Timeline for On the rank of a matrix $S$ with coefficients in $\mathbb F_{2^m}$

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Jan 20, 2014 at 23:20	comment	added	Sfarla		Ok I've just completed the proof using generating functions. It's quite easy to show the identity $\left(\sum_{n\geq 1}s_{2n-1}x^{2n-1}\right)^2\sum_{k\geq 0}\left(\sum_{i\geq 1}s_{2i}x^{2i}\right)^k=\sum_{p\geq 1}s_{2p}x^{2p}$. Thank you very much.
Jan 20, 2014 at 21:23	comment	added	Clément de Seguins Pazzis		The big sum over $A_p$ in my proof is the coefficient of index $2p$ in the formal power series $\Bigl(\sum_{n \geq 1} s_{2n-1}x^{2n-1}\Bigr)^2\sum_{k \geq 0} \Bigl(\sum_{i\geq 1} s_{2i} x^{2i}\Bigr)^k$.
Jan 20, 2014 at 18:10	comment	added	Sfarla		I tried to use generating function but the only thing that i found is that, namely $S(x)= \sum_{i=0}^n s_nx^n$, then $xS'(x)=S(x)+S^2(x)$ (where $S'(x)$ is the formal derivative of $S(x)$), but i don't know if it's useful. Could you explain better what are you trying to do and how did you obtain the identity above?
Jan 20, 2014 at 8:57	comment	added	Clément de Seguins Pazzis		Very nice remark!!
Jan 19, 2014 at 23:34	history	answered	Richard Stanley	CC BY-SA 3.0