Monday, 15 August 2016

combinatorics - A combinatorial identity with binomial coefficients and floor function.



Show that 2k+1m=02m(nm)(nm2k+1m2)=(2n+12k+1)
I tried expending the sum and using induction, but could not complete the induction step; I tried using Pascal's identity to obtain (nm2k+3m2)=(nm+12k+3m2)(nm2k+1m2) but couldn't find other identities to complete the induction step. Searching Gould's Combinatorial Identities brought up nothing, though I could easily had missed something useful there.




I'm looking to complete the induction step, but would also like to find an algebraic or a combinatorial proof. I would like to avoid using trigonometric and root functions, but would like to use complex numbers in cartesian form.


Answer



Note: This answer is the result of an analysis of the highly instructive and elegant answer by Marko Riedel. One highlight is the useful representation of (nq2) which is the introductory part of this answer.



We use the coefficient of operator [zq] to denote the coefficient of zq in a series. This way we can write e.g.
(nq)=[zq](1+z)n
and to ease comparison with Markos answer I will use his notation.





Preliminary:



The following is valid
(nq2)=[zqwn](1+z)(1+w)n1wz2



With q even, q2q we obtain
[z2qwn](1+z)(1+w)n1wz2=[wn](1+w)n[z2q](1+z)j=0wjz2j=[wn](1+w)nwq=[wnq](1+w)n=(nnq)=(nq)
With q odd, q2q+1 we obtain
[z2q+1wn](1+z)(1+w)n1wz2=[wn](1+w)n[z2q+1](1+z)j=0wjz2j=[wn](1+w)nwq=[wnq](1+w)n=(nnq)=(nq)




Comment:




  • Additionally to (1) we use a second variable z which is used as marker to select via 1+z even and odd exponent of wj.


  • In (3) we use the geometric series expansion and the linearity of the coefficient of operator.


  • In (4) we select the coefficient of z2q which is wq.



  • In (5) we apply the rule [zp]zqA(z)=[zpq]A(z).





We now apply (2) to OPs formula and obtain
2m+1k=0(nk)2k(nk2m+1k2)=nk=0(nk)2k[z2m+1kwnk](1+z)(1+w)nk1wz2=[z2m+1](1+z)[wn](1+w)n1wz2nk=0(nk)(2wz1+w)k=[z2m+1](1+z)[wn](1+w)n1wz2(1+2wz1+w)n=[z2m+1](1+z)[wn](1+w(1+2z))n1wz2=[z2m+1](1+z)[wn]nq=0(nq)wq(1+2z)qj=0wjz2j=[z2m+1](1+z)nq=0(nq)[wnq](1+2z)qj=0wjz2j=[z2m+1](1+z)nq=0(nq)(1+2z)q(z2)nq=[z2m+1](1+z)(1+2z+z2)n=[z2m+1](1+z)2n+1=(2n+12m+1)



and the claim follows.





Comment:




  • In (6) we apply the formula (2) and we change the upper limit of the sum to n without changing anything, since the formula (2) selects the proper range.


  • In (7) we collect all factors with exponent k.


  • In (8) observe the factor (1+w)n cancels nicely.



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