Are there surprisingly identical binomial coefficients? The 2019 Stack Overflow Developer...

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Are there surprisingly identical binomial coefficients?



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9












$begingroup$


Suppose $binom{n}{k}=binom{n'}{k'}$ with $k geq 2$, $k' geq 2$, $n geq 2k$ and $n' geq 2k'$. Does it follow that $n=n'$ and $k=k'$?



EDIT: Yup, $binom{16}{2}=binom{10}{3}=120$.



Now I want to ask if there are infinitely many such pairs, but I should probably ask that in a separate question. Thanks!



EDIT 2: for future reference, yes there are infinitely many such coincidences. See Singmaster's Conjecture.






number-theory binomial-coefficients







share|cite|improve this question











$endgroup$












  • $begingroup$
    Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
    $endgroup$
    – Arthur
    Nov 25 '11 at 7:01








  • 3




    $begingroup$
    A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
    $endgroup$
    – KCd
    Nov 25 '11 at 7:29
















9












$begingroup$


Suppose $binom{n}{k}=binom{n'}{k'}$ with $k geq 2$, $k' geq 2$, $n geq 2k$ and $n' geq 2k'$. Does it follow that $n=n'$ and $k=k'$?



EDIT: Yup, $binom{16}{2}=binom{10}{3}=120$.



Now I want to ask if there are infinitely many such pairs, but I should probably ask that in a separate question. Thanks!



EDIT 2: for future reference, yes there are infinitely many such coincidences. See Singmaster's Conjecture.






number-theory binomial-coefficients







share|cite|improve this question











$endgroup$












  • $begingroup$
    Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
    $endgroup$
    – Arthur
    Nov 25 '11 at 7:01








  • 3




    $begingroup$
    A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
    $endgroup$
    – KCd
    Nov 25 '11 at 7:29














9












9








9


3



$begingroup$


Suppose $binom{n}{k}=binom{n'}{k'}$ with $k geq 2$, $k' geq 2$, $n geq 2k$ and $n' geq 2k'$. Does it follow that $n=n'$ and $k=k'$?



EDIT: Yup, $binom{16}{2}=binom{10}{3}=120$.



Now I want to ask if there are infinitely many such pairs, but I should probably ask that in a separate question. Thanks!



EDIT 2: for future reference, yes there are infinitely many such coincidences. See Singmaster's Conjecture.






number-theory binomial-coefficients







share|cite|improve this question











$endgroup$




Suppose $binom{n}{k}=binom{n'}{k'}$ with $k geq 2$, $k' geq 2$, $n geq 2k$ and $n' geq 2k'$. Does it follow that $n=n'$ and $k=k'$?



EDIT: Yup, $binom{16}{2}=binom{10}{3}=120$.



Now I want to ask if there are infinitely many such pairs, but I should probably ask that in a separate question. Thanks!



EDIT 2: for future reference, yes there are infinitely many such coincidences. See Singmaster's Conjecture.






number-theory binomial-coefficients




number-theory binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '11 at 7:17







Alon Amit

















asked Nov 25 '11 at 6:43









Alon AmitAlon Amit

10.7k3768




10.7k3768












  • $begingroup$
    Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
    $endgroup$
    – Arthur
    Nov 25 '11 at 7:01








  • 3




    $begingroup$
    A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
    $endgroup$
    – KCd
    Nov 25 '11 at 7:29


















  • $begingroup$
    Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
    $endgroup$
    – Arthur
    Nov 25 '11 at 7:01








  • 3




    $begingroup$
    A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
    $endgroup$
    – KCd
    Nov 25 '11 at 7:29
















$begingroup$
Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
$endgroup$
– Arthur
Nov 25 '11 at 7:01






$begingroup$
Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few.
$endgroup$
– Arthur
Nov 25 '11 at 7:01






3




3




$begingroup$
A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
$endgroup$
– KCd
Nov 25 '11 at 7:29




$begingroup$
A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle
$endgroup$
– KCd
Nov 25 '11 at 7:29










5 Answers
5






active

oldest

votes


















3












$begingroup$

There are one or two (nontrivial) choices of $(m,n)$ where $binom{n}{2} = binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.



Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.






share|cite|improve this answer











$endgroup$





















    7












    $begingroup$

    $$
    binom{3003}{1} = binom{78}{2} = binom{15}{5} = binom{14}{6}.
    $$



    Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.






    share|cite|improve this answer









    $endgroup$





















      6












      $begingroup$

      Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.



      It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M lt infty$ such that $f(t) leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O left(frac{(log t) cdot (log log log t) }{(log log t)^3} right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.



      Reference: This MO post by Michael Hardy asks for recent progress on this problem.






      share|cite|improve this answer











      $endgroup$





















        3












        $begingroup$

        10-choose-4 = 21-choose-2.${}{}{}{}$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
          $endgroup$
          – Sasha
          Nov 25 '11 at 6:59






        • 1




          $begingroup$
          Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
          $endgroup$
          – Alon Amit
          Nov 25 '11 at 7:01










        • $begingroup$
          ${10choose 4} = {21 choose 2}$
          $endgroup$
          – Chris Taylor
          Nov 25 '11 at 7:02






        • 1




          $begingroup$
          Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
          $endgroup$
          – Gerry Myerson
          Nov 25 '11 at 10:13










        • $begingroup$
          @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
          $endgroup$
          – Srivatsan
          Nov 25 '11 at 13:42





















        2












        $begingroup$

        $$binom{10}{3}=binom{16}{2}.$$






        share|cite|improve this answer









        $endgroup$














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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          There are one or two (nontrivial) choices of $(m,n)$ where $binom{n}{2} = binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.



          Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.






          share|cite|improve this answer











          $endgroup$


















            3












            $begingroup$

            There are one or two (nontrivial) choices of $(m,n)$ where $binom{n}{2} = binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.



            Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.






            share|cite|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              There are one or two (nontrivial) choices of $(m,n)$ where $binom{n}{2} = binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.



              Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.






              share|cite|improve this answer











              $endgroup$



              There are one or two (nontrivial) choices of $(m,n)$ where $binom{n}{2} = binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.



              Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 25 '11 at 7:06

























              answered Nov 25 '11 at 6:51









              zyxzyx

              31.7k33699




              31.7k33699























                  7












                  $begingroup$

                  $$
                  binom{3003}{1} = binom{78}{2} = binom{15}{5} = binom{14}{6}.
                  $$



                  Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.






                  share|cite|improve this answer









                  $endgroup$


















                    7












                    $begingroup$

                    $$
                    binom{3003}{1} = binom{78}{2} = binom{15}{5} = binom{14}{6}.
                    $$



                    Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.






                    share|cite|improve this answer









                    $endgroup$
















                      7












                      7








                      7





                      $begingroup$

                      $$
                      binom{3003}{1} = binom{78}{2} = binom{15}{5} = binom{14}{6}.
                      $$



                      Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.






                      share|cite|improve this answer









                      $endgroup$



                      $$
                      binom{3003}{1} = binom{78}{2} = binom{15}{5} = binom{14}{6}.
                      $$



                      Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 25 '11 at 16:08









                      Michael HardyMichael Hardy

                      1




                      1























                          6












                          $begingroup$

                          Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.



                          It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M lt infty$ such that $f(t) leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O left(frac{(log t) cdot (log log log t) }{(log log t)^3} right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.



                          Reference: This MO post by Michael Hardy asks for recent progress on this problem.






                          share|cite|improve this answer











                          $endgroup$


















                            6












                            $begingroup$

                            Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.



                            It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M lt infty$ such that $f(t) leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O left(frac{(log t) cdot (log log log t) }{(log log t)^3} right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.



                            Reference: This MO post by Michael Hardy asks for recent progress on this problem.






                            share|cite|improve this answer











                            $endgroup$
















                              6












                              6








                              6





                              $begingroup$

                              Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.



                              It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M lt infty$ such that $f(t) leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O left(frac{(log t) cdot (log log log t) }{(log log t)^3} right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.



                              Reference: This MO post by Michael Hardy asks for recent progress on this problem.






                              share|cite|improve this answer











                              $endgroup$



                              Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.



                              It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M lt infty$ such that $f(t) leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O left(frac{(log t) cdot (log log log t) }{(log log t)^3} right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.



                              Reference: This MO post by Michael Hardy asks for recent progress on this problem.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Apr 13 '17 at 12:58









                              Community♦

                              1




                              1










                              answered Nov 25 '11 at 6:58









                              SrivatsanSrivatsan

                              21.1k371126




                              21.1k371126























                                  3












                                  $begingroup$

                                  10-choose-4 = 21-choose-2.${}{}{}{}$






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                    $endgroup$
                                    – Sasha
                                    Nov 25 '11 at 6:59






                                  • 1




                                    $begingroup$
                                    Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                    $endgroup$
                                    – Alon Amit
                                    Nov 25 '11 at 7:01










                                  • $begingroup$
                                    ${10choose 4} = {21 choose 2}$
                                    $endgroup$
                                    – Chris Taylor
                                    Nov 25 '11 at 7:02






                                  • 1




                                    $begingroup$
                                    Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                    $endgroup$
                                    – Gerry Myerson
                                    Nov 25 '11 at 10:13










                                  • $begingroup$
                                    @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                    $endgroup$
                                    – Srivatsan
                                    Nov 25 '11 at 13:42


















                                  3












                                  $begingroup$

                                  10-choose-4 = 21-choose-2.${}{}{}{}$






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                    $endgroup$
                                    – Sasha
                                    Nov 25 '11 at 6:59






                                  • 1




                                    $begingroup$
                                    Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                    $endgroup$
                                    – Alon Amit
                                    Nov 25 '11 at 7:01










                                  • $begingroup$
                                    ${10choose 4} = {21 choose 2}$
                                    $endgroup$
                                    – Chris Taylor
                                    Nov 25 '11 at 7:02






                                  • 1




                                    $begingroup$
                                    Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                    $endgroup$
                                    – Gerry Myerson
                                    Nov 25 '11 at 10:13










                                  • $begingroup$
                                    @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                    $endgroup$
                                    – Srivatsan
                                    Nov 25 '11 at 13:42
















                                  3












                                  3








                                  3





                                  $begingroup$

                                  10-choose-4 = 21-choose-2.${}{}{}{}$






                                  share|cite|improve this answer











                                  $endgroup$



                                  10-choose-4 = 21-choose-2.${}{}{}{}$







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Mar 21 at 21:01

























                                  answered Nov 25 '11 at 6:52









                                  Gerry MyersonGerry Myerson

                                  148k8152306




                                  148k8152306












                                  • $begingroup$
                                    Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                    $endgroup$
                                    – Sasha
                                    Nov 25 '11 at 6:59






                                  • 1




                                    $begingroup$
                                    Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                    $endgroup$
                                    – Alon Amit
                                    Nov 25 '11 at 7:01










                                  • $begingroup$
                                    ${10choose 4} = {21 choose 2}$
                                    $endgroup$
                                    – Chris Taylor
                                    Nov 25 '11 at 7:02






                                  • 1




                                    $begingroup$
                                    Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                    $endgroup$
                                    – Gerry Myerson
                                    Nov 25 '11 at 10:13










                                  • $begingroup$
                                    @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                    $endgroup$
                                    – Srivatsan
                                    Nov 25 '11 at 13:42




















                                  • $begingroup$
                                    Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                    $endgroup$
                                    – Sasha
                                    Nov 25 '11 at 6:59






                                  • 1




                                    $begingroup$
                                    Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                    $endgroup$
                                    – Alon Amit
                                    Nov 25 '11 at 7:01










                                  • $begingroup$
                                    ${10choose 4} = {21 choose 2}$
                                    $endgroup$
                                    – Chris Taylor
                                    Nov 25 '11 at 7:02






                                  • 1




                                    $begingroup$
                                    Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                    $endgroup$
                                    – Gerry Myerson
                                    Nov 25 '11 at 10:13










                                  • $begingroup$
                                    @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                    $endgroup$
                                    – Srivatsan
                                    Nov 25 '11 at 13:42


















                                  $begingroup$
                                  Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                  $endgroup$
                                  – Sasha
                                  Nov 25 '11 at 6:59




                                  $begingroup$
                                  Did you mean to write $binom{10}{4} = binom{21}{2}$ ?
                                  $endgroup$
                                  – Sasha
                                  Nov 25 '11 at 6:59




                                  1




                                  1




                                  $begingroup$
                                  Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                  $endgroup$
                                  – Alon Amit
                                  Nov 25 '11 at 7:01




                                  $begingroup$
                                  Trying to guess what you meant, since it obviously isn't $binom{10}{2}=binom{21}{4}$...
                                  $endgroup$
                                  – Alon Amit
                                  Nov 25 '11 at 7:01












                                  $begingroup$
                                  ${10choose 4} = {21 choose 2}$
                                  $endgroup$
                                  – Chris Taylor
                                  Nov 25 '11 at 7:02




                                  $begingroup$
                                  ${10choose 4} = {21 choose 2}$
                                  $endgroup$
                                  – Chris Taylor
                                  Nov 25 '11 at 7:02




                                  1




                                  1




                                  $begingroup$
                                  Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                  $endgroup$
                                  – Gerry Myerson
                                  Nov 25 '11 at 10:13




                                  $begingroup$
                                  Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant.
                                  $endgroup$
                                  – Gerry Myerson
                                  Nov 25 '11 at 10:13












                                  $begingroup$
                                  @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                  $endgroup$
                                  – Srivatsan
                                  Nov 25 '11 at 13:42






                                  $begingroup$
                                  @Gerry I took the liberty to correct the typo. Hope it's ok. Regards,
                                  $endgroup$
                                  – Srivatsan
                                  Nov 25 '11 at 13:42













                                  2












                                  $begingroup$

                                  $$binom{10}{3}=binom{16}{2}.$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    $$binom{10}{3}=binom{16}{2}.$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      $$binom{10}{3}=binom{16}{2}.$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $$binom{10}{3}=binom{16}{2}.$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 25 '11 at 6:56









                                      André NicolasAndré Nicolas

                                      455k36432821




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