Expressing power sum symmetric polynomials in terms of lower degree power sums

Combining the trace formula proposed by Gjergji Zaimi and Qiaochu Yuan,

$$p_k={rm Tr}begin{pmatrix} e_1 & 1 & cdots & 0 \ -e_2 & 0 & ddots & vdots \ vdots & vdots & ddots & 1 \ (-1)^{N-1}e_N & 0 & cdots & 0 end{pmatrix}^{k},$$
with the formula quoted by Peter Erskin,
$$e_n=frac1{n!} begin{vmatrix}p_1 & 1 & 0 & cdots\ p_2 & p_1 & 2 & 0 & cdots \ vdots&& ddots & ddots \ p_{n-1} & p_{n-2} & cdots & p_1 & n-1 \ p_n & p_{n-1} & cdots & p_2 & p_1 end{vmatrix},$$
Mathematica produces the following expansions of $p_k$:

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$$N=2$$

$$p_3=-frac{1}{2} p_1^3+frac{3}{2} p_1p_2$$

$$p_4=-frac{1}{2} p_1^4+p_1^2p_2+frac{1}{2} p_2^2$$

$$p_5=-frac{1}{4} p_1^5+frac{5}{4} p_1p_2 ^2$$

$$p_6=-frac{3}{4} p_1^4p_2+frac{3}{2} p_1^2p_2^2+frac{1}{4} p_2^3$$

$$p_7=frac{1}{8} p_1^7-frac{7}{8} p_1^5p_2+frac{7}{8} p_1^3p_2^2+frac{7}{8} p_1p_2^3$$

$$p_8=frac{1}{8} p_1^8-frac{1}{2} p_1^6p_2-frac{1}{4} p_1^4p_2^2+frac{3}{2} p_1^2p_2^3+frac{1}{8} p_2^4$$

$$p_9=frac{1}{16} p_1^9-frac{9}{8} p_1^5p_2^2+frac{3}{2} p_1^3p_2^3 +frac{9}{16} p_1p_2^4$$

$$p_{10}=frac{5}{16} p_1^8p_2-frac{5}{4} p_1^6p_2^2+frac{5}{8} p_1^4p_2^3 +frac{5}{4} p_1^2p_2^4+frac{1}{16} p_2^5$$

$$p_{11}=-frac{1}{32} p_1^{11}+frac{11}{32} p_1^9p_2-frac{11}{16} p_1^7p_2^2-frac{11}{16} p_1^5p_2^3 +frac{55}{32} p_1^3p_2^4+frac{11}{32} p_1p_2^5$$

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$$N=3$$

$$p_4=frac{1}{6} p_1^4-p_1^2p_2+frac{1}{2} p_2^2+ frac{4}{3} p_1p_3$$

$$p_5=frac{1}{6} p_1^5-frac{5}{6} p_1^3p_2+frac{5}{6} p_1^2p_3+frac{5}{6} p_2p_3$$

$$p_6=frac{1}{12} p_1^6-frac{1}{4} p_1^4p_2-frac{3}{4} p_1^2p_2^2+frac{1}{4} p_2^3+frac{1}{3} p_1^3p_2^3+p_1 p_2 p_3 +frac{1}{3} p_3^2$$

$$p_7=frac{1}{36} p_1^7-frac{7}{12} p_1^3p_2^2+frac{7}{36} p_1^4p_3+frac{7}{12} p_2^2p_3+frac{7}{9} p_1p_3^2$$

$$p_8=frac{1}{72} p_1^8-frac{1}{18} p_1^6p_2+frac{1}{12} p_1^4p_2^2-frac{1}{2} p_1^2p_2^3+frac{1}{8} p_2^4+frac{2}{9} p_1^5p_3$$
$$-frac{8}{9} p_1^3p_2p_3+frac{2}{3} p_1p_2^2p_3+frac{8}{9} p_1^2p_3^2+frac{4}{9} p_2p_3^2$$

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$$N=4$$

$$p_5=-frac{1}{24} p_1^5+frac{5}{12} p_1^3p_2-frac{5}{8} p_1p_2^2-frac{5}{6} p_1^2p_3+frac{5}{6} p_2p_3+frac{5}{4} p_1p_4$$

$$p_6=-frac{1}{24} p_1^6+frac{3}{8} p_1^4p_2-frac{3}{8} p_1^2p_2^2-frac{1}{8} p_2^3-frac{2}{3} p_1^3p_3+frac{1}{3} p_3^2+frac{3}{4} p_1^2p_4+frac{3}{4} p_2p_4$$

$$p_7=-frac{1}{48} p_1^7+frac{7}{48} p_1^5p_2+frac{7}{48} p_1^3p_2^2-frac{7}{16} p_1p_2^2-frac{7}{24} p_1^4p_3-frac{7}{12} p_1^2p_2p_3$$
$$+frac{7}{24} p_2^2p_3 +frac{7}{24} p_1^3p_4+frac{7}{8} p_1p_2p_4+frac{7}{12} p_3p_4$$

$$p_8=-frac{1}{144} p_1^8+frac{1}{36} p_1^6p_2+frac{5}{24} p_1^4p_2^2 -frac{1}{4} p_1^2p_2^2-frac{1}{16} p_2^4$$
$$-frac{1}{9} p_1^5p_3-frac{2}{9} p_1^3p_2p_3 -frac{1}{3} p_1p_2^2p_3-frac{4}{9} p_1^2p_3^2+frac{4}{9} p_2p_3^2$$
$$+frac{1}{12} p_1^4p_4+frac{1}{2} p_1^2p_2p_4+frac{1}{4} p_2^2p_4 +frac{2}{3} p_1p_3p_4+frac{1}{4} p_4^2.$$

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It seems to me that a nice and compact formula for $a_{k,rho}$ does exist. Indeed,
the coefficients in the above examples are extremely simple.

In particular, I observe that the last terms in each of $p_k$ for $N=8$
have the form

$$k prod_j frac{1}{j^{r_j}r_j!}p_j^{r_j},$$
which corresponds to
$$a_{k,rho}=kprod_j frac{1}{j^{r_j}r_j!}.$$
This formula (whose structure resembles the coefficients in the expansion of Schur functions quoted by Peter Erskin) also works for all terms of the type $p_jp_{k-j}$ at arbitrary $N$.

Apparently, this is not a general formula, as can be seen from the coefficients in front
of $p_1^k$ which do depend on $N$.
I believe, however, that the general formula for $a_{k,rho}$ with $N$ properly included should not be much more complex than the empirical one above.

Hope this helps.

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