9 Tertiary explicit estimates

9.1 The least common multiple sequence is not highly abundant for large \(n\)

9.1.1 Problem statement and notation

Definition 40

✓

\(\sigma (n)\) is the sum of the divisors of \(n\).

Definition 41

✓

A positive integer \(N\) is called highly abundant (HA) if

\[ \sigma (N) {\gt} \sigma (m) \]

for all positive integers \(m {\lt} N\), where \(\sigma (n)\) denotes the sum of the positive divisors of \(n\).

Definition 42

✓

For each integer \(n \ge 1\), define

\[ L_n := \mathrm{lcm}(1,2,\dots ,n). \]

We call \((L_n)_{n \ge 1}\) the least common multiple sequence.

Original MathOverflow question. Is it true that every value in the sequence \(L_n = \mathrm{lcm}(1,2,\dots ,n)\) is highly abundant? Equivalently,
\[ \{ L_n : n \ge 1\} \subseteq HA? \]

In this note we record the structure of an argument showing that, for all sufficiently large \(n\), the integer \(L_n\) is not highly abundant. This argument was taken from this MathOverflow answer.

9.1.2 A general criterion using three medium primes and three large primes

In this subsection we assume that \(n \geq 1\) and that \(p_1,p_2,p_3,q_1,q_2,q_3\) are primes satisfiying

\[ \sqrt{n} {\lt} p_1 {\lt} p_2 {\lt} p_3 {\lt} q_1 {\lt} q_2 {\lt} q_3 {\lt} n \]

and the key criterion

\begin{equation} \label{eq:main-ineq} \prod _{i=1}^3\Bigl(1+\frac{1}{q_i}\Bigr) \le \Biggl( \prod _{i=1}^3 \Bigl(1+\frac{1}{p_i(p_i+1)}\Bigr) \Biggr) \Bigl(1 + \frac{3}{8n}\Bigr) \Biggl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Biggr). \end{equation}

Lemma 128

We have \(4 p_1 p_2 p_3 {\lt} q_1 q_2 q_3\).

Proof ▶

Obvious from the non-negativity of the left-hand side of 1.

9.1.3 Factorisation of \(L_n\) and construction of a competitor

Lemma 129 Factorisation of \(L_n\)

There exists a positive integer \(L'\) such that

\[ L_n = q_1 q_2 q_3 \, L' \]

and each prime \(q_i\) divides \(L_n\) exactly once and does not divide \(L'\).

Proof ▶

Since \(q_i {\lt} n\), the prime \(q_i\) divides \(L_n\) exactly once (as \(q_i^2 {\gt} n\)). Hence we may write \(L_n = q_1 q_2 q_3 L'\) where \(L'\) is the quotient obtained by removing these prime factors. By construction, \(q_i \nmid L'\) for each \(i\).

Lemma 130 Normalised divisor sum for \(L_n\)

Let \(L'\) be as in Lemma 129. Then

\begin{equation} \label{eq:sigmaLn} \frac{\sigma (L_n)}{L_n} \; =\; \frac{\sigma (L')}{L'} \prod _{i=1}^3 \Bigl(1 + \frac{1}{q_i}\Bigr). \end{equation}

Proof ▶

Use the multiplicativity of \(\sigma (\cdot )\) and the fact that each \(q_i\) occurs to the first power in \(L_n\). Then

\[ \sigma (L_n) = \sigma (L') \prod _{i=1}^3 \sigma (q_i) = \sigma (L') \prod _{i=1}^3 (1+q_i). \]

Dividing by \(L_n = L' \prod _{i=1}^3 q_i\) gives

\[ \frac{\sigma (L_n)}{L_n} = \frac{\sigma (L')}{L'} \prod _{i=1}^3 \frac{1+q_i}{q_i} = \frac{\sigma (L')}{L'} \prod _{i=1}^3 \Bigl(1 + \frac{1}{q_i}\Bigr). \]

Lemma 131

There exist integers \(m \ge 0\) and \(r\) satisfying \(0 {\lt} r {\lt} 4 p_1 p_2 p_3\) and

\[ q_1 q_2 q_3 = 4 p_1 p_2 p_3 m + r \]

Proof ▶

This is division with remainder.

Definition 43

✓

With \(m,r\) as above, define the competitor

\[ M := 4 p_1 p_2 p_3 m L'. \]

Lemma 132 Basic properties of \(M\)

With notation as above, we have:

\(M {\lt} L_n\).
\[ 1 {\lt} \frac{L_n}{M} = \Bigl(1 - \frac{r}{q_1 q_2 q_3}\Bigr)^{-1} {\lt} \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}. \]

Proof ▶

The first item is by construction of the division algorithm. For the second, note that

\[ L_n = q_1 q_2 q_3 L' = (4 p_1 p_2 p_3 m + r) L' {\gt} 4 p_1 p_2 p_3 m L' = M, \]

since \(r{\gt}0\). For the third,

\[ \frac{L_n}{M} = \frac{4 p_1 p_2 p_3 m + r}{4 p_1 p_2 p_3 m} = 1 + \frac{r}{4 p_1 p_2 p_3 m} = \Bigl(1 - \frac{r}{4 p_1 p_2 p_3 m + r}\Bigr)^{-1} = \Bigl(1 - \frac{r}{q_1 q_2 q_3}\Bigr)^{-1}. \]

Since \(0 {\lt} r {\lt} 4p_1p_2p_3\) and \(4p_1p_2p_3 {\lt} q_1q_2q_3\), we have

\[ 0{\lt}\frac{r}{q_1 q_2 q_3}{\lt}\frac{4p_1p_2p_3}{q_1 q_2 q_3}, \]

\[ \Bigl(1-\frac{r}{q_1 q_2 q_3}\Bigr)^{-1} {\lt} \Bigl(1-\frac{4p_1p_2p_3}{q_1 q_2 q_3}\Bigr)^{-1}. \]

9.1.4 A sufficient condition

We give a sufficient condition for \(\sigma (M) \geq \sigma (L_n)\).

Lemma 133 A sufficient inequality

Suppose

\[ \frac{\sigma (M)}{M} \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr) \; \ge \; \frac{\sigma (L_n)}{L_n}. \]

Then \(\sigma (M) \ge \sigma (L_n)\), and so \(L_n\) is not highly abundant.

Proof ▶

By Lemma 132,

\[ \frac{L_n}{M} {\lt} \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}. \]

Hence

\[ \frac{\sigma (M)}{M} \ge \frac{\sigma (L_n)}{L_n} \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1} {\gt} \frac{\sigma (L_n)}{L_n} \cdot \frac{M}{L_n}. \]

Multiplying both sides by \(M\) gives

\[ \sigma (M) {\gt} \sigma (L_n) \cdot \frac{M}{L_n} \]

and hence

\[ \sigma (M) \ge \sigma (L_n), \]

since \(M/L_n{\lt}1\) and both sides are integers.

Combining Lemma 133 with Lemma 130, we see that it suffices to bound \(\sigma (M)/M\) from below in terms of \(\sigma (L')/L'\):

Lemma 134 Reduction to a lower bound for \(\sigma (M)/M\)

\begin{equation} \label{eq:sigmaM-lower} \frac{\sigma (M)}{M} \; \ge \; \frac{\sigma (L')}{L'} \Biggl( \prod _{i=1}^3 \Bigl(1+\frac{1}{p_i(p_i+1)}\Bigr) \Biggr) \Bigl(1 + \frac{3}{8n}\Bigr), \end{equation}

then \(L_n\) is not highly abundant.

Proof ▶

Insert 3 and 2 into the desired inequality and compare with the assumed bound 1; this is a straightforward rearrangement.

9.1.5 Conclusion of the criterion

Lemma 135 Lower bound for \(\sigma (M)/M\)

With notation as above,

\[ \frac{\sigma (M)}{M} \ge \frac{\sigma (L')}{L'} \Biggl( \prod _{i=1}^3 \Bigl(1 + \frac{1}{p_i(p_i+1)}\Bigr) \Biggr) \Bigl(1 + \frac{3}{8n}\Bigr). \]

Proof ▶

By multiplicativity, we have

\[ \frac{\sigma (M)}{M} \ge \frac{\sigma (L')}{L'} \prod _p \]