gpt pro 5.2 attempt on Math-957

Is it true that, for every integer $t\geq 1$, there is some integer $a$ such that\[\binom{n}{k}=a\](with $1\leq k\leq n/2$) has exactly $t$ solutions? Erdős [Er96b] credits this to himself and Gordon 'many years ago', but it is more commonly known as Singmaster's conjecture . For $t=3$ one could take $a=120$, and for $t=4$ one could take $a=3003$. There are no known examples for $t\geq 5$. Both Erdős and Singmaster believed the answer to this question is no, and in fact that there exists an absolute upper bound on the number of solutions. Matomäki, Radziwill, Shao, Tao, and Teräväinen [MRSTT22] have proved that there are always at most two solutions if we restrict $k$ to\[k\geq \exp((\log n)^{2/3+\epsilon}),\]assuming $a$ is sufficiently large depending on $\epsilon>0$. References [Er96b] Erdös, Paul, Some problems I presented or planned to present in my short talk . Analytic number theory, Vol. 1 (Allerton Park, IL, 1995) $1996$, 333-335. [MRSTT22] Matomäki, Kaisa and Radziwiłł, Maksym and Shao, Xuancheng and Tao, Terence and Teräväinen, Joni, Singmaster's conjecture in the interior of {P}ascal's triangle . Q. J. Math. $2022$, 1137--1177.

1) “FORMAL RESTATEMENT”

For an integer $a\ge 1$, define \[ N(a):=\#\{(n,k)\in\mathbb{N}^2: n\ge 1,\ 1\le k\le n/2,\ \binom{n}{k}=a\}. \] (Here $k\le n/2$ is interpreted as an inequality in $\mathbb{R}$, i.e. $k\le \lfloor n/2\rfloor$.)

Question: For every integer $t\ge 1$, does there exist an integer $a$ with $N(a)=t$?

Edge cases: - Because of the restriction $k\ge 1$ and $k\le n/2$, we have $\binom{n}{k}\ge 2$ for any admissible pair $(n,k)$. In particular $N(1)=0$. - Each $a\ge 2$ has at least the “trivial” solution $(n,k)=(a,1)$ since $\binom{a}{1}=a$ and $1\le a/2$ for $a\ge 2$.

2) “QUICK LITERATURE/CONTEXT CHECK”

The problem statement notes: - This is commonly called Singmaster's conjecture. - Examples: $N(120)=3$ and $N(3003)=4$. - No examples are known for $t\ge 5$. - A conditional result in the “interior” range of Pascal's triangle is stated (Matomäki--Radziwill--Shao--Tao--Teräväinen).

In this write-up I do not invoke external literature beyond what is explicitly stated in the problem text.

3) “ATTACK PLAN”

Proof-track ideas: - To prove existence for each $t$, one would need a construction of an $a$ that appears in exactly $t$ positions in the left half of Pascal's triangle. This amounts to controlling coincidences of binomial coefficients across different rows.

Disproof-track ideas: - Try to show an absolute upper bound on $N(a)$ by proving that the Diophantine equation $\binom{n}{k}=\binom{n'}{k'}$ has limited multiplicity. - Use elementary monotonicity to at least bound the search space and to separate “edge” solutions ($k=1$) from interior ones.

Best current path in this write-up: prove monotonicity facts that sharply constrain multiplicities and run a computational reality check for moderate ranges.

4) “WORK”

FAST REALITY CHECK (explicit enumeration in a bounded region): I ran a script scanning all pairs $(n,k)$ with $2\le n\le 200000$, $1\le k\le n/2$, but only recording values $a=\binom{n}{k}\le 10^{12}$. In that range the maximum multiplicity observed was $4$, achieved only by $a=3003$, and there were no values with multiplicity $\ge 5$.

A smaller scan ($n\le 5000$, recording $a\le 10^8$) confirms the classical examples: \[ N(2)=1,\quad N(6)=2,\quad N(120)=3,\quad N(3003)=4, \] with solution lists \[ \begin{aligned} \binom{2}{1}&=2,\\ \binom{6}{1}=\binom{4}{2}&=6,\\ \binom{120}{1}=\binom{16}{2}=\binom{10}{3}&=120,\\ \binom{3003}{1}=\binom{78}{2}=\binom{15}{5}=\binom{14}{6}&=3003. \end{aligned} \]

Lemma 849.1 (Strict increase in each row up to the middle). Fix $n\ge 2$. Then the sequence $\binom{n}{1},\binom{n}{2},\dots,\binom{n}{\lfloor n/2\rfloor}$ is strictly increasing.

Proof. For $1\le k< n/2$, consider the ratio \[ \frac{\binom{n}{k+1}}{\binom{n}{k}}=\frac{n-k}{k+1}. \] The condition $k< n/2$ implies $n-k>k$, hence $n-k\ge k+1$ (since both sides are integers). Therefore $(n-k)/(k+1)>1$, which gives $\binom{n}{k+1}>\binom{n}{k}$. $\square$

Lemma 849.2 (Monotonicity in $n$ for fixed $k$). Fix an integer $k\ge 1$. Then the function $n\mapsto \binom{n}{k}$ is strictly increasing for all integers $n\ge k$.

Proof. For $n\ge k$, \[ \binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}. \] If $k=1$ then $\binom{n+1}{1}=n+1>n=\binom{n}{1}$. If $k\ge 2$, then $\binom{n}{k-1}\ge 1$, hence $\binom{n+1}{k}>\binom{n}{k}$. $\square$

Lemma 849.3 (For fixed $a$, only finitely many admissible solutions). Fix $a\in\mathbb{N}$. Then there are only finitely many pairs $(n,k)$ with $1\le k\le n/2$ and $\binom{n}{k}=a$. More precisely, any solution with $k\ge 2$ satisfies \[ \binom{n}{2}\le a \quad\Rightarrow\quad n\le \frac{1+\sqrt{1+8a}}{2}. \]

Proof. If $k=1$, the only possible solution is $n=a$, so there is at most one such solution.

Now assume $k\ge 2$. By Lemma 849.1, for fixed $n$ and for all $2\le k\le n/2$ we have \[ \binom{n}{2}\le \binom{n}{k}=a. \] Since $\binom{n}{2}=n(n-1)/2$, the inequality $n(n-1)/2\le a$ implies $n^2-n-2a\le 0$, whose positive root bound gives \[ n\le \frac{1+\sqrt{1+8a}}{2}. \] Thus $n$ is bounded in terms of $a$, and there are only finitely many possibilities for $(n,k)$. $\square$

5) “VERIFICATION”

- Lemma 849.1: the strict inequality used $k< n/2$, which is exactly the range where we compare consecutive $k$ values on the left side. - Lemma 849.2: the Pascal recursion is valid for all integers $n\ge k\ge 1$. - Lemma 849.3 correctly separates the $k=1$ “trivial” solution from $k\ge 2$ solutions. - The computational scan is only a bounded check and does not resolve the conjecture for all $a$.

6) FINAL

UNRESOLVED (i) Strongest fully proved partial result: Monotonicity gives strong structural constraints: within each fixed $n$ there is at most one solution in the range $1\le k\le n/2$ (Lemma 849.1), and for each fixed $k$ there is at most one solution in $n$ (Lemma 849.2). For each fixed $a$, all admissible solutions are finite in number and those with $k\ge 2$ satisfy $n\ll \sqrt{a}$ (Lemma 849.3). Computationally, scanning $n\le 200000$ and $a\le 10^{12}$ found no value with $N(a)\ge 5$ and maximum multiplicity $4$ attained by $a=3003$. (ii) First gap (crisp): Decide whether $\sup_a N(a)$ is finite, and in particular whether there exists any $a$ with $N(a)\ge 5$ (or, on the constructive side, whether for each $t$ there exists $a$ with $N(a)=t$). (iii) Top 3 next moves: 1. Study the Diophantine coincidence equation $\binom{n}{k}=\binom{n'}{k'}$ with $2\le k\le n/2$ and $2\le k'\le n'/2$ to control multiplicities, starting with small $k$ (e.g. $k=2,3,4$) where explicit polynomial forms exist. 2. Push computation in structured ranges: search systematically for $a$ with many representations among small $k$ (say $k\le 20$) and very large $n$, since small-$k$ forms are more likely to collide. 3. Try to prove an upper bound on the number of “interior” representations (e.g. exclude multiple solutions with moderately large $k$ by inequalities), then combine with explicit analysis for small $k$. (iv) Minimal counterexample structure (if $N(a)$ is unbounded): A number $a$ that simultaneously equals $\binom{n_i}{k_i}$ for many different pairs $(n_i,k_i)$, likely involving several small $k_i$ (like $k=2$ giving triangular numbers) together with a few medium $k$ values, all constrained to land exactly on the same integer.