I found this formula. \begin{equation} \sum_{\lambda \vdash n}\frac{2^{l(\lambda)}}{z_{\lambda}}=n+1 \end{equation} where $\lambda \vdash n$ means that $\lambda$ is an integer partition of $n$, $l(\lambda)$ is the length of $\lambda$ and if $\lambda=(1^{m_{1}},2^{m_{n}},\cdots,n^{m_{n}})$, $z_{\lambda}=m_{1}!m_{2}!\cdots m_{n}! 1^{m_{1}}2^{m_{2}}\cdots n^{m_{n}}$.\
Has anyone found it before?
From symmetric function theory we know that $$ h_n(x)=\sum_{\lambda \vdash n} \dfrac{1}{z_\lambda}p_\lambda(x) $$ where $h_n(x)$ is the complete homogeneous symmetric function and $p_\lambda(x) : \lambda \vdash n$ are the power sum symmetric functions.
Now specialize $x$ to $x_1=1,x_2=1$ and $x_i = 0 $ for $i \geq 2$. Then the power sum symmetric function $p_k(x)=\sum_i x_i^k$ specializes to $2$ and hence $p_\lambda(x)=\prod_{i=1}^{l(\lambda)} p_{\lambda_i}(x)$ specializes to $2^{l(\lambda)}$.
On the other hand, $h_n(x)=\sum_{1 \leq i_1 \leq i_2 \leq ... \leq i_n} x_{i_1}...x_{i_k}$ specialize to $n+1$. This is because each term of the form $x_1^jx_2^{n-j}$ for $j=0,...,n$ contributes $1$ and the rest of the terms contribute $0$.