Is the least common multiple sequence $\text{lcm}(1, 2, \dots, n)$ a subset of the highly abundant numbers?

OK, after an hour-long conversation with an AI I was able to produce a numerical counterexample that I could verify separately. (Admittedly, the verification code was also provided by the AI (see the very end of the conversation), but it is only 29 lines long, and simple enough that I could read it line by line.)

Let $n = 200000$, and let $p_1,\dots,p_{25}$ and $q_1,\dots,q_{13}$ be the primes $$ 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641$$ and $$ 187963, 187973, 187987, 188011, 188017, 188021, 188029, 188299, 199931, 199933, 199961, 199967, 199999$$ respectively. Thus the primes $p_j$ are slightly larger than $\sqrt{n} \approx 447$ and the primes $q_m$ are slightly less than $n$. (The conversation with the AI will describe how these primes and parameters such as $25$, $13$, and $200000$ were located.)

Direct calculation shows that $$ \sum_{m=1}^{13} \log q_m - \sum_{j=1}^{25} \log p_j \approx 1.8484\dots \times 10^{-6} \tag{1}\label{501125_1}$$ is positive, but smaller than $$ \sum_{j=1}^{25} \log(1 +\frac{1}{p_j^2+p_j}) - \sum_{m=1}^{13} \log(1+\frac{1}{q_m}) \approx 1.33489\dots \times 10^{-5}. \tag{2}\label{501125_2}$$ If we define the competitor $$ M := \frac{p_1 \dots p_{25}}{q_1 \dots q_{13}} L_n$$ to $L_n$, then by the positivity of \eqref{501125_1}, $M$ is (slightly) less than $L_n$. On the other hand, since the primes $p_1,\dots,p_{25},q_1,\dots,q_{13}$, being between $\sqrt{n}$ and $n$, all divide $L_n$ exactly once, we have $$ \frac{\sigma(M)}{\sigma(L_n)} = \frac{\prod_{j=1}^{25} (1+p_j+p_j^2)/(1+p_j)}{\prod_{m=1}^{13} (1+q_m)}$$ $$ = \exp\biggl( \eqref{501125_2} - \eqref{501125_1} \biggr). $$ Since \eqref{501125_2} is larger than \eqref{501125_1}, we thus see that $\sigma(M) > \sigma(L_n)$ and hence $L_n$ is not highly abundant.

EDIT: there is significant room in the analysis, and I could imagine that a more intelligent search (without relying as much on AI assistance) could reduce the value of $n$ significantly; perhaps those who are inclined towards such computational numerical challenges would like to see how well they can "beat the AI".

Third response. By improving on Saúl RM's nice argument, I will show below that if $n>\exp(\exp(7812))$, then $L_n$ is not highly abundant. Under GRH, this range can be extended to $n>\exp(15667)$. With little tweaks the constants $7812$ and $15667$ can be reduced by a factor of about $6$ (see the Remark after the Theorem).

Subsequently, Terry Tao found a better method, and as a result we know that if $n$ is a prime power, then $L_n$ is highly abundant if and only if $$n\in\{1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49, 53, 59, 61, 64, 67, 81, 83, 89, 125, 127, 128, 131, 137, 139, 169\}.$$

Theorem. Assume that the prime number $p$ is such that $p-1\in[(400/401)n,n-1]$, and $p-1$ is divisible by the first thousand primes. Then $L_n$ is not highly abundant.

Proof. We shall prove that $$\sigma(L_n(p-1)/p)>\sigma(L_n).$$ Note that $p\in(\sqrt{n},n]$, hence $p\parallel L_n$. Let $p-1=\prod_j p_j^{s_j}$ be the prime factorization of $p-1$, and let $r_j$ be the exponent of $p_j$ in $L_n$. Then, $$\frac{\sigma(L_n(p-1)/p)}{\sigma(L_n)}=\frac{1}{p+1}\prod_j\frac{\sigma(p_j^{r_j+s_j})}{\sigma(p_j^{r_j})},$$ hence we need that $$\prod_j\frac{\sigma(p_j^{r_j+s_j})}{\sigma(p_j^{r_j})}>p+1.$$ Since $$\frac{\sigma(p_j^{r_j+s_j})}{\sigma(p_j^{r_j})}=p_j^{s_j}+\frac{p_j^{s_j}-1}{p_j^{r_j+1}-1}>p_j^{s_j}+\frac{p_j^{s_j}-1}{np_j},$$ it suffices that $$\prod_j\left(p_j^{s_j}+\frac{p_j^{s_j}-1}{np_j}\right)\geq p+1.$$ Dividing both sides by $p-1=\prod_j p^{s_j}$, this becomes $$\prod_j\left(1+\frac{1-p_j^{-s_j}}{np_j}\right)\geq 1+\frac{2}{p-1}.$$ For this it suffices that $$\sum_j\frac{1-p_j^{-s_j}}{np_j}\geq\frac{2}{p-1},$$ that is, $$\sum_j\frac{1-p_j^{-s_j}}{p_j}\geq\frac{2n}{p-1}.$$ The $p_j$'s on the left-hand side include the first thousand primes, hence the sum exceeds $401/200$ according to a numerical calculation, while the right-hand side does not exceed $401/200$. The proof is complete.

Remark. The crux of the above proof is that the sum in the last display exceeds $2$. For larger exponents $s_j$, one needs fewer primes $p_j$ for this to happen. For example, it would suffice to assume that $p-1$ is divisible by the square of the first $150$ primes, or the cube of the first $90$ primes, or the fourth power of the first $70$ primes, or the eight power of the first $60$ primes.

Corollary. If $n>\exp(\exp(7812))$, then $L_n$ is not highly abundant. If we assume the Riemann hypothesis for Dirichlet $L$-functions, then the above range can be extended to $n>\exp(15667)$.

Proof. Assume that $n>\exp(\exp(7812))$, and put $m:=n/401-1$. Let $q$ denote the product of the first thousand primes. Numerical calculation shows that $$q<\exp(7812.3)\qquad\text{and}\qquad\varphi(q)<\exp(7809.6).$$ By the Theorem above, it suffices to show that $$\theta(n-m;q,1)\neq\theta(n;q,1).$$ Assume that the two sides are equal. Then, by Theorem 1.2 of Bennett-Martin-O'Bryant-Rechnitzer (which appeared as Illinois J. Math. 62 (2018), 427–532), we conclude that $$\frac{m}{\varphi(q)}<\frac{n}{80\log m}.$$ In particular, $\log m<6\varphi(q)<\exp(7811.4)$, whence $n<\exp(\exp(7812))$, which is a contradiction. If we assume GRH and relax the assumption on $n$ to $n>\exp(15667)$, we use Corollary 1.1 of Lee (which appeared as Q. J. Math. 74 (2023), 1505-1533) to obtain the following strengthening of the previous display: $$\frac{m}{\varphi(q)}<\frac{1}{4\pi}(\log n+31008)\sqrt{n}\log n.$$ In particular, $$\sqrt{n}<\exp(7813)(\log n+31008)\log n,$$ whence $n<\exp(15667)$, which is again a contradiction.

Second response. This is an improvement of Max Alekseyev's nice counterexample. Consider $n=71$ with $$L := L_{71} = 2^6 \cdot 3^3 \cdot 5^2 \cdot 67 \cdot 71 \cdot L',$$ where the co-factor $L'$ is coprime to listed primes. Then set $$M := 2^8 \cdot 3^4 \cdot 5^3 \cdot 79 \cdot L'.$$

We have $$L = 205502400\cdot L' > 204768000\cdot L' = M,$$ while $$\sigma(L) = 771022080\cdot \sigma(L') < 771650880\cdot \sigma(L') = \sigma(M).$$ That is, $L_{71}$ is not highly abundant.

First response. I took on Terry's challenge "beat the AI". With my own little program (which is simpler than the AI's), starting from Terry's parameters, I got down to $n=13757$: $$\{p_1,\dotsc,p_{15}\}=\{127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197\},$$ $$\{q_1,\dotsc,q_8\}=\{13553, 13679, 13691, 13709, 13721, 13729, 13751, 13757\},$$ $$\sum_{m=1}^8 \log q_m - \sum_{j=1}^{15} \log p_j = 3.4179\dotsc \times 10^{-6},$$ $$\sum_{j=1}^{15} \log(1 +\frac{1}{p_j^2+p_j}) - \sum_{m=1}^8 \log(1+\frac{1}{q_m}) = 1.5531\dotsc \times 10^{-5}.$$

Added. The following text file contains variations of the data above. It shows that $L_n$ is not highly abundant for any $19919 \leq n < 201601$.

This is a community wiki answer to collate the known status of $L_n$ for various $n$, restricting to the prime power case as these are the only $n$ for which $L_n$ is different from any previous $L_{n'}$. Please update with new results as they come in.

Results are also reported in OEIS A389482.

For selected $n$, one can demonstrate that $L_n$ is not highly abundant by dividing $L_n$ by some product of primes, then multiplying $L_n$ by a slightly smaller product of primes, according to the following table.

For all $n$ in the above table, the failure of $L_n$ to be highly abundant was also verified in Lean.

For general $n$, we still divide by some primes and multiply by others, but according to the following procedures:

• For all $n \leq 10^{10}$ for which $L_n$ is not highly abundant one can use this ladder of 3477 different ways to add and remove primes, each valid for a separate range of $n$; this ladder was formed by merging together several previous partial results.
• For $n > 10^{10}$, one uses a theoretical analysis based on Dusart's effective version of the prime number theorem to verify the construction (which involves removing three large primes, then adding three medium primes, a factor of 4, and some additional terms).

Connections to the literature. The falsity of the original conjecture was already claimed in the 1944 paper of Alaoglu and Erdős, where on pages 466-467 they write "it would be easy to see that neither $n!$ nor the least common multiple of the first $n$ integers can be highly abundant infinitely often". While no proof is explicitly given, one can deduce this claim from the following (unnamed) Corollary on page 463, that describes rather precisely (most of) the prime factorization of a highly abundant number. It states the following:

Proposition. Fix $\varepsilon > 0$. There is a constant $c$ satisfying the following. Fix $N$ highly abundant. For $q$ a prime factor of $N$, let $k_q$ be the exponent of $q$ in $N$. Then less than $c \log \log N / \log \log \log N$ primes $q$ fail to satisfy the inequality: $$ (1 - \varepsilon) \log N \log \log N / (q \log q) < q^{k_q} < (1 + \varepsilon) \log N \log \log N / \log q. $$

Thus for instance, were $L_n = \exp((1+o(1))n)$ to be highly abundant, this proposition would force $L_n$ to be divisible by $p$ exactly once for almost all $(1+\varepsilon) (2n)^{1/2} \leq p \leq (1-\varepsilon) n$, exactly twice for almost all $(1+\varepsilon) (3n)^{1/3} \leq p \leq (1-\varepsilon) (2n)^{1/2}$, and so forth, where $\varepsilon>0$ is fixed, and "almost all" means that there are only $O_\varepsilon(\log n/\log\log n)$ exceptions. But this is not what actually happens: $L_n$ is in fact is divisible by $p$ exactly once for $n^{1/2} < p \leq n$, exactly twice for $n^{1/3} < p \leq n^{1/2}$, and so forth. This explains why $L_n$ is not highly abundant; a similar analysis also works for $n!$.

It is also instructive to compare $L_n$ with the collossally abundant number $C_n$ associated to the parameter $\varepsilon = \frac{1}{n \log n}$, which is a quintessential example of a highly abundant number. Standard calculations show that $C_n$ is divisible by $p$ exactly once for $(1+o(1)) (2n)^{1/2} \leq p \leq (1+o(1)) n$, exactly twice for $(1+o(1)) (3n)^{1/3} \leq p \leq (1+o(1)) (2n)^{1/2}$, and so forth, which is consistent with the proposition.

UPDATE (2025-10-06). Here is a SageMath implementation of my algorithm that for given integers $\sigma_L$ and $B$, finds the smallest integer $m\leq B$ with $\sigma(m)\geq \sigma_L$. Running it on $(\sigma_L,B)=(\sigma(L_n),L_n)$ either returns $L_n$ and thus proves that $L_n$ is HA, or returns some $M<L_n$ proving that $L_n$ is not HA.

At its core the algorithm relies on the inequality $$\frac{\sigma(m)}m \leq \prod_{\text{prime }p|m} \frac{p}{p-1}$$ and the following theorem.

Theorem. Let $p_1=2 < p_2=3 < \dots$ denote the prime numbers. Suppose that $p_k$ is the smallest prime factor of $m\leq B$ with $\sigma(m)\geq \sigma_L$. Then for $\ell := \omega(m)$, we have $$\prod_{i=k}^{k+\ell-1} p_i \leq B$$ and $$\prod_{i=k}^{k+\ell-1} \frac{p_i}{p_i-1} \geq \frac{\sigma_L}{B}.$$

The proof follows from noticing that $\frac{p}{p-1}$ is a decreasing function of $p$. My algorithm uses this theorem to effectively prune primes that may form a suitable $m$. It also uses parallelization.

To prove non-HA of $L_n$, we don't really need the smallest $m$, and the code can be stopped as soon as some $m<L_n$ is discovered. Such early termination can be requested by setting parameter just_any=True. As an example, the code checks HA-ness of $L_{67}$ and $L_{97}$.

UPDATE (2025-10-04). I've computationally verified that for an integer $n\in [67,113]$, $L_n$ is highly abundant only when $n\leq 70$ or $81\leq n\leq 96$. It follows that $n=71$ is the smallest index, for which $L_n$ is not highly abundant.

Just out of curiosity, I've also verified that $M$ constructed for $L_{71}$ by @GHfromMO is highly abundant, while $M$ for $L_{113}$ given below is not.

Changing primes that come with exponents greater than 1 enables much smaller counterexamples.

Consider $n=113$ with $$L_{113} = 2^6 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \cdot 89\cdot 113 \cdot L',$$ where the co-factor $L'$ is coprime to listed primes. Then set $$M := 2^7 \cdot 5^3 \cdot 7^3 \cdot 11^2 \cdot 13^2 \cdot L'.$$

We have $$L = 112751038400\cdot L' > 112224112000\cdot L' = M,$$ while $$\sigma(L) = 386809305120\cdot \sigma(L') < 387282168000\cdot \sigma(L') = \sigma(M).$$ That is, $L_{113}$ is not highly abundant.

Following Terry's method and GH from MO, I started from the upper bound of GH from MO and found this pretty quick: $n=4483$ $$ \{p_1,\dots,p_{13}\}=\{67,71,73,79,83,89,97,101,103,107,109,113,127\} $$

$$ \{q_1,\dots,q_7\}=\{4349,4441,4447,4451,4457,4481,4483\} $$

$$ \sum_{m=1}^{7}\log q_m-\sum_{j=1}^{13}\log p_j = 1.66862917368\times 10^{-5}, $$

$$ \sum_{j=1}^{13}\log\!\left(1+\frac{1}{p_j^2+p_j}\right) -\sum_{m=1}^{7}\log\!\left(1+\frac{1}{q_m}\right) = 5.62404029424\times 10^{-5}. $$

Update (14-Oct-2025): The range 173 to $10^{40}$ is also verified.

Update: The range 173 to 10,000,000,000 is now verified in Lean, using @Matthew Bolan's shorter ladder mentioned in another comment.

Update: I have now verified, in Lean, all the disproofs for small $n$ (below $373$) listed in the community wiki table (at time of writing). The code is available here (in the highly_abundant_small.lean file), https://gist.github.com/b-mehta/75d7f118e5bb6440c131c4c6ceaf7478#file-highly_abundant_small-lean. They can also be interacted with in the online Lean sandbox here.

Original post:

Here is a quick and direct Lean verification of the failure of this conjecture, using the idea given in the second response of GH from MO's answer.

import Mathlib

open Nat ArithmeticFunction

-- definition of highly abundant number
def IsHighlyAbundant (N : ℕ) : Prop :=
  ∀ m > 0, m < N → σ 1 m < σ 1 N

-- definition of lcm of the range {1..n}
def lcmRange (n : ℕ) : ℕ := (Finset.Icc 1 n).lcm _root_.id

-- a few small examples of lcmRange
example : lcmRange 1 = 1 := rfl
example : lcmRange 2 = 2 := rfl
example : lcmRange 6 = 60 := rfl

-- two general helper lemmas
lemma sigma_one_apply_prime {p : ℕ} (hp : p.Prime) : σ 1 p = p + 1 := by
  have : σ 1 p = σ 1 (p ^ 1) := by simp
  rw [this, sigma_one_apply_prime_pow hp]
  simp

lemma sigma_one_apply_prime_pow' {p k : ℕ} (hp : p.Prime) :
    σ 1 (p ^ k) = (p ^ (k + 1) - 1) / (p - 1) := by
  rw [sigma_one_apply_prime_pow hp, Nat.geomSum_eq]
  exact hp.two_le

namespace verify_l71

-- the choice of L to verify that the conjecture fails at 71
def L : ℕ := lcmRange 71

-- prove L is what we say it is
example : L = 5624043567677125526551547131200 := by decide +kernel

-- we follow the proof given by "GH from MO"
def L' : ℕ := 7 ^ 2 * 11 * 13 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61

def M : ℕ := 2 ^ 8 * 3 ^ 4 * 5 ^ 3 * 79 * L'

lemma L_eq : L = 205502400 * L' := by decide +kernel
lemma coprime_L'L : Nat.gcd (2 ^ 6 * 3 ^ 3 * 5 ^ 2 * 67 * 71) L' = 1 := by decide +kernel
lemma L_eq' : L = 2 ^ 6 * 3 ^ 3 * 5 ^ 2 * 67 * 71 * L' := by decide +kernel

lemma M_eq : M = 204768000 * L' := rfl
lemma coprime_L'M : Nat.gcd (2 ^ 8 * 3 ^ 4 * 5 ^ 3 * 79) L' = 1 := by decide +kernel
lemma M_eq' : M = 2 ^ 8 * 3 ^ 4 * 5 ^ 3 * 79 * L' := by decide +kernel

lemma M_lt_L : M < L := by decide +kernel

lemma sigma_L_aux : σ 1 (2 ^ 6 * 3 ^ 3 * 5 ^ 2 * 67 * 71) = 771022080 := by
  repeat rw [isMultiplicative_sigma.map_mul_of_coprime rfl]
  repeat rw [sigma_one_apply_prime_pow' (by norm_num)]
  repeat rw [sigma_one_apply_prime (by norm_num)]
  rfl

lemma sigma_M_aux : σ 1 (2 ^ 8 * 3 ^ 4 * 5 ^ 3 * 79) = 771650880 := by
  repeat rw [isMultiplicative_sigma.map_mul_of_coprime rfl]
  repeat rw [sigma_one_apply_prime_pow (by norm_num)]
  repeat rw [sigma_one_apply_prime (by norm_num)]
  rfl

-- the main result: L is not highly abundant
theorem not_HA_L71 : ¬ IsHighlyAbundant L := by
  intro h
  specialize h M (by decide +kernel) M_lt_L
  rw [L_eq', M_eq', isMultiplicative_sigma.map_mul_of_coprime coprime_L'M,
    isMultiplicative_sigma.map_mul_of_coprime coprime_L'L] at h
  simp only [sigma_M_aux, sigma_L_aux] at h
  simpa using Nat.lt_of_mul_lt_mul_right h

-- there exists an n (namely 71) such that lcmRange n is not highly abundant
theorem conjecture_fails : ∃ n, ¬ IsHighlyAbundant (lcmRange n) := ⟨71, not_HA_L71⟩

end verify_l71

I will show $L_{67}$ is highly abundant, and so $L_{71}$ is the smallest counterexample. My program is also able to recover the example proving $L_{71}$ is not highly abundant.

Following Terry's suggestion in the comments on GH from MO's answer, fix some $\alpha > 0$ and define the "score" $s(m) = \frac{\sigma(m)}{m^{\alpha}}.$ If there were some $M$ such that $M < L_{67}$ and $\sigma(M) \ge \sigma(L_{67})$, we must have $s(M) > s(L_{67})$, so it suffices to search those $M$ with $s(M) > s(L_{67})$. Terry suggests we take $\alpha = 1.0045$, which I will write this answer assuming, but I later found that $\alpha = 1.004$ made my search slightly faster.

For each prime $p_i$ there is some exponent $k_i$ such that $s(p_i^{k_i})$ is maximal, which I found with a computer search. I was able to search a finite space here as $\ln(s(p^k))$ is bounded above above by $(1-0.0045k) \ln p - \ln{(p-1)},$ which is decreasing in both $p$ and $k$. Combining these, we find that the $M$ with the largest score is $$\prod_{i \in \mathbb N} p_i^{k_i} = L_{67} \cdot 2 \cdot 3 \cdot 59^{-1} \cdot 61^{-1} \cdot 67^{-1}.$$ The fact it is so close to $L_{67}$ should be Terry's reason for the choice of $1.0045$.

The same process allows us to determine, for a given prime power $p^k$, the highest possible score of any $M$ with $p^k\mid M$ but $p^{k+1} \nmid M$ - they are just the maximal $M$ but with the power of $p$ replaced by $p^k$. This leads to the determination that no prime $p > 109$ can divide $M$, and that the other prime exponents must fall into the ranges {2: (5, 12), 3: (3, 7), 5: (2, 4), 7: (1, 3), 11: (1, 2), 13: (1, 2), 17: (1, 2), 19: (1, 2), 23: (1, 1), 29: (1, 1), 31: (1, 1), 37: (0, 1), 41: (0, 1), 43: (0, 1), 47: (0, 1), 53: (0, 1), 59: (0, 1), 61: (0, 1), 67: (0, 1), 71: (0, 1), 73: (0, 1), 79: (0, 1), 83: (0, 1), 89: (0, 1), 97: (0, 1), 101: (0, 1), 103: (0, 1), 107: (0, 1), 109: (0, 1)} in order to have a chance of beating $s(L_{67})$.

I checked these by iterating through the possible prime power factors recursively from largest to smallest, exiting the recursion early whenever the partial product exceeded $L_{67}$. Probably one can speed up this step by incorporating the score into the search and exiting early when you have lost too much score compared to $L_{67}$, or by doing integer linear programming. For the final run I used the score function $s(m) = \frac{\sigma(m)^{2000}}{m^{2009}}$ to determine all the bounds, which allowed me to do everything in exact arithmetic.

My code is here as a github gist. To find the counterexample for $L_{71}$, one simply changes N to 71 and SCORE_NUMERATOR and SCORE_DENOMINATOR to $2007$ and $2000$ respectively.

Here is a way to, assuming the conjecture below, prove that for big enough $N$, lcm$(1,\dots,N)$ is not highly abundant. This was discussed in the comments to the question, but is too long for a comment. The argument is based on a simple proof that $L_{2311}=L_{2\cdot3\cdot5\cdot7\cdot11+1}$ is not highly abundant by Eduardo Rodríguez Golvano.

Edit: The conjecture below was completely unnecessary for the argument to work (we don't need $p-1$ to be squarefree), as shown in GH from MO's answer (the third response, in particular).

Let $p_i$ be the $i^{\text{th}}$ prime. Let $r\in\mathbb{N}$ be big enough that $\sum_{i=1}^r\frac{1}{p_i}>8$.

The following would be needed to complete the argument (see the comment below by Terry Tao for ideas about how to prove it):

Conjecture: Let $k>0$. For big enough $N$ there exists a prime $p\in(2N/3,N)$ such that $p\equiv1$ mod $k$ and $p-1$ is squarefree.

Now suppose the conjecture is true and let $k=2\cdot3\cdot5\cdots p_r$. For big enough $N$ consider $p\in(2N/3,N)$ prime such that $p\equiv1$ mod $k$ and $p-1$ is squarefree. It would be enough to prove that $\sigma(L_N)<\sigma\left(L_N\cdot\frac{p-1}{p}\right)$. We can compute. Let $I=\{i\in\mathbb{N};p_i|p-1\}$ and $k_i$ the biggest exponent such that $p_i^{k_i}\leq N$. Then,

$$\frac{\sigma(L_N)}{\sigma\left(L_N\cdot\frac{p-1}{p}\right)}=(p+1)\prod_{i\in I}\frac{p_i^{k_i+1}-1}{p_i^{k_i+2}-1} =\frac{p+1}{p-1}\prod_i\frac{p_i^{k_i+1}-1}{p_i^{k_i+1}-\frac{1}{p_i}}$$ $$<\left(1+\frac{4}{N}\right)\cdot\prod_{i\in I}\left(1-\frac{1}{2p_i^{k_i+1}}\right)<\left(1+\frac{4}{N}\right)\cdot\prod_{i\in I}\left(1-\frac{1}{2p_iN}\right).$$ So taking logs, $$\ln\left(\frac{\sigma(L_N)}{\sigma\left(L_N\cdot\frac{p-1}{p}\right)}\right)<\frac{4}{N}-\sum_{i\in I}\frac{1}{2p_iN}\leq\frac{4}{N}-\sum_{i=1}^r\frac{1}{2p_iN}<0.$$

This is an answer to the following question from the comments:

Is every factorial a highly abundant number?

My claim is, for $n>15$, there exists $k$ such that $\sigma(\frac{2^k−1}{2^k}n!)>\sigma(n!)$.

It can be easily checked for these intervals (see below):

$k=9$ for $16 \leq n \leq 72$

$k=60$ for $73 \leq n \leq 1320$

At this point, with such high numbers, we can pick any $k$ around $n/2$ with the only requirement $2^k−1$ is not a Mersenne prime (EDIT: In the proof below, the exact requirement is $k$ is divisible by $3$, that guarantees $2^k-1$ is divisible by $7$).

To check the previous intervals we can do the following. For $n=16$ and $k=9$ we have $2^9−1=511=7 \times 73$, and $16! = 2^{15}\cdot7^2\cdot m$, where $m$ is coprime to $2$, $7$ and $73$. Then:

$\sigma(\frac{511}{512}\cdot16!)=\frac{2^7−1}{2^{16}−1}\cdot\frac{7^4−1}{7^3−1}\cdot74 \cdot \sigma(16!)>\frac{2^7−1}{2^{16}−1}\cdot518 \cdot \sigma(16!)>\sigma(16!)$

And this is not going to change until at least $n=73$, so the claim holds for the whole interval. The other interval can be checked with similar calculations.

EDIT: The proof for big $n$ could be like this. Suppose $s$ is the exponent of $7$ in $n!$, so $s<\frac{n}{6}$, and $r$ is the exponent of $2$ in $n!$. For $n>1023$ we have $r>\frac{99n}{100}$. We choose $k$ divisible by $3$ (that implies $2^k-1$ is divisible by $7$) such that $r-4 \leq 2k \leq r+1$, so:

$2^k>2^{\frac{99n-400}{200}}$

Then we have:

$\frac{\sigma(\frac{2^k−1}{2^k}n!)}{\sigma(n!)} > \frac{2^{r+1-k}-1}{2^{r+1}-1} \cdot \frac{7^{s+2}-1}{7^{s+1}-1} \cdot \frac{2^k-1}{7} = \frac{2^{r+1-k}-1}{2^{r+1}-1} \cdot (1+\frac{6}{7^{s+2}-7}) \cdot (2^k-1) = \frac{2^{r+1-k}-1}{2^{r+1}-1} \cdot (2^k-1+\frac{6}{7} \cdot \frac{2^k-1}{7^{s+1}-1}) > \frac{2^{r+1-k}-1}{2^{r+1}-1} \cdot (2^k-1+\frac{6}{7} \cdot \frac{2^{\frac{99n-400}{200}}-1}{7^{\frac{n+6}{6}}-1})$

The last term is increasing with $n$ and, for $n>1023$, is much bigger than $2$:

$\frac{\sigma(\frac{2^k−1}{2^k}n!)}{\sigma(n!)} > \frac{2^{r+1-k}-1}{2^{r+1}-1} \cdot (2^k+1) = \frac{2^{r+1}-2^k+2^{r+1-k}-1}{2^{r+1}-1} = 1 + \frac{2^{r+1-k}-2^k}{2^{r+1}-1} \geq 1$

This works for $n>1023$ so the proof is complete.