On the sizes of the maximal prime powers divisors of factorials

Dan Levy

arxiv: 2601.03414 · v3 · submitted 2026-01-06 · 🧮 math.NT

On the sizes of the maximal prime powers divisors of factorials

Dan Levy This is my paper

Pith reviewed 2026-05-16 16:26 UTC · model grok-4.3

classification 🧮 math.NT

keywords factorialsprime powersp-adic valuationsmaximal divisorstwin primesasymptotic comparisonnumber theory

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The pith

For any prime p there exists n0(p) such that p to the valuation power exceeds all larger q to their valuation for n at least n0.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper shows that no matter which prime p you pick, after some point n0 that depends only on p, the largest power of p dividing n factorial is bigger than the largest power of any bigger prime q dividing the same n factorial. This is established using the usual formula that sums how many multiples of p, p squared and so on fit into n. A reader might care because it tells us that the prime factorization of large factorials is eventually led by the smallest primes in terms of the size of their highest powers. In the special case where q is the twin prime p plus two, the smallest possible n0 turns out to be p times p plus p all over two.

Core claim

Let p be any prime, and p to the power of nu_p of n factorial the maximal power of p dividing n factorial. It is proved that there exists a positive integer n0 which depends only on p such that q to the power of nu_q of n factorial is less than p to the power of nu_p of n factorial for all n greater than or equal to n0 and all primes q greater than p. For twin primes p and q equals p plus two it is proved that the minimal n0 is p squared plus p all over two.

What carries the argument

The floor-sum formula for computing the exponent nu_r of a prime r in n factorial, which determines the size of the maximal r-power divisor.

Load-bearing premise

The asymptotic growth rates of the valuation sums are enough to guarantee that the exponential inequality becomes and stays true for all sufficiently large n.

What would settle it

Pick a small prime like p equals 5, calculate the proposed n0, then verify for n equals n0 and several larger values whether any prime q greater than 5 has a larger maximal power in n factorial than 5 does, looking for a violation at any scale.

read the original abstract

Let p be any prime, and $p^(\nu_p(n!))$ the maximal power of $p$ dividing $n!$. It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ and all primes $q > p$. For twin primes $p$ and $q = p + 2$ it is proved that the minimal $n_0$ satisfying $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

The explicit n0 formula for twin primes fails for p=2, undercutting the main new claim.

read the letter

The one thing to know is that the paper's explicit formula for the minimal n0 in the twin prime case is incorrect for p=2. The general result is that for any fixed prime p, the p-power in n! eventually overtakes the q-power for every larger prime q. This comes from the p-adic valuation v_p(n!) being roughly n/(p-1), which is larger than n/(q-1) for q > p, and although the base is smaller, the exponent difference wins out for large n. The paper gives an explicit n0 = (p^2 + p)/2 for the case when the next prime is p+2. That explicit formula is the new part, and it looks like a precise statement not directly in earlier work. However, it doesn't hold up for the smallest case. For p=2, the formula gives n0=3. But in 3!, the valuation v_2 is 1 and v_3 is 1, so we have 2^1 = 2 and 3^1 = 3. The claimed inequality 3 < 2 is false. So at the proposed minimal n0 the property already fails. This means either the formula is only for odd p, or there's a miscalculation in deriving the minimal value. The rest of the argument seems to rest on standard estimates for the sum in the valuation formula. If the full proof checks the inequality for all q > p once n is large enough, that part might be fine, but the twin prime explicit part is the one with the visible problem. For someone interested in bounds on the prime power factors of n!, this gives some control, but the error in the main new claim is a problem. It is worth checking the full proof to see if the general existence is proved cleanly or if there are other gaps. I would bring this to a reading group to see how the proof handles small n and the twin prime case specifically. I would not cite it as is because of the counterexample. And I would not send it to peer review until the formula is corrected and verified with explicit checks for small primes.

Referee Report

1 major / 0 minor

Summary. The paper claims to prove that for any prime p there exists n0 = n0(p) such that q^{v_q(n!)} < p^{v_p(n!)} holds for all n ≥ n0 and every prime q > p. For twin-prime pairs p and q = p + 2 it further claims that the minimal such n0 is exactly (p² + p)/2.

Significance. If the claims held, the result would give a precise asymptotic comparison of the largest prime-power divisors of n!, showing that the contribution from the smallest prime p eventually dominates all larger prime powers in size, with an explicit threshold in the twin-prime case.

major comments (1)

[Abstract] Abstract (and the corresponding theorem on twin primes): the explicit formula n0 = (p² + p)/2 is false. For the twin-prime pair p = 2, q = 3 one obtains n0 = 3, yet v_2(3!) = ⌊3/2⌋ = 1 and v_3(3!) = ⌊3/3⌋ = 1, so 3^1 = 3 ≯ 2^1 = 2 and the required strict inequality fails at n = n0 itself.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their careful reading of the manuscript and for identifying an error in the claimed minimal n0 for the twin-prime case. We agree that the explicit formula requires correction and will revise the relevant theorem and abstract accordingly.

read point-by-point responses

Referee: [Abstract] Abstract (and the corresponding theorem on twin primes): the explicit formula n0 = (p² + p)/2 is false. For the twin-prime pair p = 2, q = 3 one obtains n0 = 3, yet v_2(3!) = ⌊3/2⌋ = 1 and v_3(3!) = ⌊3/3⌋ = 1, so 3^1 = 3 ≯ 2^1 = 2 and the required strict inequality fails at n = n0 itself.

Authors: We agree with the referee that the proposed formula does not hold for p=2. At n=3 we have v_2(3!)=1 and v_3(3!)=1, yielding 2^1=2 and 3^1=3, so the strict inequality 3^1 < 2^1 fails. The minimal n0 for this pair is therefore at least 4 (where v_2(4!)=3 and 8>3). We will revise the statement of the twin-prime theorem to give the correct minimal n0, either by adjusting the formula or by treating p=2 as a separate case. The general existence result (for arbitrary primes p) is unaffected by this correction. revision: yes

Circularity Check

0 steps flagged

No circularity; derivation is self-contained

full rationale

The paper establishes the existence of n0(p) and the explicit formula for twin-prime pairs via the standard de Polignac/Legendre formula for v_p(n!) = sum floor(n/p^k) together with direct comparison of the resulting exponential sizes. No fitted parameters are renamed as predictions, no self-citation supplies a uniqueness theorem or ansatz, and the claimed minimal n0 is not defined in terms of itself. The argument is an ordinary existence proof in elementary number theory and remains independent of its own conclusions.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The central claim rests on the standard definition of the p-adic valuation of n! via the floor-sum formula and on the comparison of the resulting exponential quantities; no free parameters or new entities are introduced.

axioms (1)

standard math The p-adic valuation is given by v_p(n!) = sum_{k >= 1} floor(n / p^k)
This is the classical definition invoked to express the exponents whose powers are compared.

pith-pipeline@v0.9.0 · 5421 in / 1317 out tokens · 52301 ms · 2026-05-16T16:26:02.780388+00:00 · methodology

On the sizes of the maximal prime powers divisors of factorials

Core claim

What carries the argument

Load-bearing premise

What would settle it

discussion (0)