}{p_\lambda^{\sigma}(1) \cdot p_\lambda^{\sigma}(3)} < 1 \,, \text{ and}\qquad{(1)}\]
for any \(m<\lambda\), the polynomial \(P_m^\sigma(X)\) is log-concave at \(k=2\).
We prove the two parts of Theorem 2 separately. For part i), the first step is to refine the asymptotics Starr gives. In Section 2 we independently establish explicit bounds on the 2-fold and 3-fold convolutions of \(\sigma_{-1}\) and thereby on \(p_n^\sigma(2)\) and \(p_n^\sigma(3)\). These bounds involve only simple combinations of sum-of-divisor functions (no convolutions), so can be computed efficiently. The claim in Eqn. ?? is then a direct verification. Part ii) requires a structured approach and significant computer time to exhaustively check no smaller values give a counterexample. We outline the strategy in Section 3.
Finally, the methods used to bound the fraction in Eqn. ?? culminate in bounding the density of counterexamples.
Theorem 3. The following bounds on the asymptotic density of counterexamples hold: \[\begin{align} \delta_\mathrm{sup} & \mathrel{\vcenter{:}}= \limsup_{x \to \infty} \frac{\#\left\{n \leq x : \frac{(p_n^{\sigma}(2))^2}{p_n^{\sigma}(1) \cdot p_n^{\sigma}(3)} < 1 \right\}}{x} \leq 0.000679406 \intertext{and} \delta_\mathrm{inf} & \mathrel{\vcenter{:}}= \liminf_{x \to \infty} \frac{\#\left\{n \leq x : \frac{(p_n^{\sigma}(2))^2}{p_n^{\sigma}(1) \cdot p_n^{\sigma}(3)} < 1 \right\}}{x} \geq 2.47323 \ldots \times 10^{-16} \,. \end{align}\]
Remark 1. This implies that, if the asymptotical density \(\delta\) of counterexamples to the log-concavity of \(P_n^\sigma(X)\) at \(k = 2\) exists, then it will satisfy \[2.47323 \ldots \times 10^{-16} \leq \delta \leq 0.000679406 \, .\] In particular, it will be both positive and strictly less than 1. In any case, \(P_n^\sigma(X)\) is log-concave at \(k=2\) for infinitely many \(n\) and not log-concave at \(k=2\) for infinitely many other \(n\).
We end the introduction by recalling some notation and basic results.
For \(m, n \in \mathbb{N}\) we denote their greatest common divisor by \((m,n)\) and their least common multiple by \([m,n]\). Some of the constants we use are derived from quotients Riemann \(\zeta\)-values: fix parameters \(r, s > 1\) and \(\mu \in \mathbb{N}\), and write \[C_\mu^{(r,s)} \mathrel{\vcenter{:}}= \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{\mu-1} \frac{1}{d'^r e'^s} \,,\] for the partial sum to \((\mu-1)\)-terms. From [13], we recall that \[\label{eqn:gcd1sum} \lim_{\mu \to \infty} C_\mu^{(r,s)} = \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{\infty} \frac{1}{d'^r e'^s} = \frac{\zeta(r)\zeta(s)}{\zeta(r+s)} \,.\tag{2}\] We also recall the following basic estimate from integral comparison \[\label{eqn:harmonic} \log(n) \leq \sum_{i=1}^{n-1} i^{-1} \leq \log(n) + 1 \,.\tag{3}\]
The main objective in this section lies in proving part i) of Theorem 2 which we accomplish by using the following proposition, with suitable parameters.
Proposition 4. Let \(\mu \in \mathbb{N}\) be arbitrary and define \[\begin{align} p_n^\sigma(2)_{\mathrm{up}} &\mathrel{\vcenter{:}}= \frac{1}{2!} \left(\frac{5}{2} (n-1) \sigma_{-3}(n) + (1 + \log(n))^2\right) \,;\\[1ex] p_n^\sigma(2)_{\mathrm{low},\mu} &\mathrel{\vcenter{:}}= \frac{1}{2!} \left(C_\mu^{(2,2)} \cdot (n-1) \sigma_{-3}(n) - C_\mu^{(2,2)} \cdot (n-1) \cdot 2n^{{\frac{1}{2}}} \Big(\frac{\mu}{n}\Big)^3 - (\log(n) + 1)^2\right) \,;\\[1ex] p_n^\sigma(3)_{\mathrm{up}} &\mathrel{\vcenter{:}}= \begin{aligned}[t]\frac{1}{3!}\bigg( \frac{35(n-1)^2}{16} \sigma_{-5}(n) & {} + \frac{15(n-1)}{4} \sigma_{-4}(n) \zeta(3) (1 + \log(n)) \\[-1ex] &{} + \frac{5}{2} (n-1) \sigma_{-3}(n) \zeta(2) + n (1 + \log(n))^3\bigg)\, ;\end{aligned}\\[1ex] p_n^\sigma(3)_{\mathrm{low},\mu} &\mathrel{\vcenter{:}}= \frac{1}{3!}\bigg(\begin{align}[t] \frac{(n-1)^2}{2} \sigma_{-5}(n) & C_\mu^{(2,2)} C_\mu^{(4,2)} - (n-1)^2 n^{{\frac{1}{2}}} \Big(\frac{\mu}{n}\Big)^5 C_\mu^{(2,2)} C_\mu^{(4,2)}\\[-0.5ex] & {} - \frac{3(n-1)}{2} C_\mu^{(2,2)} \sigma_{-4}(n) \zeta(3) (1 + \log(n))\\[-0.5ex] &{} - C_\mu^{(2,2)} \mu^2 \zeta(2) (1 + \log(n)) - n(1+\log(n))^3\bigg) \,.\end{align} \end{align}\] Then, for all \(n \in \mathbb{N}\), we have \[\begin{align} p_n^\sigma(2)_{\mathrm{up}} \geq p_n^\sigma(2) &\geq p_n^\sigma(2)_{\mathrm{low},\mu}\, ,\\* p_n^\sigma(3)_{\mathrm{up}} \geq p_n^\sigma(3) &\geq p_n^\sigma(3)_{\mathrm{low},\mu} \,. \end{align}\]
Remark 2. Obtaining effective bounds requires careful manipulation of the given Cauchy convolutions. A simpler lower bound for \(p_n^\sigma(2)\) may be obtained via the ansatz \[p_n^\sigma(2) = \frac{1}{2!} \sum_{k = 1}^{n-1} \sigma_{-1}(k)\sigma_{-1}(n - k) \geq \frac{2}{n^2} \sum_{k = 1}^{n - 1}\sigma(k)\sigma(n - k) = \frac{1}{6n^2} \left(5\sigma_3(n) - (6n - 1)\sigma(n)\right)\] where we have used that \(k(n - k)\) attains its maximum at \(k = \frac{n}{2}\) and have invoked a well-known identity of Eisenstein series. However, its leading term is \(\sim {\frac{5}{6}}n\sigma_{-3}(n)\) while the leading term of \(p_n^\sigma(2)_{\mathrm{low},\mu}\) is \(\sim \smash{C_\mu^{(2,2)}}n\sigma_{-3}(n)\), where the constant \(\smash{C_\mu^{(2,2)}}\) can be taken arbitrarily close to \(\frac{5}{2}\), the constant given by Lemma 1.
The proof of Proposition 4 is split into smaller pieces within the subsections to follow.
Fix \(f: \mathbb{N}\longrightarrow \mathbb{R}\) a general function. We are interested in sums of the type \[\sum_{\substack{k = 1 \\ d \mid k \,, e \mid n - k}}^{n-1} f(k) \, .\] An argument in [9] shows that if \((d,e) \nmid n\), then no \(k\) satisfy the conditions \(d \mid k, e \mid n -k\). Otherwise \((d,e) \mid n\), and all such \(k\) satisfying \(d \mid k , e \mid n-k\) are given by \(k \equiv k_0 \, \mathrm{mod} \, [d,e]\), where \(k_0 = xnd / (d,e)\) for \((d,e) = x d + y e\) a solution given by the extended Euclidean algorithm. Letting \(\mathbb{1}_\mathbb{N}\) be the indicator function of \(\mathbb{N}\) on the rationals, this means \[\sum_{\substack{k = 1 \\ d \mid k \,, e \mid n - k}}^{n-1} f(k) \, = \sum_{\substack{k=1 \\ k \equiv k_0 \, \mathrm{mod} \, [d,e]}}^{n-1} f(k)\mathbb{1}_\mathbb{N}\left(\frac{n}{(d,e)}\right)\,.\]
Even though the examination of these sums would be an interesting field of study in itself, we will focus on the two specific cases appearing in our computations.
Lemma 2. Define \[\mathcal{F}(n,d,e) \mathrel{\vcenter{:}}= \sum_{\substack{k = 1 \\ d \mid k \,, e \mid n - k}}^{n-1} 1 \, .\] We have \[\left(\frac{n-1}{[d,e]} - 1\right) \cdot \mathbb{1}_\mathbb{N}\left( \frac{n}{(d, e)} \right) \leq \mathcal{F}(n, d, e) \leq \left( \frac{n-1}{[d,e]} + 1 \right) \mathbb{1}_\mathbb{N}\left( \frac{n}{(d, e)} \right) \,.\]
Proof. This is because there are \(\lfloor \frac{n-1}{[d,e]} \rfloor\) complete intervals of length \([d,e]\), each of which contributes a solution, plus a potential solution from the final incomplete interval. Since \(\frac{n-1}{[d,e]} -1 \leq \lfloor \frac{n-1}{[d,e]} \rfloor\) and \(\lfloor \frac{n-1}{[d,e]} \rfloor + 1 \leq \frac{n-1}{[d,e]} + 1\), we obtain the claimed bounds. ◻
In what follows, it is convenient to use the following slightly weaker bounds \[\label{eqn:p2inner:bounds} \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) - 1 \leq \mathcal{F}(n, d, e) \leq \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) + 1 \,.\tag{4}\]
Lemma 3. Define \[\mathcal{G}(n,d,e) \mathrel{\vcenter{:}}= \sum_{\substack{k = 1 \\ d \mid k \,, e \mid n - k}}^{n-1} k - 1 \, .\] We have \[\begin{align} \biggl( -\frac{3(n-1)}{2} + \frac{(n-1)^2}{2[d,e]} \biggr) \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) & \leq \mathcal{G}(n,d,e) \\ &\leq \biggr( [d,e] + \frac{3(n-1)}{2} + \frac{(n-1)^2}{2[d,e]} \biggr) \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) \, . \end{align}\]
Proof. Again \[\begin{align} \mathcal{G}(n,d,e) = \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) \cdot \sum_{\substack{k=1 \\ k \equiv k_0 \, \mathrm{mod} \, [d,e]}}^{n-1} (k-1) \end{align}\] Let \(M \mathrel{\vcenter{:}}= [d,e]\), and \(\alpha = \lfloor (n-1) / M \rfloor\). If \(M \geq n\), it is obvious that \[\begin{align} \sum_{\substack{k=1 \\ k \equiv k_0 \, \mathrm{mod} \, [d,e]}}^{n-1} (k-1) \leq k_0 - 1 \leq [d,e] \leq [d,e] + \frac{3(n-1)}{2} + \frac{(n-1)^2}{2[d,e]} \, . \end{align}\] Otherwise, by taking the maximum value of \(k_0 = M\) in the interval \([1, \ldots, M]\), we find \[\begin{align} \sum_{\substack{k=1 \\ k \equiv k_0 \, \mathrm{mod} \, [d,e]}}^{n-1} (k-1) & \leq \sum_{\substack{k=1 \\ k \equiv [d,e] \, \mathrm{mod} \, [d,e]}}^{n} (k-1) \\ & \leq \sum_{i=1}^{\alpha+1} (i M - 1) \\ & = \frac{(\alpha+1)(\alpha+2)}{2} M \\ & \leq M + \frac{3(n-1)}{2} + \frac{(n-1)^2}{2M} \\ & = [d,e] + \frac{3(n-1)}{2} + \frac{(n-1)^2}{2[d,e]} \end{align}\] Conversely, by taking the minimum value of \(k_0 = 1\) in the interval \([1,\ldots,M]\), we can find a lower bound. Since \(\alpha > (n-1)/M - 1\), a lower bound is certainly \[\begin{align} \geq \sum_{i=1}^{\alpha+1} ((i-1)M) - (n-1) & = \frac{\alpha(\alpha+1)}{2} M - (n-1) \\ & \geq \frac{(n-1)^2}{2M} - \frac{3(n-1)}{2} \\ & = \frac{(n-1)^2}{2[d,e]} - \frac{3(n-1)}{2} \,. \end{align}\] This yields the claim. ◻
For \(k = 2\), we write out the convolution and apply explicit estimates. We have \[2! \cdot p_n^\sigma(2) \, = \, \sum_{k=1}^{n-1} \sigma_{-1}(k) \sigma_{-1}(n-k) \, = \, \sum_{k=1}^{n-1} \sum_{\substack{d \smash{\mid} k \\ e \smash{\mid} n-k}} \frac{1}{d e} \, = \, \sum_{d,e=1}^{n-1} \frac{1}{d e} \mathcal{F}(n,d,e)\] as in Lemma 2. Applying Eqn. 4 gives \[\begin{align} 2! \cdot p_n^\sigma(2) & \leq \sum_{d,e=1}^{n-1} \frac{1}{d e} \Bigg( \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) + 1 \Bigg)\\ & \leq \sum_{d,e=1}^{n-1} \frac{1}{d e} \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) + (1 + \log(n))^2 \,, \end{align}\] by Eqn. 3 . Set \(w = (d,e) \mid n\), with \(d = d' w\) and \(e = e' w\), so that \((d',e') = 1\) and \([d,e] = w d' e'\). Therefore \[\begin{align} 2! \cdot p_n^\sigma(2) & \leq (n-1) \sum_{w\mid n} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \frac{1}{w^3 d'^2 e'^2} + (1 + \log(n))^2 \\ & \leq (n-1) \sum_{w\mid n} \frac{1}{w^3} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{\infty} \frac{1}{d'^2 e'^2} + (1 + \log(n))^2 \\ & = \lim_{\mu \to \infty} C_{\mu}^{(2,2)} \cdot (n-1) \sum_{w\mid n} \frac{1}{w^3} + (1 + \log(n))^2 \\ & = \frac{5}{2} (n-1) \sigma_{-3}(n) + (1 + \log(n))^2 \,, \end{align}\] where \(\lim_{\mu\to\infty} C_\mu^{(2,2)} = \frac{5}{2}\) from Eqn. 2 .
For the lower bound on \(p_n^\sigma(2)\), note that \(C_\mu^{(2,2)} < \frac{5}{2}\), and is monotonically increasing as \(\mu \to \infty\). By Eqns. 4 and 3 , we again have \[\begin{align} 2! \cdot p_n^\sigma(2) &= \sum_{d,e=1}^{n-1} \frac{1}{d e} \mathcal{F}(n,d,e) \\ &\geq \sum_{d,e=1}^{n-1} \frac{1}{d e} \Bigg( \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) - 1 \Bigg)\\ &\geq \sum_{d,e=1}^{n-1} \frac{1}{d e} \frac{n-1}{[d,e]} \cdot \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) - ( \log(n) + 1)^2 \end{align}\] Set \(w = (d,e) \mid n\), with \(d = d' w\) and \(e = e' w\), so that \((d',e') = 1\) and \([d,e] = w d' e'\). Then \[\begin{align} 2! \cdot p_n^\sigma(2) &\geq (n-1) \sum_{w\mid n} \frac{1}{w^3} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \frac{1}{d'^2 e'^2} - (\log(n) + 1)^2 \notag \\ &\geq (n-1) \sum_{w\mid n} \frac{C_\mu^{(2,2)}}{w^3} - (n-1) \sum_{\substack{w\mid n \\ n/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} - (\log(n) + 1)^2 \label{eqn:p2penult} \\ &\geq C_\mu^{(2,2)} \cdot (n-1) \sigma_{-3}(n) - C_\mu^{(2,2)} \cdot (n-1) \cdot 2n^{{\frac{1}{2}}} \Big(\frac{\mu}{n}\Big)^3 - (\log(n) + 1)^2 \,, \notag \end{align}\tag{5}\] by splitting the sum over \(w\) according to whether \(n/w \geq \mu\) or \(n/w < \mu\). More precisely, we have \[\begin{align} \sum_{w \mid n} \frac{C_{n/w}^{(2,2)}}{w^3} \geq \sum_{\substack{w \mid n \\ n/w \geq \mu}} \frac{C_{\mu}^{(2,2)}}{w^3} = \sum_{\substack{w \mid n}} \frac{C_{\mu}^{(2,2)}}{w^3} - \sum_{\substack{w \mid n \\ n/w < \mu}} \frac{C_{\mu}^{(2,2)}}{w^3} \,. \end{align}\] Note above, we also use the elementary estimate \(\sum_{w\mid n} 1 = \sigma_0(n) \leq 2 n^{\frac{1}{2}}\).
For \(k = 3\), we write out the convolution and apply estimates on \(p_{n-k}^\sigma(2)\). We have \[\begin{align} 3! \cdot p_n^\sigma(3) & = \sum_{k=1}^{n-1} 2! \cdot p_{k}^\sigma(2) \sigma_{-1}(n-k) \\ & \leq \sum_{k=1}^{n-1} \Big( \frac{5}{2} (k-1) \sigma_{-3}(k) + (1 + \log(k))^2\Big) \sigma_{-1}(n-k) \\ & = \frac{5}{2} \sum_{k=1}^{n-1} (k-1) \sigma_{-3}(k) \sigma_{-1}(n-k) + \sum_{k=1}^{n-1} (1 + \log(k))^2 \sigma_{-1}(n-k) \,. \end{align}\]
We treat each term separately. For the latter \[\begin{align} \sum_{k=1}^{n-1} (1 + \log(k))^2 \sigma_{-1}(n-k) &= \sum_{d=1}^{n-1} \sum_{\substack{k=1 \\ d \mid n-k}}^{n-1} (1 + \log(k))^2 \frac{1}{d} \\ &\leq \sum_{k=1}^{n-1} (1 + \log(k))^2 \sum_{d=1}^{n-1} \frac{1}{d} \\ &\leq \sum_{k=1}^{n-1} (1 + \log(n))^2 \sum_{d=1}^{n-1} \frac{1}{d} \\ & \leq n (1 + \log(n))^3 \,. \end{align}\] For the former, invoke Lemma 3. It holds that \[\begin{align} \sum_{k=1}^{n-1} (k-1) \sigma_{-3}(k) \sigma_{-1}(n-k) & = \sum_{d,e=1}^{n-1} \sum_{\substack{k=1 \\ d \mid k \\ e \mid n-k}}^{n-1} \frac{k-1}{d^3 e} \\ & = \sum_{d,e=1}^{n-1} \frac{1}{d^3 e} \mathcal{G}(n,d,e) \\ & \leq \sum_{d,e=1}^{n-1} \frac{1}{d^3 e} \Big( \frac{(n-1)^2}{2[d,e]} + \frac{3(n-1)}{2} + [d,e] \Big) \mathbb{1}_\mathbb{N}\biggl( \frac{n}{(d, e)} \biggr) \,. \end{align}\] Again set \(w = (d,e) \mid n\), with \(d = d'w\) and \(e = e' w\), where \((d',e') = 1\) and \([d,e] = d' e' w\). Then we have \[\begin{align} & \sum_{k=1}^{n-1} (k-1) \sigma_{-3}(k) \sigma_{-1}(n-k) \\ &\leq \sum_{w \mid n} \sum_{d',e'=1}^{n/w-1} \frac{1}{d'^3 e' w^4} \bigg( \frac{(n-1)^2}{2d' e' w} + \frac{3(n-1)}{2} + d' e' w\bigg) \\ & \leq \sum_{w \mid n} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \bigg( \frac{(n-1)^2}{2 d'^4 e'^2 w^5} + \frac{3(n-1)}{2 d'^3 e' w^4}+ \frac{1}{d'^2 w^3} \bigg) \\ & \leq \frac{(n-1)^2}{2} \lim_{\mu \to \infty} C_\mu^{(4,2)} \cdot \sum_{w \mid n} \frac{1}{w^5} + \frac{3(n-1)}{2} \sum_{w \mid n} \frac{1}{w^4} \sum_{d'=1}^{\infty} \frac{1}{d'^3} \sum_{e'=1}^{n-1} \frac{1}{e'} + (n-1) \sum_{w \mid n} \frac{1}{w^3} \sum_{d'=1}^\infty \frac{1}{d'^2} \\ &= \frac{(n-1)^2}{2} \cdot \frac{7}{4} \sigma_{-5}(n) + \frac{3(n-1)}{2} \sigma_{-4}(n) \zeta(3) (1 + \log(n)) + (n-1) \sigma_{-3}(n) \zeta(2) \,, \end{align}\] where \(\lim_{\mu \to \infty} C_\mu^{(4,2)}= \frac{7}{4}\) from Eqn. 2 .
Overall, we obtain \[\begin{align} 3! \cdot p_n^\sigma(3) \leq {} \frac{35(n-1)^2}{16} \sigma_{-5}(n) + \frac{15(n-1)}{4} \sigma_{-4}(n) \zeta(3) (1 + \log(n)) & \\ \qquad {} + \frac{5}{2} (n-1) \sigma_{-3}(n) \zeta(2) + n (1 + \log(n))^3 & \end{align}\]
For \(k = 3\), we write out the convolution and apply estimates on \(p_{n-k}^\sigma(2)\). Fix a parameter \(\mu > 0\), we take the second to last equality for \(p_k^\sigma(2)\) in Eqn. 5 , to obtain a better estimate here, giving \[\begin{align} 3! \cdot p_n^\sigma(3) & = \sum_{k=1}^{n-1} 2! \cdot p_{k}^\sigma(2) \sigma_{-1}(n-k) \\ & \geq \sum_{k=1}^{n-1} \Big( C_\mu^{(2,2)} \cdot (k-1) \sigma_{-3}(k) - (k-1) \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} - (\log(k) + 1)^2 \Big) \sigma_{-1}(n-k) \\ & \geq \begin{aligned}[t] C_\mu^{(2,2)} \sum_{k=1}^{n-1} & (k-1) \sigma_{-3}(k) \sigma_{-1}(n-k) - \sum_{k=1}^{n-1} (k-1) \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} \sigma_{-1}(n-k)\\[-2ex] &- \sum_{k=1}^{n-1} (\log(k) + 1)^2 \sigma_{-1}(n-k) \end{aligned} \end{align}\] Handling each sum separately, we find \[\begin{align} & \sum_{k=1}^{n-1} (\log(k) + 1)^2 \sigma_{-1}(n-k) \leq n (1 + \log(n))^3 \,, \end{align}\] as before. Then \[\begin{align} \sum_{k=1}^{n-1} (k-1) \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} \sigma_{-1}(n-k) & = \sum_{d=1}^{n-1} \sum_{\substack{k=1 \\ d \mid n-k}}^{n-1} (k-1) \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} \frac{1}{d} \\ & \leq \sum_{\substack{k=1}}^{n-1} k \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)}}{w^3} \sum_{d=1}^{n-1} \frac{1}{d} \\ & = \sum_{\substack{k=1}}^{n-1} \sum_{\substack{w\mid k \\ k/w < \mu}} \frac{C_\mu^{(2,2)} \cdot k}{w^3} \sum_{d=1}^{n-1} \frac{1}{d} \\ & \leq \sum_{\substack{w=1}}^{n-1} \sum_{\substack{k = 1 \\ w\mid k}}^{w \mu} \frac{C_\mu^{(2,2)} \cdot k}{w^3} \sum_{d=1}^{n-1} \frac{1}{d}\, . \end{align}\] Setting \(k = \alpha w\), we get \[\begin{align} & \leq\sum_{\substack{w=1}}^{n-1} \sum_{\substack{\alpha = 1}}^{\mu} \frac{C_\mu^{(2,2)} \cdot \alpha w}{w^3} \sum_{d=1}^{n-1} \frac{1}{d} \\ & \leq \sum_{\substack{w=1}}^{n-1} \frac{C_\mu^{(2,2)} \cdot \mu^2}{w^2} \sum_{d=1}^{n-1} \frac{1}{d} \\ & \leq C_\mu^{(2,2)} \mu^2 \zeta(2) (1 + \log(n)) \end{align}\]
Finally, from Lemma 3 as before, we have \[\begin{align} & \sum_{k=1}^{n-1} (k-1) \sigma_{-3}(k) \sigma_{-1}(n-k) \\ & = \sum_{d,e=1}^{n-1} \frac{1}{d^3 e} \mathcal{G}(n,d,e) \\ & \geq \sum_{d,e=1}^{n-1} \frac{1}{d^3 e} \bigg( \frac{(n-1)^2}{2[d,e]} - \frac{3(n-1)}{2} \bigg) \cdot \mathbb{1}_{\mathbb{N}}\Big( \frac{n}{[d,e]} \Big) \,. \end{align}\] Set \(w = (d,e) \mid n\), with \(d = d' w\) and \(e = e' w\) with \((d',e') = 1\). Note \(\lim_{\mu\to\infty} C_\mu^{(4,2)} = \frac{7}{4}\) and \(C_\mu^{(4,2)}\) is monotonically increasing, from Eqn. 2 . Then \[\begin{align} & \geq \sum_{w \mid n} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \frac{1}{d'^3 e' w^4} \bigg( \frac{(n-1)^2}{2e'd'w} - \frac{3(n-1)}{2} \bigg) \\ & = \frac{(n-1)^2}{2} \sum_{w \mid n} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \frac{1}{d'^4 e'^2 w^5} - \frac{3(n-1)}{2} \sum_{w \mid n} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n/w-1} \frac{1}{d'^3 e' w^4} \\ & \geq \frac{(n-1)^2}{2} \sum_{w \mid n} \frac{1}{w^5} C_\mu^{(4,2)} - \frac{(n-1)^2}{2} \sum_{\substack{w \mid n \\ n/w < \mu}} \frac{1}{w^5} C_\mu^{(4,2)} - \frac{3(n-1)}{2} \sum_{w \mid n} \frac{1}{w^4} \sum_{\substack{d',e'=1 \\ (d',e') = 1}}^{n-1} \frac{1}{d'^3 e'} \\ & \geq \frac{(n-1)^2}{2} \sigma_{-5}(n) C_\mu^{(4,2)} - (n-1)^2 n^{\frac{1}{2}}\Big(\frac{\mu}{n}\Big)^5 C_\mu^{(4,2)} - \frac{3(n-1)}{2} \sigma_{-4}(n) \zeta(3) (1 + \log(n)) \,, \end{align}\] by splitting the sum according to whether \(n/w \geq \mu\) or \(n/w < \mu\), as before.
Overall, we obtain \[\begin{align} 3! \cdot p_n^\sigma(3) &\geq \begin{aligned}[t] \frac{(n-1)^2}{2} \sigma_{-5}(n) & C_\mu^{(2,2)} C_\mu^{(4,2)} - (n-1)^2 n^{\frac{1}{2}}\Big(\frac{\mu}{n}\Big)^5 C_\mu^{(2,2)} C_\mu^{(4,2)}\\ &{} - \frac{3(n-1)}{2} C_\mu^{(2,2)} \sigma_{-4}(n) \zeta(3) (1 + \log(n))\\[1ex] &{} - C_\mu^{(2,2)} \mu^2 \zeta(2) (1 + \log(n)) - n(1+\log(n))^3 \, ,\end{aligned} \end{align}\] concluding the proof of Proposition 4.
Having proven Proposition 4, we are in a position to advance to the proof of Theorem 2. For this, we fix \(\mu = 500\). Then we obtain \[\begin{align} C_{500}^{(2,2)} = 2.49452421836436753653\ldots \,, \\ C_{500}^{(4,2)} = 1.74791042060522673981\ldots \,. \end{align}\] Note that \(\lambda = 65\,214\,507\,758\,400 = 29\sharp \cdot 2^5 \, 3^2 \, 5^1 \, 7^1\), where \(m\sharp\) is \(m\) primorial. One can directly compute that \[\begin{align} & \sigma_{-1}(\lambda) = \tfrac{68031360}{11350339} = 5.99377340183407737865\ldots \\ & \sigma_{-3}(\lambda) = \tfrac{3367719899875296294987}{2802008927062274116000} = 1.20189477890284014791\ldots \\ & \sigma_{-5}(\lambda) = \tfrac{6731466498568936993292460214710631881036243488437907}{ 6491741578415157169646584347037579301727795363840000} = 1.03692767453203280218\ldots \,. \end{align}\] One then has \[\begin{align} p_{\lambda}^\sigma(2)_\mathrm{up}& = 9.79762204799613549624\ldots \times 10^{13} \\ p_{\lambda}^\sigma(3)_{\mathrm{low},500} &= 1.60236908240243543394\ldots \times 10^{27} \\ p_{\lambda}^\sigma(1) &= 5.99377340183407737865\ldots \,. \end{align}\] Whence \[\frac{p_\lambda^{\sigma}(2)^2}{p_\lambda^{\sigma}(1) p_\lambda^{\sigma}(3)} < \frac{(p_\lambda^{\sigma}(2)_\mathrm{up})^2}{p_\lambda^{\sigma}(1) \, p_\lambda^{\sigma}(3)_{\mathrm{low},500}} = 0.99949007855677265035\ldots < 1 \,.\] So as claimed \(P_\lambda^\sigma(X)\) is not log-concave at \(k = 2\). 0◻
We establish the minimality in Theorem 2, part ii) via three steps. Firstly we check explicitly that there is no counterexample up to \(n = 10^7\). This requires using more efficient algorithms for computing polynomial products (or convolutions of coefficients), to be tractable. Then we use interval arithmetic to establish that \[\frac{p_n^{\sigma}(2)^2}{p_n^{\sigma}(3)} \,,\] is \(>4\) for \(n \geq 10^7\) (indeed even for \(n \geq 2 \times 10^6\)). This implies any such counterexample \(n\) must have \(\sigma_{-1}(n) > 4\). Finally we enumerate all such \(1 \leq n \leq \lambda\) with the property \(\sigma_{-1}(n) > 4\) and verify none of them give a counterexample.
One can implement the following procedures in any of SageMath, gp/pari or Mathematica, with various advantages and disadvantages in each case. In particular, gp/pari is significantly faster for number
theoretic routines like computing divisor sums, but lacks a native interface to list convolution routines (ListConvolve in Mathematica or convolve in SageMath) which allow the asymptotically faster \(O(n \log(n))\) multiplication of polynomials using the Fast Fourier transform.
On a high-end consumer grade laptop (13th Gen Intel(R) Core(TM) i9-13980HX @ 2.2 GHz, with 64 GB RAM), the computation of \((p_n^\sigma(1))_{1 \leq n \leq 10^7}\) takes the following time in each computer
algebra system with a single thread.
| Software | Command | Time |
|---|---|---|
SageMath |
[ sigma(x,1)/x for x in range(1,10^7)] |
45.7 seconds |
Mathematica |
Map[DivisorSigma[-1, #]&, Range[1,10^7]] |
19.1 seconds |
gp/pari |
[ sigma(i,-1) | i <- [1..10^7]] |
5.3 seconds |
Since \(p_n^\sigma(2) = \frac{1}{2!} (\sigma_{-1} \ast \sigma_{-1})(n)\) is the Cauchy convolution of \(p_n^\sigma(1) = \sigma_{-1}(n)\), the naive method requires \(O(n^2)\) steps to compute directly, making computation beyond \(n = 10^4\) already very time-consuming. Using Fast Fourier Transform based algorithms one can compute polynomial multiplication
(hence such Cauchy convolutions) in time \(O(n \log(n))\). This is available in SageMath via \(\texttt{convolve}\), while in Mathematica it is available via
ListConvolve (with suitable options). Exact rational arithmetic to compute the values of \(p_n^\sigma(2)\) and \(p_n^\sigma(3)\) is very time-consuming, and produces results
with very large height, for example \[p_{50}^\sigma(3) = \frac{16481674253589243490606751802749287}{19881006138756642992377224960000} \,.\] Therefore, for the computations of \(p_n^\sigma(2)\) and \(p_n^\sigma(3)\), we first evaluate \(p_n^\sigma(1)\) as a real number to 100 decimal places (which adds some overhead to the previous
timings), and then carry out the (fast) polynomial multiplication.
On a high-end consumer grade laptop (13th Gen Intel(R) Core(TM) i9-13980HX @ 2.2GHz, with 64 GB RAM), the follow are representative timings for the computation of list convolution with lists of size \(10^7\), in each computer algebra system. Note compute two convolutions, the first to calculate \(p_n^\sigma(2)\) from \(p_n^\sigma(1)\), and a second to compute
\(p_n^\sigma(3)\) from \(p_n^\sigma(2)\) and \(p_n^\sigma(1)\), doubling the overall time. Note Mathematica has an efficient implementation
allowing computation up to \(10^8\) in reasonable time (with sufficient memory); the final line reports the resources required for this computation on the Universität zu Köln departmental server
(Intel(R) Xeon(R) CPU E5-2687W v4 @ 3.00GHz, with 380 GB of RAM).
| Software | Size | Schematic version of command | Time | Memory |
|---|---|---|---|---|
gp/pari |
\(10^7\) | Vec(Pol(list1) * Pol(list2)) |
3.9 hrs | 10.5 GB |
SageMath |
\(10^7\) | convolve(list1, list2) |
1.6 hrs | 37.5 GB |
Mathematica |
\(10^7\) | ListConvolve[list1, list2, {1, -1}, 0] |
12.4 min | 11.6 GB |
Mathematica |
\(10^8\) | ListConvolve[list1, list2, {1, -1}, 0] |
12.7 hrs | 180 GB |
Once the values of \(p_n^\sigma(1)\), \(p_n^\sigma(2)\) and \(p_n^\sigma(3)\) for \(1 \leq n \leq 10^7\) are computed, it
is straightforward to check that no counterexample occurs in this range. As noted, we computed up to \(10^8\) in Mathematica at an earlier stage of the research, and can report \[\min_{1 \leq n \leq 10^8} \frac{(p_n^\sigma(2))^2}{p_n^\sigma(1) p_n^\sigma(3)} = 1.199090503\ldots \,,\] which occurs at \(n = 73\,513\,440\). The complete list of record minima \(1 \leq n \leq 10^8\) for this ratio, and the ratio value is as follows
| Minimum | Ratio value |
|---|---|
| 3 | 10.12500… |
| 4 | 4.60449… |
| 6 | 3.23210… |
| 12 | 2.50636… |
| 24 | 2.30262… |
| 36 | 2.29878… |
| 48 | 2.23419… |
| 60 | 2.06896… |
| 120 | 1.94358… |
| 180 | 1.92494… |
| 240 | 1.88976… |
| 360 | 1.80650… |
| 720 | 1.75360… |
| 840 | 1.72334… |
| Minimum | Ratio value |
|---|---|
| 1 260 | 1.70484… |
| 1 680 | 1.67076… |
| 2 520 | 1.59772… |
| 5 040 | 1.54807… |
| 10 080 | 1.52438… |
| 15 120 | 1.51057… |
| 25 200 | 1.49966… |
| 27 720 | 1.46924… |
| 55 440 | 1.42263… |
| 110 880 | 1.40023… |
| 166 320 | 1.38734… |
| 277 200 | 1.37709… |
| 332 640 | 1.36544… |
| 554 400 | 1.35533… |
| Minimum | Ratio value |
|---|---|
| 665 280 | 1.35472… |
| 720 720 | 1.32241… |
| 1 441 440 | 1.30150… |
| 2 162 160 | 1.28950… |
| 3 603 600 | 1.27994… |
| 4 324 320 | 1.26910… |
| 7 207 200 | 1.25969… |
| 8 648 640 | 1.25912… |
| 10 810 800 | 1.24806… |
| 21 621 600 | 1.22832… |
| 36 756 720 | 1.21836… |
| 61 261 200 | 1.20933… |
| 73 513 440 | 1.19909… |
Remark 3. Except for the first entry (which must be \(n = 3\), as the ratio is not defined for \(n = 1, 2\)), the sequence of minima appears to be given by OEIS sequence A004394, the superabundant numbers. Heuristically, one might see Cauchy convolutions of a function with itself as mollifications of increasing order. Therefore, extrema mostly depend on the lowest order convolution, which in our case equals the function \(\sigma_{-1}\) itself, whose maxima occur at the superabundant numbers.
Below is a table of the upper bound and lower bound on the ratio \(\frac{p_n^\sigma(2)^2}{p_n^\sigma(1) p_n^\sigma(3)}\) computed with \(\mu = 2\,000\), on the sequence \((S_i)_{i=42}^{80}\) of superabundant numbers, heuristically giving the first counterexamples as \(\lambda = S_{80} = 65\,214\,507\,758\,400\).
| \(i\) | \(S_i\) | lower | upper |
|---|---|---|---|
| 42 | 73513440 | 1.1977… | 1.2002… |
| 43 | 122522400 | 1.1889… | 1.1912… |
| 44 | 147026880 | 1.1883… | 1.1907… |
| 45 | 183783600 | 1.1779… | 1.1802… |
| 46 | 367567200 | 1.1593… | 1.1616… |
| 47 | 698377680 | 1.1565… | 1.1588… |
| 48 | 735134400 | 1.1502… | 1.1524… |
| 49 | 1102701600 | 1.1497… | 1.1519… |
| 50 | 1163962800 | 1.1479… | 1.1502… |
| 51 | 1396755360 | 1.1382… | 1.1404… |
| 52 | 2327925600 | 1.1298… | 1.1320… |
| 53 | 2793510720 | 1.1293… | 1.1315… |
| 54 | 3491888400 | 1.1194… | 1.1215… |
| 55 | 6983776800 | 1.1016… | 1.1038… |
| 56 | 13967553600 | 1.0930… | 1.0951… |
| 57 | 20951330400 | 1.0925… | 1.0947… |
| 58 | 27935107200 | 1.0887… | 1.0908… |
| 59 | 41902660800 | 1.0839… | 1.0861… |
| 60 | 48886437600 | 1.0823… | 1.0844… |
| 61 | 80313433200 | 1.0729… | 1.0750… |
| \(i\) | \(S_i\) | lower | upper |
|---|---|---|---|
| 62 | 160626866400 | 1.0559… | 1.0580… |
| 63 | 321253732800 | 1.0476… | 1.0496… |
| 64 | 481880599200 | 1.0472… | 1.0492… |
| 65 | 642507465600 | 1.0435… | 1.0455… |
| 66 | 963761198400 | 1.0390… | 1.0410… |
| 67 | 1124388064800 | 1.0374… | 1.0394… |
| 68 | 1927522396800 | 1.0349… | 1.0369… |
| 69 | 2248776129600 | 1.0292… | 1.0312… |
| 70 | 3373164194400 | 1.0288… | 1.0308… |
| 71 | 4497552259200 | 1.0252… | 1.0272… |
| 72 | 4658179125600 | 1.0208… | 1.0228… |
| 73 | 6746328388800 | 1.0207… | 1.0227… |
| 74 | 9316358251200 | 1.0128… | 1.0147… |
| 75 | 13974537376800 | 1.0124… | 1.0143… |
| 76 | 18632716502400 | 1.0088… | 1.0108… |
| 77 | 27949074753600 | 1.0044… | 1.0064… |
| 78 | 32607253879200 | 1.0029… | 1.0049… |
| 79 | 55898149507200 | 1.0005… | 1.0024… |
| 80 | 65214507758400 | 0.9950… | 0.9969… |
We will show that for \(n \geq 10^7\), any such counterexample must have \(\sigma_{-1}(n) > 4\). This will allow us to reduce the search for counterexamples to a rather sparse set of integers.
Fix \(\mu = 500\), with \(C_{500}^{(2,2)} = 2.49452\ldots\). Since \(\sigma_{-3}(n) > 1\), we note that \[p_n^\sigma(2)_{\mathrm{low},500} > a(n) \mathrel{\vcenter{:}}= \frac{1}{2!} \left(C_{500}^{(2,2)} \cdot (n-1) - C_{500}^{(2,2)} \cdot (n-1) \cdot 2n^{{\frac{1}{2}}} \bigg(\frac{500}{n}\bigg)^3 - (\log(n) + 1)^2\right) \,.\] Likewise, since \(\sigma_{-k}(n) < \zeta(k)\) for \(k = 3, 4, 5\), we have \[p_n^\sigma(3)_{\mathrm{up}} < b(n) \mathrel{\vcenter{:}}= \begin{align}[t]\frac{1}{3!}\bigg( \frac{35(n-1)^2}{16} \zeta(5) & {} + \frac{15(n-1)}{4} \zeta(4) \zeta(3) (1 + \log(n)) \\[-1ex] &{} + \frac{5}{2} (n-1) \zeta(3) \zeta(2) + n (1 + \log(n))^3\bigg)\, .\end{align}\\[1ex]\] We have that \[\lim_{n\to\infty} \frac{a(n)^2}{b(n)} = \frac{24 \cdot (C_{500}^{(2,2)})^2}{35 \zeta(5)} = 4.1150029176747844809\ldots \,,\] so eventually the ratio lies in an arbitrarily small interval around the limit, and in particular is \(>4\). For \(n \geq 2 \times 10^6\), we compute the resulting interval explicitly.
Working with the bound \(a(n)\) on \(p_n^\sigma(2)_{\mathrm{low},500}\): for \(n \geq 2 \times 10^6\), we have that \[\begin{align} \big\lvert \tfrac{1}{n} \cdot \tfrac{1}{2!} C_{500}^{(2,2)} \cdot (n-1) - \tfrac{1}{2!} C_{500}^{(2,2)} \big\rvert &< \tfrac{1}{1\,500\,000} \,, \\ \big\lvert \tfrac{1}{n} \cdot \tfrac{1}{2!} C_{500}^{(2,2)} \cdot (n-1) \cdot 2n^{{\frac{1}{2}}} \big(\tfrac{500}{n}\big)^3 - 0 \big\rvert &< \tfrac{1}{15\,000\,000} \,, \\ \big\lvert \tfrac{1}{n} \cdot \tfrac{1}{2!} \cdot (\log(n) + 1)^2 - 0 \big\rvert &< \tfrac{1}{15\,000} \,. \end{align}\] Working with the bound \(b(n)\) on \(p_n^\sigma(3)_{\mathrm{up}}\): for \(n \geq 2 \times 10^6\), we have that \[\begin{align} \big\lvert \tfrac{1}{n^2} \cdot \tfrac{1}{3!} \cdot \tfrac{35(n-1)^2}{16} \zeta(5) - \tfrac{1}{3!} \cdot \tfrac{35}{16} \zeta(5) \big\rvert &< \tfrac{1}{2\,500\,000} \,, \\ \big\lvert \tfrac{1}{n^2} \cdot \tfrac{1}{3!} \cdot \tfrac{15(n-1)}{4} \zeta(4) \zeta(3) (1 + \log(n)) - 0 \big\rvert &< \tfrac{1}{150\,000} \,, \\ \big\lvert \tfrac{1}{n^2} \cdot \tfrac{1}{3!} \cdot \tfrac{5}{2} (n-1) \zeta(3) \zeta(2) - 0 \big\rvert &< \tfrac{1}{2\,400\,000} \,, \\ \big\lvert \tfrac{1}{n^2} \cdot \tfrac{1}{3!} \cdot n (1 + \log(n))^3 - 0 \big\rvert &< \tfrac{1}{3\,000} \,. \end{align}\]
Interval arithmetic now shows that for \(n \geq 2 \times 10^6\), we have \[\frac{a(n)^2}{b(n)} \in (4.11085\ldots, 4.11916\ldots) \,,\] and in particular is \(> 4\). The upshot now is that for \(n \geq 2\times10^6\), \[\frac{p_n^\sigma(2)^2}{p_n^\sigma(1)p_n^\sigma(3)} \geq \frac{a(n)^2}{p_n^\sigma(1) \cdot b(n)} \geq \frac{4}{p_n^\sigma(1)} \,.\] So if \[\frac{4}{p_n^\sigma(1)} \geq 1 \,,\] i.e. \(p_n^\sigma(1) = \sigma_{-1}(n) \leq 4\), then \(n\) certainly cannot be a counterexample to the log-concavity of \(p_n^\sigma(k)\) at \(k=2\). In other words, any counterexample to the log-concavity of \(p_n^\sigma(k)\) at \(k=2\) in the range \(10^7 \leq n \leq \lambda\), must have \(\sigma_{-1}(n) > 4\).
Counterexamples must have \(\sigma_{-1}(n) > 4\). This is OEIS sequence A068404, whose first entry is \(27\,720\). The density of such numbers is reported to be between \(0.000176363\) and \(0.000679406\), see the data on [14] obtained using the method of Wall et al. [15] and method of Deléglise [16], respectively. This leaves between roughly \(0.000176363 \lambda = 11\,501\,426\,231\) and \(0.000679406 \lambda = 44\,307\,127\,859\) candidates to check, saving a factor of \(10^3\) over checking the whole range.
We outline a procedure below to enumerate all such \(n\) with \(\sigma_{-1}(n) > 4\); this procedure can be trivially parallelised and shows there are actually \(16\,565\,226\,666\) candidates to check. The parallel version of our computation in gp/pari took approximately 5697.75 hours = 237.4 days of CPU time, across 48+ threads (on multiple machines), equating to around 5
days of real time.
To start, recall that \(\sigma_{-1}(n)\) is multiplicative, so given the prime factorisation \(n = \prod_{i} p_i^{a_i}\), we have \[\sigma_{-1}(n) = \prod\nolimits_{i} \sigma_{-1}(p_i^{a_i}) \,.\] For a prime \(p\), we also have \[\label{eqn:Sigma-1UpperBound} \sigma_{-1}(p^a) = \sum_{i=0}^a p^{-i} = \frac{p - p^{-a}}{p-1} < \frac{p}{p-1} \,.\tag{6}\] Note that this is a decreasing function for \(p > 1\).
Remark 4. Given the inequality in Eqn. 6 , it is straightforward to show that any number \(n\) satisfying \(\sigma_{-1}(n) > 4\) has at least 4 distinct prime divisors. When restricting to odd numbers, this amount even increases to at least 21 distinct prime divisors.
Fix a list of the first \(m\) primes \(\{ p_1 = 2, p_2 = 3, \ldots, p_m \}\) and choose exponents \(a_i\) such that \(p_1^{a_1} \cdots p_m^{a_m} \leq \lambda\). Write \(s = p_1^{a_1} \cdots p_m^{a_m}\). The question is whether a number of the form \(n = sn'\), \((s, n') = 1\) can have \(\sigma_{-1}(n) > 4\) and \(n \leq \lambda\). We can build such a number by selecting the smallest remaining prime (which simultaneously has the largest possible contribution to \(\sigma_{-1}\)) until either \(\sigma_{-1}(s) \cdot \prod_{j=m+1}^{m'} \frac{p_j}{p_j-1} > 4\), in which case \(s\) is a potentially good starting value, in that some multiple \(n = sn'\), with \((s,n') = 1\), below \(\lambda\) can potentially have \(\sigma_{-1}(n) > 4\). Or we continue until \(s \cdot \prod_{j=m+1}^{m'} p_j > \lambda\), in which case no multiple \(n = sn'\), with \((s,n') = 1\), of \(s\) below \(\lambda\) can have \(\sigma_{-1}(n) > 4\).
Remark 5. Since \(\prod_{i=1}^m \sigma_{-1}(p_m) = \prod_{i=1}^m (1 + p_m^{-1}) > \sum_{i=1}^m p_m^{-1}\) diverges, one can always find some multiple \(sn'\), with \((s,n') = 1\), which has \(\sigma_{-1}(sn') > 4\). For our enumeration question, the important point is whether this multiple satisfies \(sn' \leq \lambda\) or not.
For a fixed list of the first \(m\) primes, there are only finitely many possible exponents, i.e. \(0 \leq a_i \leq \log_{p_i}(\lambda)\), to test for good starting values \(s \leq \lambda\). Once this list is determined, we enumerate all multiples \(n = sn'\), \((s, n') = 1\), with \(n' \leq \lambda/s\), and check whether or not the lower bound on the ratio (with given \(\mu\)) satisfies \[\frac{(p_n^\sigma(2)_{\mathrm{low},\mu})^2}{p_n^\sigma(1)p_n^\sigma(3)_{\mathrm{up}}} > 1 \,.\] If the lower bound is \(> 1\), then \(n\) cannot be a counterexample to the log-concavity of \(P_n^\sigma(X)\) at \(k=2\). Checking these bounds for sufficiently large \(\mu\) should determine whether or not it really is the case that \[\frac{p_n^\sigma(2)^2}{p_n^\sigma(1)p_n^\sigma(3)} < 1 \,,\] and so whether or not \(n\) is a counterexample to log-concavity of \(P_n^\sigma(X)\) at \(k=2\).
Remark 6. The step of iterating through all multiples \(n = sn'\), \((s,n') = 1\) can be done in parallel as each \(s\) is independent. Alternatively, for small values of \(s\), it is actually better to search for \(sn'\) in parallel over the individual residue classes \(n' \, \mathrm{mod} \, \mathop{\mathrm{lcm}}(p_1, \ldots, p_m)\) with \((n',p_1 p_2 \cdots p_m) = 1\), as this makes the parallelisation finer grained.
We implemented the procedure with \(m = 6\), and initial list of primes \(\{ 2, 3, 5, 7, 11, 13 \}\), using gp/pari. We found a total of \(123366\) potential starting values \(s\). After enumerating all multiples \(sn' \leq \lambda\), and testing with \(\mu =
500\), we found only 4 potential counterexamples \(n \leq \lambda\) \[32\,607\,253\,879\,200, \; 55\,898\,149\,507\,200, \;59\,753\,194\,300\,800, \; 65\,214\,507\,758\,400 \,.\]
Testing again with \(\mu = 5\,000\) eliminates the first 3, showing that \(\lambda\) is indeed the first counterexample. This completes the proof of Theorem 2, part ii).
For our last result, we will make use of another simple bound.
Lemma 4. For \(n \geq 2 \times 10^6\), it holds that \[\frac{p_n^\sigma(2)^2}{p_n^\sigma(1) p_n^\sigma(3)} < \frac{6.20927\ldots}{\sigma_{-1}(n)} \, .\]
Proof. Fix \(\mu = 1000\), with \(C_{1000}^{(2,2)} = 2.49726\ldots\), and \(C_{1000}^{(4,2)} = 1.74895\ldots\). Since \(\sigma_{-3}(n) < \zeta(3)\), we note that \[p_n^\sigma(2)_{\mathrm{up}} < c(n) \mathrel{\vcenter{:}}= \frac{1}{2!} \left(\frac{5}{2} (n-1) \zeta(3) + (1 + \log(n))^2\right) \,.\] Likewise, since \(\sigma_{-5}(n) > 1\), and \(\sigma_{-4}(n) < \zeta(4)\) \[p_n^\sigma(3)_{\mathrm{low},1000} > d(n) \mathrel{\vcenter{:}}= \frac{1}{3!}\bigg(\begin{align}[t] \frac{(n-1)^2}{2} & C_{1000}^{(2,2)} C_{1000}^{(4,2)} - (n-1)^2 n^{{\frac{1}{2}}} \Big(\frac{1000}{n}\Big)^5 C_{1000}^{(2,2)} C_{1000}^{(4,2)}\\ & {} - \frac{3(n-1)}{2} C_{1000}^{(2,2)} \zeta(4) \zeta(3) (1 + \log(n))\\ &{} - C_{1000}^{(2,2)} \cdot (1000)^2 \zeta(2) (1 + \log(n)) - n(1+\log(n))^3\bigg) \,. \end{align}\] As in Section 3.2, we compute that \[\lim_{n\to\infty} \frac{c(n)^2}{d(n)} = \frac{75 \zeta(3)^2}{4 C_{1000}^{(2,2)} C_{1000}^{(4,2)}} = 6.20309329901722084305\ldots \,.\] In particular the ratio is eventually in an arbitrarily small interval around this limit. We can check for \(n \geq 2 \times 10^6\), that the the following holds \[\frac{c(n)^2}{d(n)} \in (6.19692\ldots, 6.20927\ldots) \,.\] So for \(n \geq 2 \times 10^6\), we have \[\frac{p_n^\sigma(2)^2}{p_n^\sigma(1) p_n^\sigma(3)} < \frac{c(n)^2}{p_n^\sigma(1) \cdot d(n)} < \frac{6.20927\ldots}{\sigma_{-1}(n)} \,,\] as claimed. ◻
We are now in a position to finalise our proofs.
Proof of Theorem 3. The reasoning in Section 3 established that any possible counterexample \(n\) to the log-concavity has to satisfy \(\sigma_{-1}(n) > 4\). As mentioned there, the asymptotical density of these numbers is at most 0.000679406.
Conversely, due to Lemma 4, for any \(n\) satisfying \(\sigma_{-1}(n) > 6.20927\ldots\) the polynomial \(P_n^\sigma(X)\) is not log-concave at \(k = 2\). The smallest natural numbers \(n\) to satisfy this is the 93rd superabundant number \(S_{93} = 4\,043\,299\,481\,020\,800\) and hence the claim follows because \(\sigma_{-1}(kn) > \sigma_{-1}(n)\) for any \(k \in \mathbb{N}\). In particular, \(\delta_{\mathrm{inf}} \geq \frac{1}{S_{93}} = 2.47323 \ldots \times 10^{-16}\). ◻
Remark 7. As \(\mu \to \infty\), we can reduce the constant in upper bound in Lemma 4 to a small interval around \[\frac{75\zeta(3)^2}{4 \lim_{\mu \to \infty} C_\mu^{(2,2)} C_\mu^{(4,2)}} = \frac{30 \zeta(3)^2}{7} = 6.19260342185843243105\ldots \,.\] As the 92nd superabundant number \(S_{92} = 3\,032\,474\,610\,765\,600\) satisfies \[\sigma_{-1}(S_{92}) = 6.18956\ldots < \frac{30 \zeta(3)^2}{7} \,,\] we cannot readily improve the asymptotic density bound further, without more involved work.