Binomial Coefficient Coincidences

These seven equations between binomial coefficients are ‘coincidences’: they aren’t among the four known infinite families. De Weger conjectured that there are no more such coincidences:

• Benjamin M. M. de Weger, Equal binomial coefficients: some elementary considerations, Journal of Number Theory 63, no. 2 (1997), 373–386.

At that time, he and his collaborators checked there were no others involving binomial coefficients less than 10³⁰. Later they checked that there are none involving binomial coefficients less than 10⁶⁰:

• Aart Blokhuis, Andries Brouwer and Benne de Weger, Binomial collisions and near collisions.

So, De Weger’s conjecture stands open. The four infinite families, by the way, are these:

$\displaystyle{ \binom{n}{k} = \binom{n}{\,n-k\,}, \qquad 0 \le k \le n}$

$\displaystyle{ \binom{n}{0} = 1, \qquad n \ge 0 }$

$\displaystyle{ \binom{\binom{n}{k}}{1} = \binom{n}{k}, \qquad \qquad 0 \le k \le n}$

and the only nontrivial one: the Lind–Singmaster family involving the Fibonacci numbers $F_i$ where $F_0 = 0,\ F_1 = 1$ :

$\displaystyle{ \binom{F_{2i+2}F_{2i+3}}{\,F_{2i}F_{2i+3}\,} \;=\; \binom{F_{2i+2}F_{2i+3}-1}{\,F_{2i}F_{2i+3}+1\,}, \qquad i = 1,2,3,\dots }$

The first three equations in the Lind–Singmaster family are these:

$\begin{array}{ccc} \displaystyle{ \binom{15}{5} } &= &\displaystyle{\binom{14}{6}} \\ \\ \displaystyle{\binom{104}{39} } &= &\displaystyle{\binom{103}{40}} \\ \\ \displaystyle{ \binom{714}{272} } &=& \displaystyle{\binom{713}{273}} \end{array}$

I’ll explain the Lind–Singmaster family later. But here’s the question I’m most interested in:

Is there any good explanation for the seven binomial coefficient coincidences?

Today my collaborator Paul Schwahn found a beautiful explanation of the first one, namely

$\displaystyle{ \binom{10}{3} = \binom{16}{2} }$

His explanation uses representation theory. The Lie algebra $\mathfrak{so}(10)$ has a 10-dimensional representation, the ‘vector’ representation $V_{10}$ , and also two 16-dimensional representations, the ‘left and right-handed spinor’ representations $S^\pm_{10}.$ There’s an isomorphism of representations

$\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }$

and similarly for $S^-_{10},$ but we might as well work with $S^+_{10}.$ Here $\Lambda^k$ means the kth exterior power. For any vector space $X$ we have

$\displaystyle{ \dim(\Lambda^k X) = \binom{\dim X}{k} }$

Thus, taking dimensions, the isomorphism of representations

$\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }$

instantly gives

$\displaystyle{ \binom{16}{2} = \binom{10}{3} }$

It is not super-easy to prove this isomorphism of representations, but it’s still nice to find a deeper layer of meaning underlying what might otherwise seem like a meaningless coincidence!

Can we find representation-theoretic explanations—or other explanations—for the other six coincidences?

I have not succeeded, but let me tell you about two failed tries.

We can look for isomorphisms like

$\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }$

involving representations of $\mathfrak{so}(n)$ for larger $n.$ In fact this isomorphism is part of a pattern! The next one involves the vector and left-handed spinor representations of $\mathfrak{so}(12).$ But it’s this:

$\displaystyle{ \Lambda^2 S^+_{12} \cong \Lambda^4 V_{12} \oplus \Lambda^0 V_{12} }$

so it gives

$\displaystyle{ \binom{32}{2} = \binom{12}{4} + 1 }$

$\displaystyle{ 496 = 495 + 1 }$

So we fail to get an equation between binomial coefficients: we’re off by one.

The second paper I cited, Binomial collisions and near collisions, presents a list of cases where two binomial coefficients differ by one. This is on the list. So we failed to explain an equation between binomial coefficients, but explained a near-miss.

Here’s another failed attempt at explaining an equation between binomial coefficients. The equation

$\displaystyle{ \binom{78}{2} = \binom{14}{6} = 3003 }$

is fascinating to anyone who knows their exceptional Lie groups. 78 is the dimension of $\mathrm{E}_6,$ while 14 is the dimension of $\mathrm{G}_2.$ $\mathrm{G}_2$ is a subgroup of $\mathrm{E}_6$ because $\mathrm{G}_2$ is the automorphism group of the octonions and $\mathrm{E}_6$ is the isometry group of the bioctonionic plane. We’d get the above equation if the 2nd exterior power of the adjoint representation of $\mathrm{E}_6,$ upon being restricted to $\mathrm{G}_2,$ were isomorphic to the 6th exterior power of the adjoint representation of $\mathrm{G}_2.$

Amazingly, it seems these two representations of $\mathrm{G}_2$ are not isomorphic even though their dimensions are the same: both 3003.

Even more amazingly, $\mathrm{E}_6$ and $\mathrm{G}_2$ both have irreducible representations of dimension 3003, but they are not the representations I just mentioned.

I would be happy for someone to check these two claims.

If anyone knows good explanations of the remaining six binomial coefficient coincidences, please let me know!

The Lind–Singmaster family

Lind and Singmaster were trying to find all $n,k$ with

$\displaystyle{ \binom{n}{k} = \binom{n-1}{k+1} }$

I’ll rapidly sketch the key steps of their argument. Simplifying the equation above we get

$n(k+1) = (n-k)(n-k-1)$

$n^2 - (3k+2)n + (k^2+k) = 0$

Solve for $n$ using the quadratic formula. This formula turns out to have

$\sqrt{5k^2 +8k+4}$

in it. So we need $5k^2 +8k+4$ to be a perfect square!

Now we’re trying to find integer solutions of

$5k^2+8k+4 = m^2$

A quadratic diophantine equation! Multiply by 5 and complete the square:

$5m^2 = (5k+4)^2 + 4$

$y = 5k+4$ is an integer when $k$ is, so we need to find integer solutions of

$y^2 - 5m^2 = -4$

This is a ‘Pell equation’, and people know how to solve these. In this particular case we get all the solutions from this fact:

$L_n^2 - 5F_n^2 = 4(-1)^n$

where $F_n$ are the Fibonacci numbers 0, 1, 1, 2, 3, … and $L_n$ are the Lucas numbers 2, 1, 3, 4, 7, …. These are two sequences satisfying the same famous recurrence relation, just with different initial conditions.

We want $n$ odd, to get

$L_n^2 - 5F_n^2 = -4$

It turns out $y = L_n, m = F_n$ with $n$ odd give all solutions of the Pell equation

$y^2 - 5m^2 = -4$

However, remember I said $y = 5k+4$ is an integer when $k$ is. But the converse isn’t always true, and we need $k$ to be an integer! This clearly happens iff $y \equiv 4 \bmod 5.$

So we need to know when $L_n \equiv 4 \bmod 5$ Apparently this happens iff $n \equiv 3 \bmod 4.$ I won’t think about this now… but this is the last hard step.

In summary, we’ve seen

$\displaystyle{ \binom{n}{k} = \binom{n-1}{k+1} }$

if and only if $y = 5k+4$ is a Lucas number $L_n$ with $n \equiv 3 \bmod 4.$ We could quit here, but people like to use the identity

$L_{4i+3} - 4 = 5F_{2i} F_{2i+3}$

to get a formula for $k$ in terms of Fibonacci numbers. This is gilding the lily, I’d say, but that eventually leads to the formula I showed you:

$\displaystyle{ \binom{F_{2i+2}F_{2i+3}}{\,F_{2i}F_{2i+3}\,} \;=\; \binom{F_{2i+2}F_{2i+3}-1}{\,F_{2i}F_{2i+3}+1\,}, \qquad i = 1,2,3,\dots }$

The takeaway message is: our problem can easily be reduced to a quadratic diophantine equation, then put in Pell form… and it’s known that the sequence of integer solutions of a Pell equation obeys a linear recurrence relation! We luck out in this case and get solutions connected to Lucas and Fibonacci numbers.

There may be a simpler argument, but this is what I’ve seen.

This entry was posted on Sunday, September 20th, 2026 at 10:29 pm and is filed under mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.