# Sum of squares and partitions

This is an off-shot from my previous post on MO.

Given an integer partition $$\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$$ of $$n$$, denote $$\ell(\lambda)$$ to be the length of $$\lambda$$.

Let $$r_2(n)$$ denote the number of ways of expressing $$n$$ as a sum of two squares of integers, look it up on OEIS.

Here is perhaps a "new formulation" of $$r_2(n)$$:

QUESTION. Is this true? If it is, please either provide a reference or a proof. $$r_2(n) =\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$ where $$\lambda_{\ell(\lambda)+1}=0$$ and the product excludes $$\lambda_j=\lambda_{j+1}$$.

Example. Take $$n=4$$. The solutions to $$4=x^2+y^2$$ are $$(\pm2,0), (0,\pm2)$$ and hence $$r_2(4)=4$$. On the other hand, $$\lambda=(4,0), (3,1,0), (2,2,0), (2,1,1,0), (1,1,1,1,0)\vdash 4$$ so that $$(-1)^{4-4}4\cdot4+(-1)^{4-3}4\cdot2\cdot4\cdot1+(-1)^{4-2}4\cdot2 +(-1)^{4-2}4\cdot1\cdot4\cdot1+(-1)^{4-1}4\cdot1=4.$$

• How far did you test it? – Brendan McKay Feb 29 '20 at 3:12
• At least as far as $n=50$, but I can bet it is true. – T. Amdeberhan Feb 29 '20 at 3:16

Start by checking that the following formal product can be expanded as a sum over partitions $$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{\lambda_j-\lambda_{j+1}}\right)\left(\prod_{j\geq 1}x_j^{\lambda_j}\right)$$ with the convention that $$a_0=1$$. The proof of this is really just a weighted version of the usual generating function for partitions.
If we set $$x_1=t$$ and $$x_i=-t$$ for $$i\geq 2$$, and all $$a_r=4r$$ for $$r\geq 1$$ then the right side becomes $$\sum_{n\geq 0} t^n\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$ and the identity says that this is equal to the product $$\prod_{i\geq 1} \left(1+\frac{4(-1)^{i-1}t^i}{(1-(-1)^{i-1}t^i)^2}\right)=\frac{(1+t)^2}{(1-t)^2}\cdot \frac{(1-t^2)^2}{(1+t^2)^2}\cdot \frac{(1+t^3)^2}{(1-t^3)^2}\cdots$$ $$=\prod_{k\geq 1}\frac{(1-t^{2k})^{10}}{(1-t^k)^4(1-t^{4k})^4}.$$ This last final expression is a well known product formula for $$r_2(n)$$: $$\sum_nr_2(n)t^n=\frac{\eta(t^2)^{10}}{\eta(t)^4\,\eta(t^4)^4}, \qquad \text{where \eta(t)=t^{\frac1{24}}\prod_{k=1}^{\infty}(1-t^k) is the Dedekind eta function}.$$ So your identity follows.