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$\{f_i(x,y,z)\}$ is a sequence of polynomials with $f_0=1$. They satisfy $$f_{n+1}(x,y,z)=(x+z)(y+z)f_n(x,y,{z+1})-z^2f_n(x,y,z)$$ Pro. $f_i$s are symmetric.

Easy observation I made as following: $f_{n+1}(x,y,z)=(x+z)(y+z)( f_n(x,y,z+1)-f_n(x,y,z) ) + \sigma_2(x,y,z) f_n(x,y,z)$ It's sufficient to prove that $f_n(x,y,z+1)-f_n(x,y,z)$s are in the form $(x+y)g_n(x,y,z)$ with $g_n$s symmetric. Thus by induction I met difficulty with things like $g_n(x,y,z+1)$.

I don't know whether this way would work or not. I need your help, thx.

Kirby Lee
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  • Your idea might work. What you will probably need to do is calculate the first few $f_n$ by hand to see what's going on. It will probably suggest a more specific form for $g_n$ that will allow you to complete the induction. – Matt Samuel Aug 17 '16 at 00:33
  • @MattSamuel $f_n$s become harder to calculate with tedious terms as $n$ goes up. I checked $f_4$ on MMA, symmetric. But I got nothing more on structure of $g_n$s, which I think would show up adding some sym terms into them. Seems hard to deal... – Kirby Lee Aug 17 '16 at 03:53
  • Don't know if it will help, but for what you mention above it might help to write the first few $f_n$ in terms of elementary symm. polynomials or homogenous symm. polynomials to see if one can find any pattern. For the first few values of $n$ I find: $f_1 = e_2$, $f_2 = e_2^2 + e_2e_1-e_3$ and $f_3 = e_2^3 + (e_2e_1 - e_3)(1 + 2 e_1 + 3 e_2)$. Likewise for the homogenous polynomials: $f_1 = h_1^2 - h_2$ and $f_2 = (h_1^2 - h_2)^2 + ( h_2 h_1 - h_3)$ – Winther Aug 17 '16 at 04:54

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