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Diffstat (limited to 'README.md')
-rw-r--r-- | README.md | 27 |
1 files changed, 18 insertions, 9 deletions
@@ -1,7 +1,9 @@ # 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 $${n\choose k}={m\choose l},$$ @@ -23,19 +25,22 @@ solutions with $l \leq 4$. </table> Also an infinite family of other solutions is known -$$ -{F_{2i+2} F_{2i+3}\choose F_{2i}F_{2i+3}} -= {F_{2i+2} F_{2i+3} -1 \choose F_{2i}F_{2i+3} +1} -$$ + +$${F_{2i+2} F_{2i+3}\choose F_{2i}F_{2i+3}} = {F_{2i+2} F_{2i+3} -1 \choose F_{2i}F_{2i+3} +1}$$ 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 ${n\choose k}<10^{42}$ are the ones found by de Weger: namely, the ones in the infinite family above, -$$ {15 \choose 5} = {14 \choose 6} = 3003 = {78 \choose 2}$$ -$$ {104 \choose 39} = {103 \choose 40} = 61218182743304701891431482520$$ + +$${15 \choose 5} = {14 \choose 6} = 3003 = {78 \choose 2}$$ + +$${104 \choose 39} = {103 \choose 40} = 61218182743304701891431482520$$ + And the “sporadic” solutions: + $$ {153 \choose 2} ={19\choose 5} = 11628$$ + $$ {221\choose 2}={17 \choose 8}=24310$$ This program searches up to ${n\choose k} <10^X$, when given the arguments `entry-limit X`. @@ -50,8 +55,12 @@ But still searching to $10^{42}$ with this method already requires 11 GB of memo Using this modular trick we can also search up to the 60,000th row (use arguments `row-limit 60000`), and (sadly) confirm that the only repeated entries are the ones listed above and the new entries in the infinite family, -$$ {713\choose 273} = {714\choose 272} \approx 3.5 \times 10^{204}$$ -$$ {4894\choose 1870} = {4895\choose 1869} \approx 4.6 \times 10^{1141}$$ -$$ {33551 \choose 12816} = {33552 \choose 12815} \approx 6.0 \times 10^{9687}$$ + +$${713\choose 273} = {714\choose 272} \approx 3.5 \times 10^{204}$$ + +$${4894\choose 1870} = {4895\choose 1869} \approx 4.6 \times 10^{1141}$$ + +$${33551 \choose 12816} = {33552 \choose 12815} \approx 6.0 \times 10^{9687}$$ + Again we run into a memory bottleneck — searching this far required 14 GB of memory. |