Numerical experiments suggest that $\binom{2m}{m + k}\cdot\frac{3m - 1 - 2k^2}{2m - 1}$ is integer for all $-m \le k\le +m$. It means that expression evaluation could be implemented very efficiently, only using integer addition and multiplication.

However, I've failed to derive computationally efficient expression so far. The two ideas I have are linear combination of binomials, and recursion -- but more insight is needed to go further.


1 Answer 1


It equals $$ \binom{2m}{m+k}\frac{3m-1-2k^2}{2m-1}=-(m-1)\binom{2m}{m+k}+4m\binom{2m-2}{m+k-1}. $$ I got it by expanding $3m-1-2k^2=2(m^2-k^2)-(2m^2-3m+1)=2(m-k)(m+k)-(2m-1)(m-1)$.

  • $\begingroup$ Awesome! Thank you Fedor for your lightning-fast and razor-sharp response. $\endgroup$ Jul 12, 2015 at 15:37
  • $\begingroup$ Wow... Stunning! $\endgroup$
    – MR_BD
    Aug 31, 2016 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.