The number of symmetric polynomials of n degree

Question

How many symmetric polynomials of n degree with all their coefficients $\ =1 $ are there?Is there a type that computes their number?

Answer 1

The number of different symmetric monomials of degree $n$ in $k$ variables with all the coefficients 1 is equal to the number $p(n,k)$ of partitions of $n$ in up to $k$ parts, for every such partition say $a_1+a_2+\dots + a_k = n$ with $a_1 \ge a_2 \ge \dots \ge a_k \ge 0$ we have the symmetric "monomial": $$ x_1^{a_1} x_2^{a_2} \dots x_k^{a_k} + \dots $$

The number of different symmetric homogeneous polynomials of degree $n$ in $k$ variables with all the coefficients 1 is as a consequence $ 2^{p(n,k)}-1 $

The number of different symmetric polynomials of degree (exactly) $n$ in $k$ variables with all the coefficients 1 is then $$ (2^{p(n,k)}-1)2^{p(n-1,k)+p(n-2,k)+ \dots+ p(1,k)+p(0,k)} $$

So if you write $$P(n,k) = p(n,k) + p(n-1,k)+ \dots +p(0,k) $$ the number of partitions of numbers up to $n$ in at most $k$ parts then the number you ask for (if I have understood the question and I haven't made a silly mistake is) $$ 2^{P(n,k)} -2^{P(n-1,k)} $$ EDIT: This was wrong. Sorry.

EDIT 2: The answer above is answering the question if we consider the total degree of the symmetric functions, in view of the comments we are looking instead for polynomials with largest degree $n$ in any of the variables then we have the following:

1. The number of different symmetric monomials with degree up to $n$ in all the variables is $n^k$.
2. The total number of symmetric polynomials composed with this monomials is $2^{n^k}$.
3. So the number of symmetric polynomials having at least one term of degree $n$ in one of the variables is $$ 2^{n^k}-2^{(n-1)^k} $$