Erdős problem #1184 — wave 6x
Date: 2026-07-27 UTC
0. Mandatory live-page gate
I fetched https://www.erdosproblems.com/1184 and its linked discussion
thread through the Bright Data browser. I did not infer the statement from
the tracker metadata or use a datacenter curl response as a substitute.
The live page showed:
- status: OPEN;
0 claimed proofs for this problem;Currently working on this problem: None;Interested in collaborating: None;- likes, “looks difficult”, “looks tractable”, “results could be
formalisable”, and “working on formalising”: all None;
- last page edit: 06 April 2026.
Thus the mandatory stop condition did not fire.
Verbatim current statement
> Let \(f(n,k)\) count the number of \(1\leq i\leq k\) such that
> \(P(n+i)>k\) (where \(P(m)\) is the largest prime divisor of \(m\)).
> Is it true that, if \(\alpha>1\) is such that
> \(n=k^{\alpha+o(1)}\), then
> \[ > f(n,k)=(1-\rho(\alpha)+o(1))k, > \]
> where \(\rho\) is the Dickman function?
Verbatim listed known results
> Erdős [Er76e] proved that, for every \(\alpha>1\), if \(k\) is
> sufficiently large and \(n>k^\alpha-k\) then
> \[ > f(n,k)>\left(1-\frac1\alpha+c_\alpha\right)k > \]
> for some constant \(c_\alpha>0\), and if \(1<\alpha<2\) and
> \(n\leq k^\alpha-k\) then
> \[ > f(n,k)<(\alpha-1+o(1))k. > \]
> He knew of no non-trivial bounds when \(\alpha\geq2\).
>
> Ramachandra, Shorey, and Tijdeman [RST75b] proved that if
> \(n>\exp(c(\log k)^2)\) for a constant \(c>0\) then
> \(f(n,k)\geq k-\pi(k)\).
The one current comment
The linked thread is
https://www.erdosproblems.com/forum/thread/1184. Its only comment is:
> I believe the problem statement should read as
> \(f(n,k)=(1-\rho(\alpha)+o(1))k\) instead.
It was posted by JakeMallen at 01:02 on 05 April 2026. The revision
history shows that the preceding page version had
\(\rho(1/\alpha)\); the current page has incorporated the correction to
\(\rho(\alpha)\). The site labels comments as unverified user content.
There are no claimed proofs or worker markers in the thread.
1. Claim labels
Every mathematical assertion below is labelled:
- [a] elementary-rigorous;
- [b] rigorous modulo the explicitly named theorem(s);
- [c] plausible/structural-unverified;
- [d] computational-only.
Live-page observations and bibliographic metadata are factual audit
notes, rather than mathematical claims.
2. Exact reformulation as a short-interval smooth-number problem
Write
\[ \Psi(x,y)=\#\{m\leq x:P(m)\leq y\}. \]For integers \(n>k\geq2\), every one of \(n+1,\ldots,n+k\) is either
\(k\)-smooth or has largest prime factor greater than \(k\). Therefore
\[ \boxed{\quad f(n,k)=k-\bigl(\Psi(n+k,k)-\Psi(n,k)\bigr). \quad} \tag{2.1} \]This is an exact identity. [a]
Consequently, the live question is exactly the all-starting-points
short-interval asymptotic
\[ \Psi(n+k,k)-\Psi(n,k)=(\rho(\alpha)+o(1))k \quad\text{when}\quad n=k^{\alpha+o(1)}. \tag{2.2} \]It is not merely the classical long-interval Dickman theorem. [a]
3. Primary-source literature audit
3.1 Original sources listed by the page
The official journal scan of P. Erdős,
Problems and results on consecutive integers,
Publ. Math. Debrecen 23 (1976), 271–282, exists. Page 272 states
equation (3) in the equivalent form
\[ \log n=(\alpha+o(1))\log k,\qquad f(n,k)=(1-c_\alpha+o(1))k, \]where the preceding paragraph defines \(c_\alpha\) as the density of
\(k\)-smooth integers up to \(k^\alpha\), and notes
\(c_\alpha=1-\log\alpha\) for \(1\leq\alpha\leq2\). The same page states
the two displayed Erdős bounds and attributes the
\(k-\pi(k)\) result to Ramachandra, Shorey, and Tijdeman. This directly
confirms that the current \(\rho(\alpha)\), not
\(\rho(1/\alpha)\), is the intended parameter. [a]
The cited Grimm paper also exists:
K. Ramachandra, T. N. Shorey, and R. Tijdeman,
[*On Grimm's problem relating to factorisation of a block of consecutive
integers. II*](https://doi.org/10.1515/crll.1976.288.192),
J. Reine Angew. Math. 288 (1976), 192–201. I use the live page's stated
consequence as ground truth and do not need to strengthen it here.
3.2 A 2024 theorem not listed on the live page
Khalid Younis,
[*Asymptotics for smooth numbers in short
intervals*](https://arxiv.org/abs/2409.05761),
arXiv:2409.05761v1 (9 September 2024), is a real 30-page primary
preprint. Its Theorem 1.1 says that for fixed
\(17/30<\theta\leq1\), there is \(C=C(\theta)>0\) such that
\[ \frac{\Psi(x+h,y)-\Psi(x,y)}{h} =\frac{\Psi(x,y)}x \left(1+O_\theta\!\left(\frac{\log(u+1)}{\log y}\right)\right), \qquad u=\frac{\log x}{\log y}, \tag{3.1} \]uniformly for
\[ x^\theta\leq h\leq x,\qquad \exp\!\left(C(\log x)^{2/3}(\log\log x)^{4/3}\right) \leq y\leq2x. \tag{3.2} \]Thus (3.1) is a relative asymptotic. The arXiv source file itself was
downloaded and checked, rather than reconstructing the formula from a
search snippet. [b: Younis, Theorem 1.1]
Theorem 1.3 of the same paper states, assuming the Riemann hypothesis,
an analogous result for every fixed \(1/2<\theta\leq1\), in the much
larger range \((\log x)^{K(\theta)}\leq y\leq2x\); its main term has an
extra saddle-point factor which is \(1+o(1)\) when \(u\) is bounded.
**[b: RH and Younis, Theorem 1.3, together with its displayed
saddle-point estimate]**
The recent peer-reviewed paper of Sarvagya Jain,
[*Existence of Smooth Numbers in Short
Intervals*](https://doi.org/10.1093/qmath/haag010),
Q. J. Math. 77 (2026), 397–422, explicitly records Younis's
\(x^{17/30+o(1)}\) all-interval asymptotic as the current asymptotic
input. Jain proves stronger existence results in other ranges, not the
density asymptotic (2.2), so it does not extend the result below.
The needed global estimate is classical. For example, A. Hildebrand,
[*On the number of positive integers \(\leq x\) and free of prime
factors \(>y\)*](https://doi.org/10.1016/0022-314X(86)90013-2),
J. Number Theory 22 (1986), 289–307, gives in the present power-sized
\(y\) regime
\[ \frac{\Psi(x,y)}x =\rho\!\left(\frac{\log x}{\log y}\right)+o(1) \]uniformly when \(\log x/\log y\) remains in a fixed compact subset of
\((1,\infty)\). [b: the Dickman–de Bruijn/Hildebrand theorem]
I also searched exact strings from #1184, the expression
P(n+i)>k, the title and citations of the original paper, and
2024–2026 primary papers on asymptotics for smooth numbers in all short
intervals. I found no newer primary theorem improving Younis's
\(17/30\) all-interval asymptotic exponent. This is a documented search
miss, not a claim that no such work can exist.
4. Main result: an unconditional solved range
Proposition
For every fixed
\[ \boxed{1<\alpha<\frac{30}{17}}, \]and every integer sequence \(k\to\infty\),
\(n=k^{\alpha+o(1)}\), the conclusion of Erdős problem #1184 holds:
\[ f(n,k)=(1-\rho(\alpha)+o(1))k. \tag{4.1} \]**[b: Younis, Theorem 1.1, and the classical
Dickman–de Bruijn/Hildebrand theorem]**
Proof
Set
\[ x=n,\qquad y=k,\qquad h=k,\qquad u=\frac{\log n}{\log k}=\alpha+o(1). \tag{4.2} \][a]
Because \(\alpha<30/17\), choose a fixed number
\[ \frac{17}{30}<\theta<\frac1\alpha. \tag{4.3} \]Then
\[ \frac{\log h}{\log x} =\frac{\log k}{\log n} =\frac1\alpha+o(1)>\theta, \]so \(x^\theta\leq h\leq x\) for all sufficiently large \(k\).
Also \(y=k\leq2n=2x\). [a]
For the lower condition on \(y\), with the fixed constant
\(C=C(\theta)\) from Younis,
\[ \frac{C(\log n)^{2/3}(\log\log n)^{4/3}}{\log k} =O_\alpha\!\left( \frac{(\log\log n)^{4/3}}{(\log n)^{1/3}} \right)=o(1). \]Hence
\[ k\geq \exp\!\left(C(\log n)^{2/3}(\log\log n)^{4/3}\right) \]eventually. Thus every hypothesis of Younis's Theorem 1.1 is met.
[a]
Younis therefore gives
\[ \frac{\Psi(n+k,k)-\Psi(n,k)}k =\frac{\Psi(n,k)}n \left(1+O_\theta\!\left( \frac{\log(u+1)}{\log k}\right)\right) =\frac{\Psi(n,k)}n(1+o(1)), \tag{4.4} \]where the middle parenthesis is multiplicative. **[b: Younis,
Theorem 1.1]**
Since \(u=\alpha+o(1)\) remains bounded, the global Dickman theorem and
continuity of \(\rho\) give
\[ \frac{\Psi(n,k)}n=\rho(u)+o(1)=\rho(\alpha)+o(1). \tag{4.5} \][b: Dickman–de Bruijn/Hildebrand]
Combining (4.4), (4.5), and the exact complement identity (2.1) proves
(4.1). [a]
The inequalities are strict: Younis requires
\(\theta>17/30\), so this argument does not include
\(\alpha=30/17\). [a]
Conditional extension
Assume the Riemann hypothesis. For any fixed \(1<\alpha<2\), choose
\(1/2<\theta<1/\alpha\). Then \(h=k\geq n^\theta\) eventually and
\(k\geq(\log n)^{K(\theta)}\). Younis's Theorem 1.3 applies; its
saddle-point factor is \(1+o(1)\) for
\(\log n/\log k=\alpha+o(1)\). The same Dickman and complement steps
give (4.1). Thus RH implies #1184 for every fixed
\[ \boxed{1<\alpha<2}. \][b: RH, Younis Theorem 1.3, and Dickman–de Bruijn/Hildebrand]
Again the endpoint \(\alpha=2\) is not included because the theorem
requires \(\theta>1/2\). [a]
5. Exact subquadratic prime-sum identity
When
\[ n+k<(k+1)^2, \tag{5.1} \]there is also the exact formula
\[ \boxed{ f(n,k)= \sum_{a=1}^{\left\lfloor (n+k)/(k+1)\right\rfloor} \left[ \pi\!\left(\left\lfloor\frac{n+k}{a}\right\rfloor\right) -\pi\!\left(\max\!\left\{k,\left\lfloor\frac n a\right\rfloor\right\}\right) \right]. } \tag{5.2} \][a]
Indeed, an integer \(m\leq n+k\) cannot have two prime factors greater
than \(k\), counted with multiplicity, because their product would be at
least \((k+1)^2>m\). Thus every \(m\in(n,n+k]\) counted by \(f\) has a
unique representation
\[ m=ap,\qquad p>k\ \text{prime},\qquad a\leq\frac{n+k}{k+1}\leq k. \]Conversely, every such product has largest prime factor \(p>k\).
For fixed \(a\), the allowable primes are exactly
\[ \max\!\left(k,\left\lfloor\frac n a\right\rfloor\right)For a fixed \(1<\alpha<2\), condition (5.1) holds eventually whenever
\(n=k^{\alpha+o(1)}\). On \(1\leq\alpha\leq2\), the defining
differential-delay equation of the Dickman function integrates to
\[ \rho(\alpha)=1-\log\alpha. \]Hence in this range the conjectured conclusion is equivalently
\[ f(n,k)=(\log\alpha+o(1))k. \tag{5.3} \][a]
Formula (5.2) is a clean reduction, but applying a prime number theorem
separately to all its summands is too demanding. Near
\(a\asymp n/k\), the prime interval has base point
\(n/a\asymp k\) and length
\[ \frac{k}{a}\asymp\frac{k^2}{n}=k^{2-\alpha+o(1)}. \tag{5.4} \]Thus for \(\alpha\) near \(2\), a termwise approach asks for primes in
every extremely short interval. It also loses the aggregate structure
that Younis's smooth-number method exploits. [a]
6. Reproducible exact computation
The standalone checker is:
runs/erdos1184_wave6x_reverify.py
Run it with:
python3 runs/erdos1184_wave6x_reverify.py
It uses only the Python standard library. It constructs largest-prime-
factor and prime-counting arrays from scratch and checks that array
against independent trial division through 10,000.
First, it exhaustively verified (5.2) for every
\[ 2\leq k\leq60,\qquad kSecond, set \(A=\lfloor k^{3/2}\rfloor\). For each displayed \(k\), the
checker examined every integer \(n\in[A,2A]\), not a sample. It
computed the exact minimum and maximum of \(f(n,k)\), their first
locations, and the mean of \(f(n,k)/k\). This dyadic box satisfies
\(n=k^{3/2+O(1/\log k)}\). [a]
\[ \begin{array}{r|r|r|r@{\ }l|r@{\ }l|c} k&A&\#n& \min f&(\text{first }n)& \max f&(\text{first }n)& \operatorname{mean}(f/k)\\ \hline 100&1000&1001&46&(1021)&59&(1891)&0.529940060\\ 300&5196&5197&135&(5326)&165&(9639)&0.505784748\\ 1000&31622&31623&452&(32229)&526&(61105)&0.492288651\\ 3000&164316&164317&1343&(165674)&1529&(320679)&0.481347714\\ 10000&1000000&1000001&4435&(1003204)&4947&(1941502)&0.472267884 \end{array} \][d]
For comparison, the limiting prediction at \(\alpha=3/2\) is
\[ 1-\rho(3/2)=\log(3/2)=0.405465108108\ldots. \tag{6.1} \][a]
The first scan obtains each rolling-window count from the largest-factor
array. Independently, the checker uses a separate Eratosthenes prime
list, divides every integer in the full union of the scanned windows by
all prime powers at most \(k\), constructs a second indicator array, and
repeats each entire scan. The two methods agree on the total, minimum,
maximum, and first extremizers in every row. The complete run finished
in about 3.5 seconds on this VM and printed ALL CHECKS PASSED. [d]
These finite values are still far from (6.1) and prove no asymptotic
statement; they are reported only as exact reproducible data. [d]
7. What remains, precisely
Unconditionally, the argument above leaves
\[ \alpha\in[30/17,\infty); \]under RH it leaves \(\alpha\in[2,\infty)\). These are statements about
the reach of the cited theorems and this deduction, not assertions that
no other method can work. [b]
The exact missing analytic input for a given remaining fixed
\(\alpha\) is an all-interval, not almost-all, estimate
\[ \Psi(x+h,y)-\Psi(x,y) =\bigl(\rho(\alpha)+o(1)\bigr)h \tag{7.1} \]uniformly along
\[ y=h,\qquad x=y^{\alpha+o(1)}. \tag{7.2} \]For \(30/17\leq\alpha<2\), it would suffice to lower Younis's admissible
all-interval exponent from \(\theta>17/30\) to some
\(\theta<1/\alpha\). At \(\alpha\geq2\), one needs an asymptotic at or
below the square-root scale. [a]
Almost-all short-interval theorems do not establish (7.1), because the
sequence \(n=n(k)\) in #1184 is arbitrary and could, in principle, lie
inside the exceptional sets for every \(k\). Existence of one smooth
number is also far weaker than the required
\((\rho(\alpha)+o(1))k\) count. [a]
Brute force cannot supply the uniform limiting step. As a scale
estimate, a monolithic exact scan of the full dyadic range at
\(\alpha=2\), \(k=10^6\), would address about \(2\cdot10^{12}\)
integers. A four-byte largest-factor array alone would require about
8 TB, and a classical sieve would perform on the order of
\(7\cdot10^{12}\) factor updates—roughly \(10^2\)–\(10^3\) core-hours
after realistic segmented-factorization and memory overhead. Even that
finite calculation would not prove (7.1). [d]
PARTIAL: Younis's 2024 all-interval theorem plus the classical Dickman theorem proves #1184 for every fixed \(1<\alpha<30/17\) unconditionally (and for \(1<\alpha<2\) under RH); the endpoints and larger \(\alpha\) remain beyond the cited machinery, with an exact subquadratic prime-sum reduction and independently verified finite scans supplied.