- Let $a(n)$ be A089039 (i.e., number of circular permutations of $2n$ letters that are free of jealousy). Here

$$ a(n) = \sum\limits_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!(n-k-1)!^2}{(k-1)!^2(n-2k)!k}, \\ a(1) = 1 $$

- Let $b(n)$ be A001040. Here

$$ b(n) = (n-1)b(n-1) + b(n-2), \\ b(0) = 0, b(1) = 1 $$

- Let $c(n)$ be A001053. Here

$$ c(n) = (n-1)c(n-1) + c(n-2), \\ c(0) = 1, c(1) = 0 $$

- Let $d(n)$ be an integer sequence such that

$$ d(n) = (n-1)!(b(n-1)+c(n)), \\ d(1) = 1 $$

I conjecture that $$d(n)=a(n).$$

Note that Feb 10 2019 formula in the OEIS entry of A089039 is mine. I just checked the result numerically and added it to the encyclopedia as correct.

Here is the *PARI/GP* program to check it numerically:

```
a(n) = if(n == 1, 1, sum(k = 1, n\2, n!*(n-k-1)!^2/((k-1)!^2*(n-2*k)!*k)))
d(n) = if(n == 1, 1, my(v1, v2); v1 = [0, 1]; v2 = [1, 0]; for(i=2, n, v1 = [v1[2], (i-1)*v1[2] + v1[1]]; v2 = [v2[2], (i-1)*v2[2] + v2[1]]); (n-1)!*(v1[1] + v2[2]))
test(n) = d(n) == a(n)
```

Is there a way to prove it?

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