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?



Is the permanent of the matrix $[(\frac{i+j}{2n+1})]_{0\le i,j\le n}$ always positive?
This is the sequence A322898 in OEIS.
I used a program in PARI and calculated the values of a(26) to a(34).



```
a(26) = -7000008163328
a(27) = 22712032822272
a(28) = 2244036651776
a(29) = 4363027965018112
a(30) = 30229121955004416
a(31) = -46693326700068864
a(32) = -23328907207088128
a(33) = 3173005987716005888
a(34) = 136427303851761536

```

The conjecture is **disproved**.


