# Is the permanent of the matrix $[(\frac{i+j}{2n+1})]_{0\le i,j\le n}$ always positive?

Recall that the permanent of an $$n\times n$$ matrix $$A=[a_{i,j}]_{1\le i,j\le n}$$ is defined by $$\operatorname{per}A=\sum_{\sigma\in S_n}\prod_{i=1}^n a_{i,\sigma(i)}.$$

In 2004, R. Chapman [Acta Arith. 115(2004), 231-244] determined the value of $$\det\left[\left(\frac{i+j-1}p\right)\right]_{1\le i,j\le(p+1)/2}=\det\left[\left(\frac{i+j}p\right)\right]_{0\le i,j\le(p-1)/2},$$ where $$(\frac{\cdot}p)$$ is the Legendre symbol. Motivated by this, here I pose the following conjecture involving permanents.

Conjecture. For each $$n=0,1,2,\ldots$$, we have $$\operatorname{per}\left[\left(\frac{i+j}{2n+1}\right)\right]_{0\le i,j\le n}>0,\tag{*}$$ where $$(\frac{\cdot}{2n+1})$$ is the Jacobi symbol.

Let $$a_n$$ denote the permanent in $$(*)$$. Via Mathematica I find that $$\begin{gather}a_0=a_1=1,\ a_2=a_3=2,\ a_4=20,\ a_5=16,\ a_6=48,\ a_7=55, \\a_8=128,\ a_9=320,\ a_{10}=1206,\ a_{11}=768,\ a_{12}=406446336, \\a_{13}=43545600,\ a_{14}=141312,\ a_{15}=2267136,\ a_{16}=389112, \\a_{17}=1624232,\ a_{18}=138739712,\ a_{19}=122605392,\ a_{20}=2262695936, \\a_{21}=20313407488,\ a_{22}=17060393728,\ a_{23}=189261676544, \\a_{24}=374345132371011500507136,\ a_{25}=669835780976. \end{gather}$$

I don't know how to solve the conjecture. Any ideas towards the solution?

• I changed \text{per}\,A to \operatorname{per}A, in particular, deleting the manually added white space to the right of $\operatorname{per}$. With \operatorname{}, the spacing to the left and right depends on the context, so that $\operatorname{per}(A)$ has less space to the right of $\operatorname{per}$ than $\operatorname{per} A$ has. That is standard usage. – Michael Hardy Dec 30 '18 at 17:27