Singmaster’s conjecture

Pascal's triangle starts with 1. Each new row has 1 at both ends, and each interior entry adds the two entries above it. The row $1,4,6,4,1$ contains 6; two rows later, 6 appears near both ends again.

These entries count choices: there are six ways to choose two objects from four, one from six, or five from six. Singmaster asks whether one fixed ceiling bounds how many positions any number greater than one can occupy across the entire triangle. Reflected positions count separately; the endlessly repeated boundary 1s are excluded.[1] [2]

How many different choice counts can have exactly the same numerical value? Pascal's triangle records the number of k-element subsets of an n-element set as $\binom{n}{k}$. Singmaster's target is a fixed ceiling on the number of positions occupied by any one integer $t\geq2$, throughout all rows.

Count ordered pairs with $1\leq k<n$ satisfying $$\binom{n}{k}=t.$$ One absolute constant $C$ must work for every t. The reflected pair $(n,n-k)$ counts separately unless it is the same position.

For t equal to 6, the positions $(4,2)$, $(6,1)$ and $(6,5)$ give three occurrences. The row $n=4$ illustrates a central position counted once, while row $n=6$ contributes a mirrored pair. These are exact small calculations, not evidence for a universal bound.

Reflection follows from the choice-complement identity $\binom{n}{k}=\binom{n}{n-k}$. Further repetitions equate binomial counts from different rows. The cited curve study converts specified displacement equations between such positions into algebraic equations in integer coordinates, while keeping those restricted equations separate from the global conjecture.[1] [2]

The conjecture predicts one fixed ceiling. A full proof is missing.[1]

The 2022 theorem gives at most four occurrences in a specified interior region, for sufficiently large t depending on a fixed parameter $\varepsilon$. For example, fix $0<\varepsilon<1/3$ and define the required distance from either side by $$d(n)=\exp((\log n)^{2/3+\varepsilon}).$$ Here $\log$ is the natural logarithm and $\exp(x)$ means $e^{x}$, using its standard base e. The theorem counts positions satisfying $d(n)\leq k$ and $k\leq n-d(n)$. It does not cover every near-side position or supply the full universal bound across all positions. Finiteness for a fixed displacement between two positions is also weaker than one global multiplicity ceiling.[1] [2]

Equal counts can arise from different choice problems: choosing two from four and one from six give the same answer. Repetitions therefore expose arithmetic relations beneath a basic counting rule. The question is whether such coincidences have a universal limit even though the triangle itself never ends.[1] [2]

The same number can encode choices with different population sizes and subset sizes. Understanding all such equalities probes whether prime-factor constraints in binomial coefficients restrict coincidences much more strongly than their size alone suggests.

The cited interior proof combines comparisons of how coefficients vary with arithmetic information about their prime factors. The curve approach studies integer points arising from equal coefficients at specified displacements. These connections give substantive reasons to study repetitions: a simple counting pattern brings analytic number theory and arithmetic geometry into contact.[1] [2]