Expansion of $(x+y)^n+(x+z)^n+(y+z)^n-x^n-y^n-z^n$ in terms of elementary symmetric polynomials

Question

Consider the symmetric polynomial in $3$ variables $$ f_n(x,y,z)=(x+y)^n + (x+z)^n+(y+z)^n - x^n-y^n-z^n $$ where $n\geq 0$ is an integer. I'm inquiring if there is a closed formula for the coefficients of $f_n$ in the basis of elementary symmetric polynomials? In other words, writing $$ f_n(x,y,z)=\sum_{\nu_1+2\nu_2+3\nu_3=n} c(\nu_1, \nu_2,\nu_3)e_1^{\nu_1}e_2^{\nu_2}e_3^{\nu_3} $$ with $e_1=x+y+z$, $e_2=xy+xz+yz$ and $e_3=xyz$, is there a formula for $c(\nu_1, \nu_2, \nu_3)$?

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Addendum: As per requested, here are the $f_n$'s for $n\leq 10$ $$ \begin{aligned} f_0 &= 0\\ f_1&=e_1\\ f_2&=e_1^2\\ f_3&=e_1^3-6e_3\\ f_4&=e_1^4-12e_{1}e_3\\ f_5&=e_1^5-20e_1^2e_3+10e_2e_3\\ f_6&=e_1^6-30e_1^3e_3+30e_1e_2e_3\\ f_7&=e_1^7 - 42 e_1^4 e_3 + 63 e_1^2 e_2 e_3 - 14 e_2^2 e_3 + 7 e_1 e_3^2\\ f_8&=e_1^8 - 56 e_1^5 e_3 + 112 e_1^3 e_2 e_3 - 56 e_1 e_2^2 e_3 + 28 e_1^2 e_3^2\\ f_9&=e_1^9 - 72 e_1^6 e_3 + 180 e_1^4 e_2 e_3 - 144 e_1^2 e_2^2 e_3 + 18 e_2^3 e_3 + 72 e_1^3 e_3^2 - 18 e_1 e_2 e_3^2 - 6 e_3^3\\ f_{10}&=e_1^{10} - 90 e_1^7 e_3 + 270 e_1^5 e_2 e_3 - 300 e_1^3 e_2^2 e_3 + 90 e_1 e_2^3 e_3 + 150 e_1^4 e_3^2 - 90 e_1^2 e_2 e_3^2 - 30 e_1 e_3^3 \end{aligned} $$

Answer 1

The key idea is that there is a simple generating function identity $$ \frac1{1-xt} = \sum_{n=0}^\infty x^n t^n \tag{1} $$ for a geometric series. Add two instances of this identity to get $$ \frac1{1-xt}+\frac1{1-yt} = \frac{2-(x+y)t}{(1-xt)(1-yt)} = \sum_{k=0}^\infty (x^n+y^n) t^n. \tag{2} $$ This implies that the sequence $\,a_n:=x^n+y^n\,$ has a recursion $$ a_n = (x+y)a_{n-1} - (xy)a_{n-2},\; \forall\, n\in\mathbb{Z} \tag{3} $$ where the coefficients are from the product expansion $$ (1-xt)(1-yt)=1-(x+y)t+(xy)t. \tag{4} $$ This generalizes to any finite sum of geometric series.

In your case, define $$ f_n(x,y,z):=(x\!+\!y)^n\!+\!(x\!+\!z)^n\!+\!(y\!+\!z)^n \!-\! x^n\!-\!y^n\!-\!z^n. \tag{5} $$ Use the same reasoning to get the generating function $$ \sum_{n=0}^\infty f_n(x,y,z)t^n = \frac{N_n(e_1,e_2,e_3)}{D_n(e_1,e_2,e_3)} \tag{6} $$ where the denominator is $$ 1\!-\!(3e_1)t\!+\!(3e_1^2\!+\!2e_2)t^2\!-\! (e_1^3\!+\!4e_1e_2)t^3\!+\!(2e_1^2e_2\!+\!e_2^2)t^4\!-\! (e_1e_2^2\!+\!e_1^2e_3)t^5\!+\!(e_1e_2e_3\!-\!e_3^2)t^6. \tag{7} $$ The coefficients of this polynomial give the coefficients of the linear recursion for $\,f_n(x,y,z).\,$ Of course, we also need the initial values of $\,f_0(x,y,z)\,$ up to $\,f_5(x,y,z)\,$ to start the recursion. These values now supplied in the question.

Note that I used the SymmetricReduction[] function of Mathematica to get the expression in terms of elementary symmetric functions.

Answer 2

In addition to @Somos' answer, is the following useful? $p\left( w \right) = \left( {w - x} \right)\left( {w - y} \right)\left( {w - z} \right) = {w^3} - {e_1}{w^2} + {e_2}w - {e_3}$

${g_n}\left( {x,y,z} \right) = {x^n} + {y^n} + {z^n} = \sum\limits_{{\nu _1} + 2{\nu _2} + 3{\nu _3} = n} {{a_n}\left( {{\nu _1},{\nu _2},{\nu _3}} \right)} \;e_1^{{\nu _1}}e_2^{{\nu _2}}e_3^{{\nu _3}}$

${h_n}\left( {x,y,z} \right) = {\left( {y + z} \right)^n} + {\left( {z + x} \right)^n} + {\left( {x + y} \right)^n} = \sum\limits_{{\nu _1} + 2{\nu _2} + 3{\nu _3} = n} {{b_n}\left( {{\nu _1},{\nu _2},{\nu _3}} \right)} \;e_1^{{\nu _1}}e_2^{{\nu _2}}e_3^{{\nu _3}}$

$\sum\limits_{n = 0}^\infty {{g_n}{t^n}} = \frac{1}{t}\frac{{d\log p}}{{dw}}\left( {\frac{1}{t}} \right)$

$\sum\limits_{n = 0}^\infty {{h_n}{t^n}} = - \frac{1}{t}\frac{{d\log p}}{{dw}}\left( {{e_1} - \frac{1}{t}} \right)$

So,

${c_n} = {b_n} - {a_n}$

Then

$${a_n} = {\left( { - 1} \right)^{{\nu _1} + {\nu _3}}}\left( {3\frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3}} \right)!}}{{{\nu _1}!{\nu _2}!{\nu _3}!}} + 2\frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3} - 1} \right)!}}{{\left( {{\nu _1} - 1} \right)!{\nu _2}!{\nu _3}!}} + \frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3} - 1} \right)!}}{{{\nu _1}!\left( {{\nu _2} - 1} \right)!{\nu _3}!}}} \right)$$ $$\begin{array}{c} {b_n} = {\left( { - 1} \right)^{{\nu _2} + {\nu _3}}}{2^{{\nu _1} - 2}}\left( {4\sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3}} \right) \wedge \left\lfloor {\frac{{{\nu _1}}}{2}} \right\rfloor } {{{\left( { - 1} \right)}^r}\left( {2{\nu _2} + 3{\nu _3} + 3r} \right)\frac{{\left( {{\nu _2} + {\nu _3} + r - 1} \right)!\left( {{\nu _1} + {\nu _3} - r} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r} \right)!{\nu _2}!{\nu _3}!r!}}} } \right.\\ - 8\sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3} - 1} \right) \wedge \left\lfloor {\frac{{{\nu _1} - 1}}{2}} \right\rfloor } {{{\left( { - 1} \right)}^r}\frac{{\left( {{\nu _2} + {\nu _3} + r} \right)!\left( {{\nu _1} + {\nu _3} - r - 1} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r - 1} \right)!{\nu _2}!{\nu _3}!r!}}} \\ \left. { + \sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3} - 2} \right) \wedge \left( {\left\lfloor {\frac{{{\nu _1}}}{2}} \right\rfloor - 1} \right)} {{{\left( { - 1} \right)}^r}\frac{{\left( {{\nu _2} + {\nu _3} + r} \right)!\left( {{\nu _1} + {\nu _3} - r - 2} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r - 2} \right)!{\nu _2}!{\nu _3}!r!}}} } \right) \end{array}$$