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7 added 6 characters in body

Is there an explicit formula expressing the power sum symmetric polynomials $$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ of degree $k$ in $N < k$ variables entirely through the power sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $j \le N$?

Examples: $$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$

$$N=2,\ k=3: \quad p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$

What is the general formula?

I am looking for a formula similar to that for the expansion of the Schur functions $s_\lambda$ in terms of the symmetric power sums:

$$s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$ where the coefficients $\chi^\lambda_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)$, which contains $r_j$ parts of length $j$.

Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner: $$p_k=\sum_{\rho}a_{k;\rho}\prod_{j=1}^N p_j^{r_j},$$ where $\rho=(1^{r_1},2^{r_2},\dots,N^{r_N})$ is the partition of $k$ such that $k=r_1+2r_2+3r_3+...+Nr_N$.

In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and $a_{3;\ (1^{3},2^{0})}=-1/2$.

My question can be thus reformulated as follows: given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?

Comment

Actually, Wikipedia tells us how to construct a certain explicit formula for $p_k$. It gives the following expressions for $p_n$ with $n=N$ in terms of $e_j,$

$$p_n = \begin{vmatrix} e_1 & 1 & 0 & \cdots & \\ 2e_2 & e_1 & 1 & 0 & \cdots & \\ 3e_3 & e_2 & e_1 & 1 & \cdots & \\ \vdots &&& \ddots & \ddots & \\ ne_n & e_{n-1} & \cdots & & e_1 & \end{vmatrix},$$

and for $e_n$ with $n=N$ in terms of $p_j,$

$$e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}.$$

As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p_n$ with $n > N$ and for $e_n$ with $n < N$.

For $p_n$ with $n>N$ one should take into account that all $e_k=0$ for $k > N$, so that the resulting matrix has zeros in both right-upper and left-lower corners.

Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the explicit formula which seems to solve the problem.

However, I still don't know how to obtain the coefficients $a_{k;\rho}$ in the expansion of $p_k$ in terms of the first $N$ power sums which would be the desired (really explicit) formula.

6 improved formatting

Is there an explicit formula expressing the power sum symmetric polynomials $$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ of degree $k$ in $N < k$ variables entirely through the power sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $j \le N$?

Examples: $$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$

$$N=2,\ k=3: \quad p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$

What is the general formula?

I am looking for a formula similar to that for the expansion of the Schur functions $s_\lambda$ in terms of the symmetric power sums:

$$s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$ where the coefficients $\chi^\lambda_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)$, which contains $r_j$ parts of length $j$.

Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner: $$p_k=\sum_{\rho}a_{k;\rho}\prod_{j=1}^N p_j^{r_j},$$ where $\rho=(1^{r_1},2^{r_2},\dots,N^{r_N})$ is the partition of $k$ such that $k=r_1+2r_2+3r_3+...+Nr_N$.

My question can be thus reformulated as follows:

given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?

In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and $a_{3;\ (1^{3},2^{0})}=-1/2$.

My question can be thus reformulated as follows: given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?

Comment

Actually, Wikipedia tells us how to construct a certain explicit formula for $p_k$. It gives the following expressions for $p_n$ with $n=N$ in terms of $e_j,$

$$p_n = \begin{vmatrix} e_1 & 1 & 0 & \cdots & \\ 2e_2 & e_1 & 1 & 0 & \cdots & \\ 3e_3 & e_2 & e_1 & 1 & \cdots & \\ \vdots &&& \ddots & \ddots & \\ ne_n & e_{n-1} & \cdots & & e_1 & \end{vmatrix},$$

and for $e_n$ with $n=N$ in terms of $p_j,$

$$e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}.$$

As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p_n$ with $n > N$ and for $e_n$ with $n < N$.

For $p_n$ with $n>N$ one should take into account that all $e_k=0$ for $k > N$, so that the resulting matrix has zeros in both right-upper and left-lower corners.

Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the explicit formula which seems to solve the problem.

However, I still don't know how to obtain the coefficients $a_{k;\rho}$ in the expansion of $p_k$ in terms of the first $N$ power sums which would be the desired (really explicit) formula.