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+# Repeated numbers in Pascal's Triangle
+
+Every number ${n \choose k}, 1<k<n-1$ appears at least four times in Pascal's triangle:
+$${n\choose k} = {n\choose n-k} = {{n\choose k}\choose 1}={{n\choose k}\choose {n\choose k}}$$
+This gives rise to the natural question of which numbers appear more than four times. Or, taking
+out the symmetry and trivial case of ${n\choose 1}$, what are the solutions to
+$${n\choose k}={m\choose l},$$
+for $1<k\leq \frac n2$, $k<l\leq \frac m2$.
+
+According to ["A Note on the Diophantine equation (x 4)=(y 2)"](https://www.researchgate.net/publication/235418296_A_Note_on_the_Diophantine_equation_x_4y_2)
+(Ákos Pintér, 1995),
+the cases of $(k,l)=(2,3)$ and $(k,l)=(2,4)$ have been solved completely,
+and from ["Equal binomial coefficients: some elementary considerations"](https://repub.eur.nl/pub/1356/1356_ps.pdf)
+(Benjamin M.M. de Weger, 1995), we also know that there are no solutions with $(k,l)=(3,4)$; so here is the full list of
+solutions with $l \leq 4$.
+
+<table>
+<tr><th>k</th><th>l</th><th>n</th><th>m</th><th>n choose k = m choose l</th></tr>
+<tr><td>2</td><td>3</td><td>16</td><td>10</td><td>120</td></tr>
+<tr><td>2</td><td>3</td><td>56</td><td>22</td><td>1540</td></tr>
+<tr><td>2</td><td>3</td><td>120</td><td>36</td><td>7140</td></tr>
+<tr><td>2</td><td>4</td><td>21</td><td>10</td><td>210</td></tr>
+</table>
+
+Also an infinite family of other solutions is known
+$$
+{F_{2i+2} F_{2i+3}\choose F_{2i}F_{2i+3}}
+= {F_{2i+2} F_{2i+3} -1 \choose F_{2i}F_{2i+3} +1}
+$$
+
+De Weger did a computer search up to ${n\choose k} < 10^{30}$ in 1995. Now that we have
+faster computers we can go higher — and we can say for certain that the only solutions
+up to ${n\choose k}<10^{41}$ are the ones found by de Weger: namely, the ones in the infinite
+family above,
+$$ {15 \choose 5} = {14 \choose 6} = 3003 = {78 \choose 2}$$
+$$ {104 \choose 39} = {103 \choose 40} = 61218182743304701891431482520$$
+And the “sporadic” solutions:
+$$ {153 \choose 2} ={19\choose 5} = 11628$$
+$$ {221\choose 2}={17 \choose 8}=24310$$
+
+This program searches up to $10^X$, when given the argument $X$.
+The search works by putting all ${n\choose k}<10^X$ with $5\leq k\leq \frac n 2$ into
+an array and sorting it. Then we check for adjacent elements which are equal,
+and use a binary search to check whether elements in the array are ${n\choose 2},{n\choose 3},{n\choose 4}$.
+This finds all solutions with $l>4$ (since the solutions with $l\leq 4$ are known).
+
+Unfortunately searching to $10^{41}$ with this method already requires 13 GB of memory.