Erdős Problem #386

Erdős Problem #386

1) ROUND-2 OBJECTIVE

I pursue (C) obstruction/partial reduction. Round-1 isolated only the very weak necessary condition that $\binom{n}{k}$ must be squarefree. The most promising next step is to prove that any hypothetical infinite family cannot live in the interior of Pascal's triangle. In this round I prove a rigorous asymptotic reduction: \[ \text{if } \binom{n}{k}\text{ is a product of consecutive primes, then } k=o(n) \] (after using symmetry to assume $2\leq k\leq n/2$). Thus any infinite family must occur in a thin boundary layer.

2) Round-1 FOUNDATION USED

I use the following Round-1 inputs.

1. Round-1 squarefree condition: if $\binom{n}{k}$ is a product of consecutive primes, then every prime exponent is $1$.
2. Round-1 finite search up to $n\leq 200$, which found only the eight known examples.

I also use the elementary symmetry \[ \binom{n}{k}=\binom{n}{n-k}, \] so it suffices to work with $2\leq k\leq n/2$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is an interval-by-interval valuation analysis of prime divisors of $\binom{n}{k}$, combined with an entropy-versus-primorial comparison.

For $k$ in the interior, the prime support of $\binom{n}{k}$ is forced to contain whole prime intervals and to miss whole neighboring prime intervals. If the support were a block of consecutive primes, these forced holes create strong restrictions. When these restrictions are combined with the estimates \[ \theta(x)=x+o(x) \qquad\text{and}\qquad \log\binom{n}{k}=nH(k/n)+o(n), \] where $H(c)=-c\log c-(1-c)\log(1-c)$, one gets a contradiction unless $k=o(n)$.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was that squarefreeness did not explain where in Pascal's triangle examples could occur.

The new plan has two stages.

1. First, rule out $n/3<k\leq n/2$ for all sufficiently large $n$ by showing that the prime support would have to be both too large combinatorially and too small logarithmically.
2. Second, once $k\leq n/3$, show that any positive-proportion regime $k\sim cn$ with $0<c\leq 1/3$ also leads to an entropy contradiction. This yields $k=o(n)$.

This directly overcomes the main Round-1 deficiency: it replaces a local squarefree condition with a global asymptotic localization theorem.

5) WORK (ROUND-2)

Write \[ B_{n,k}:=\binom{n}{k}, \] and assume throughout that $2\leq k\leq n/2$. If $B_{n,k}$ is a product of consecutive primes, let $S_{n,k}$ be the set of prime divisors of $B_{n,k}$; by definition, $S_{n,k}$ is a consecutive block in the ordered list of all primes.

Lemma 386.1. Assume $n\geq 10$ and $n/3<k\leq n/2$. Then for every prime $p$ we have:

1. if $n/3<p\leq k$, then $p\nmid B_{n,k}$;
2. if $k<p\leq n/2$, then $p\mid B_{n,k}$;
3. if $n/2<p\leq n-k$, then $p\nmid B_{n,k}$;
4. if $n-k<p\leq n$, then $p\mid B_{n,k}$.

Proof. Since $p>n/3$ and $n\geq 10$, we have $p^2>n$, so only the first $p$-adic level contributes. Therefore \[ \nu_p(B_{n,k})=\left\lfloor \frac np\right\rfloor-\left\lfloor \frac kp\right\rfloor-\left\lfloor \frac{n-k}{p}\right\rfloor. \] Now evaluate this quantity on each interval.

If $n/3<p\leq k$, then $\lfloor n/p\rfloor=2$, $\lfloor k/p\rfloor=1$, and $\lfloor (n-k)/p\rfloor=1$, so $\nu_p(B_{n,k})=0$.

If $k<p\leq n/2$, then $\lfloor n/p\rfloor=2$, $\lfloor k/p\rfloor=0$, and $\lfloor (n-k)/p\rfloor=1$, so $\nu_p(B_{n,k})=1$.

If $n/2<p\leq n-k$, then $\lfloor n/p\rfloor=1$, $\lfloor k/p\rfloor=0$, and $\lfloor (n-k)/p\rfloor=1$, so $\nu_p(B_{n,k})=0$.

If $n-k<p\leq n$, then $\lfloor n/p\rfloor=1$, $\lfloor k/p\rfloor=0$, and $\lfloor (n-k)/p\rfloor=0$, so $\nu_p(B_{n,k})=1$. This proves all four assertions. $\square$

Theorem 386.2. If $B_{n,k}$ is a product of consecutive primes, then for all sufficiently large $n$ one has \[ k\leq \frac n3. \]

Proof. Suppose instead that $n/3<k\leq n/2$ for infinitely many large $n$. Set \[ I_2=(k,n/2],\qquad I_3=(n/2,n-k]. \] By Lemma 386.1, every prime in $I_2$ divides $B_{n,k}$ and no prime in $I_3$ divides $B_{n,k}$.

If both $I_2$ and $I_3$ contained primes, then the prime support $S_{n,k}$ would contain primes below and above a missing prime interval, contradicting the assumption that $S_{n,k}$ is consecutive. Hence at least one of $I_2$ or $I_3$ is prime-free.

But $I_2$ and $I_3$ have the same length, \[ |I_2|=|I_3|=\frac n2-k. \] Fix any $\delta>0$. If $k\leq (1/2-\delta)n$, then both intervals have length at least $\delta n$, and hence both contain primes for all sufficiently large $n$ by the prime number theorem. This is impossible. Therefore, for every $\delta>0$, one must have \[ k>(1/2-\delta)n \] for all sufficiently large counterexamples in the range $n/3<k\leq n/2$. Equivalently, \[ k=\frac n2+o(n). \]

Now consider \[ I_1=(n/3,k],\qquad I_4=(n-k,n]. \] Because $k=n/2+o(n)$, the interval $I_1$ has length \[ k-\frac n3=\frac n6+o(n), \] so $I_1$ contains primes for all sufficiently large $n$. Let $q$ be the largest prime in $I_1$. By Lemma 386.1, $q\notin S_{n,k}$. Also $I_4$ has length $k\sim n/2$, so it contains primes, and every such prime lies in $S_{n,k}$ by Lemma 386.1.

Since $S_{n,k}$ contains a prime larger than $q$ but does not contain $q$, it cannot contain any prime below $q$; otherwise $S_{n,k}$ would not be consecutive. By maximality of $q$, there are no primes in $(q,k]$, and Lemma 386.1 rules out all primes in $I_3$. Hence all prime divisors of $B_{n,k}$ lie in $I_2\cup I_4$. Because $B_{n,k}$ is a product of distinct consecutive primes, we obtain the upper bound \[ B_{n,k} \leq \prod_{k<p\leq n/2} p\; \prod_{n-k<p\leq n} p. \] Taking logarithms gives \[ \log B_{n,k} \leq \theta(n/2)-\theta(k)+\theta(n)-\theta(n-k). \] Using $\theta(x)=x+o(x)$ and $k=n/2+o(n)$, \[ \log B_{n,k}\leq \frac n2+o(n). \] On the other hand, Stirling's formula yields \[ \log B_{n,k}=nH(k/n)+o(n)=n\log 2+o(n), \] since $k/n\to 1/2$. Because $\log 2>1/2$, this is a contradiction. Therefore $k\leq n/3$ for all sufficiently large $n$. $\square$

Theorem 386.3. If $B_{n,k}$ is a product of consecutive primes, then \[ k=o(n) \] as $n\to\infty$ (again with $2\leq k\leq n/2$).

Proof. By Theorem 386.2 we may assume $k\leq n/3$ for all sufficiently large solutions. Suppose, for contradiction, that $k\neq o(n)$. Then there exists a subsequence with \[ \frac{k}{n}\to c \qquad\text{for some } c\in(0,1/3]. \] Set \[ J_3=(n/2,n-k],\qquad J_4=(n-k,n]. \] For primes $p>n/2$, the same floor-count argument gives:

1. if $p\in J_3$, then $p\nmid B_{n,k}$;
2. if $p\in J_4$, then $p\mid B_{n,k}$.

Now $J_3$ has length \[ \frac n2-k\geq \frac n6+o(n), \] while $J_4$ has length $k\sim cn$. Hence both $J_3$ and $J_4$ contain primes for all sufficiently large $n$ by the prime number theorem. Let $r$ be the largest prime in $J_3$. Then $r\notin S_{n,k}$, while $S_{n,k}$ contains some prime larger than $r$ coming from $J_4$. As before, the consecutivity of $S_{n,k}$ implies that $S_{n,k}$ cannot contain any prime below $r$. Since there are no primes in $(r,n-k]$ by maximality of $r$, every prime divisor of $B_{n,k}$ must lie in $J_4$. Therefore \[ B_{n,k}\leq \prod_{n-k<p\leq n} p, \] and so \[ \log B_{n,k}\leq \theta(n)-\theta(n-k)=k+o(n)=cn+o(n). \] But Stirling's formula gives \[ \log B_{n,k}=nH(c)+o(n), \] where \[ H(c)=-c\log c-(1-c)\log(1-c). \] Since $0<c\leq 1/3<e^{-1}$, we have $-\log c>1$, and hence \[ H(c)>-c\log c>c. \] Thus $nH(c)+o(n)>cn+o(n)$ for large $n$, contradicting the upper bound. This contradiction proves that $k=o(n)$. $\square$

Empirical note. An exhaustive search up to $n\leq 500$ recovers only the eight known examples \[ \binom{4}{2},\ \binom{6}{2},\ \binom{7}{3},\ \binom{10}{4},\ \binom{14}{4},\ \binom{15}{2},\ \binom{15}{6},\ \binom{21}{2}, \] and no additional ones.

6) ADVERSARIAL VERIFICATION

The critical points are the following.

1. Small examples with $k/n$ not small exist. Yes: $\binom{7}{3}$, $\binom{10}{4}$, $\binom{14}{4}$, and $\binom{15}{6}$ all lie deep enough in Pascal's triangle to show that no non-asymptotic statement like $k\leq n/3$ can hold for every $n$. My theorem explicitly says "for all sufficiently large $n$" and is therefore consistent with these examples.
2. Could the support contain primes below the chosen missing prime? No. In both Theorems 386.2 and 386.3 I pick the largest missing prime in a forbidden interval. Because the support contains a larger prime from the top interval, any smaller support prime would force the missing prime to lie inside the consecutive block, which is impossible.
3. Did I use squarefreeness correctly? Yes. The assumption "product of consecutive primes" means exactly that each prime divisor occurs with exponent $1$, so the numerical upper bounds by products of primes are valid.
4. What analytic input is really needed? Only two standard asymptotic facts: the prime number theorem in the form $\theta(x)=x+o(x)$ and the existence of primes in every interval of length $\delta x$ for fixed $\delta>0$ and large $x$, plus Stirling's formula.

No hidden symmetry or endpoint issue remains: all arguments are made after replacing $k$ by $\min(k,n-k)$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance beyond Round-1 is the rigorous asymptotic localization theorem: \[ \text{if } \binom{n}{k}\text{ is a product of consecutive primes, then } k=o(n) \] (after symmetry reduction to $k\leq n/2$). In particular, all sufficiently large solutions are forced into a boundary layer of Pascal's triangle.

This is a substantial strengthening of the Round-1 squarefree observation and sharply narrows the remaining search space. The unresolved part is exactly the edge regime $k=o(n)$, especially the delicate case of very small $k$ such as $k=2$.

8) COMPLETION ESTIMATE

COMPLETION: 58%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #386, erdosproblems.com/386, accessed 2026-03-07.
2. T. F. Bloom, Erdős Problem #386 discussion thread, erdosproblems.com/forum/thread/386, accessed 2026-03-07.

Problem 386: a prime-gap condition in the thin boundary layer

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. I read the complete 386.tex, whose target is to classify binomial coefficients $\binom nk$, $2\le k\le n/2$, that are products of distinct consecutive primes. Its entropy argument confines any infinite family to $k=o(n)$. I investigate a smaller part of that remaining layer and prove a further necessary condition. The supplied numerical searches are not inherited as independently verified computations.

2) WORK

For all sufficiently large $n$, if $2\le k\le n^{1/3}$ and $\binom nk$ is a product of consecutive primes, then the interval \[ (n-k,n] \] contains no prime.

Indeed, by the prime number theorem there is a prime $q$ in $(n/2,2n/3]$ for every sufficiently large $n$. Our range has $k\le n/3$, so $q\le n-k$, $q>k$, and $q^2>n$. Legendre's formula gives \[ v_q\binom nk=1-0-1=0. \] If a prime $p$ lies in $(n-k,n]$, the same formula gives $v_p\binom nk=1$. Consecutivity of the prime support, together with the missing $q<p$, then forbids all prime factors at most $q$. Every prime between $q$ and $n-k$ also has valuation zero. Therefore all prime factors lie in $(n-k,n]$.

There are at most $\lceil k/2\rceil$ primes in those $k$ consecutive integers, since they are odd for large $n$. Squarefreeness would give \[ \binom nk\le n^{\lceil k/2\rceil}\le n^{2k/3}, \] where the last inequality holds for every $k\ge2$. But \[ \binom nk=\prod_{i=0}^{k-1}\frac{n-i}{k-i} >\left(\frac nk\right)^k\ge n^{2k/3}. \] The strict inequality follows because $n>k$ and at least the factor $i=1$ is strictly greater than $n/k$. This contradiction proves the claim.

3) ADVERSARIAL VERIFICATION

The result is necessary, not sufficient. Prime-free intervals of fixed length occur often enough that excluding intervals containing a prime does not resolve even the fixed-$k$ problem. The proof uses an ordinary fixed-proportion consequence of the prime number theorem, not an unproved statement about primes in intervals of length $k$. The odd-prime count is valid because the interval is far above two. Repeated prime factors would invalidate the product bound, so the original distinct-prime hypothesis is essential.

4) FINAL

UNRESOLVED. In the regime $2\le k\le n^{1/3}$, every sufficiently large candidate must end inside a prime-free interval of length $k$. No exclusion of all such intervals or full classification is obtained.

5) SOURCES CHECKED

The complete supplied version, read 2026-09-23. Its prime number theorem input is used only to obtain a prime between $n/2$ and $2n/3$; the additional parity/product argument is proved above. No claim of literature priority is made.

6) COMPLETION ESTIMATE

Subjective progress toward classifying all consecutive-prime binomial coefficients. COMPLETION: 15%

Problem #386

FORMAL RESTATEMENT

Let $p_1=2<p_2=3<p_3=5<\cdots$ denote the increasing sequence of primes. Call an integer $N\ge 2$ a consecutive-prime product if there exist indices $1\le i\le j$ such that \[ N=\prod_{r=i}^j p_r. \] Equivalently, $N$ is squarefree and its prime divisors form an interval in the prime sequence.

The problem asks: are there infinitely many pairs of integers $(n,k)$ with $n\ge 4$ and $2\le k\le n-2$ such that the binomial coefficient $\binom{n}{k}$ is a consecutive-prime product?

QUICK LITERATURE/CONTEXT CHECK

This is Erdős problem #386 in the Erdős Problems database, presented as open. The page gives the example $\binom{21}{2}=2\cdot 3\cdot 5\cdot 7$ and notes several small examples communicated by Weisenberg: \[ \binom{7}{3}=5\cdot 7,\quad \binom{10}{4}=2\cdot 3\cdot 5\cdot 7,\quad \binom{14}{4}=7\cdot 11\cdot 13,\quad \binom{15}{6}=5\cdot 7\cdot 11\cdot 13. \] For $k=2$, the known values of $n$ such that $\binom{n}{2}$ is a product of consecutive primes are $n=4,6,15,21,715$ (OEIS A280992).

ATTACK PLAN

A consecutive-prime product is extremely rigid: it is squarefree and its prime factors lie in a very short interval on the logarithmic scale (no "missing primes"). Potential routes:

• Use known results on the least and greatest prime factors of $\binom{n}{k}$ to force the set of prime factors to be "too spread out" to be consecutive, at least for large $n$ in many $(n,k)$ ranges.
• Exploit squarefreeness: show that for large $n$ and nontrivial $k$ some small prime must appear with exponent $\ge 2$ in $\binom{n}{k}$, contradicting squarefreeness.
• Conversely, attempt constructive families (if they exist) by forcing $\binom{n}{k}$ to match a primorial segment (e.g. via diophantine constraints), though this seems highly nontrivial.

WORK

We can at least formalise necessary conditions and record computational data.

Necessary condition 1 (squarefree). If $\binom{n}{k}$ is a consecutive-prime product, then $\binom{n}{k}$ is squarefree.

Computational check for small $n$ (not a proof). Searching all $2\le k\le n-2$ for $n\le 200$ yields exactly the following solutions (up to the symmetry $k\leftrightarrow n-k$): \[ \begin{array}{rcl} \binom{4}{2}&=&2\cdot 3,\\ \binom{6}{2}&=&3\cdot 5,\\ \binom{7}{3}&=&5\cdot 7,\\ \binom{10}{4}&=&2\cdot 3\cdot 5\cdot 7,\\ \binom{14}{4}&=&7\cdot 11\cdot 13,\\ \binom{15}{2}&=&3\cdot 5\cdot 7,\\ \binom{15}{6}&=&5\cdot 7\cdot 11\cdot 13,\\ \binom{21}{2}&=&2\cdot 3\cdot 5\cdot 7. \end{array} \] No further examples occur with $n\le 200$. (In particular, for $k=2$ this recovers $n=4,6,15,21$; the next known $k=2$ example is at $n=715$.)

VERIFICATION

The "Necessary condition 1" is immediate. The listed small solutions were verified by direct factorisation and checking that the prime factors are distinct and consecutive in the prime sequence; symmetry $\binom{n}{k}=\binom{n}{n-k}$ explains paired solutions.

FINAL

UNRESOLVED.

Most promising partial results obtained above:

• The property forces $\binom{n}{k}$ to be squarefree, providing a concrete obstruction to exploit.
• A complete enumeration for $n\le 200$ produces only finitely many sporadic examples (listed), consistent with rarity.

Specific barrier:

• Current factor-distribution theorems for $\binom{n}{k}$ do not seem strong enough to rule out (or construct) infinitely many cases where the entire prime-support is a consecutive block.

Smallest missing step to a full resolution:

• Either (disproof) construct an infinite family $(n_i,k_i)$ with $\binom{n_i}{k_i}$ squarefree and with prime factors forming a consecutive block; or (proof of finiteness) show that for all sufficiently large $(n,k)$ with $2\le k\le n-2$, $\binom{n}{k}$ has either a repeated prime factor or at least two "separated" prime factors (missing a prime between them), contradicting the consecutive-prime condition.

COMPLETION ESTIMATE

COMPLETION: 25%