Let $\lambda\vdash n$ denote the integer partition of $n$. Define the product $\mathcal{N}(\lambda)=\lambda_1\lambda_2\cdots\lambda_r$ when $\lambda=(\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_r>0)$.
Let $\gamma$ be the Euler's constant. Lehmer proved that
$$\lim_{n\rightarrow\infty}\frac1n\sum_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)}=e^{-\gamma}.$$
I like to ask:
QUESTION. Does the following limit exist? Or, what is the asymptotic growth of the sum?
$$\lim_{n\rightarrow\infty}\sum_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}.$$
ADDED. Glad to see that Fedor Petrov's answer would (potentially) work for
$$\lim_{n\rightarrow\infty}\sum_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^r}.$$
Best Answer
The limit equals 2. We have $$ \sum_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}=[x^n]\prod_{k=1}^\infty\frac1{1-x^k/k^2}=\sum_{m=0}^n[x^m]\prod_{k=2}^\infty\frac1{1-x^k/k^2}\\ =b_0+b_1+\ldots+b_n,$$ where $$ \sum b_i x^i=\prod_{k=2}^\infty \frac1{1-x^k/k^2}=:f(x), $$ the standard uniform convergence argument shows that $$\sum b_i=\sup_{0<x<1} f(x)=f(1)=\prod_{k=2}^\infty\frac{k^2}{k^2-1}=2.$$