I found a proof I guess: First note that a Nakayama algebra with a circle as a quiver and $n \geq 2$ simple modules has Kupisch series of the form $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=c_0+1 \geq 3$ and $c_i-1 \leq c_{i+1}$ else. Now in case one has $c_0 \geq n+1$ then the simple modules $S_0$ should have infinite projective dimension and thus the algebra has infinite global dimension. But the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$ has finite global dimension. (this should also prove a slight generalisation of a result of gustafson in http://ac.els-cdn.com/0021869385900699/1-s2.0-0021869385900699-main.pdf?_tid=cdb3970c-8c15-11e7-be52-00000aab0f27&acdnat=1503941273_ffdc739fc29a56625414330e0bcbe36f, namely a Nakayama algebra of finite global dimension with $n$ simple modules has Loewy length at most $2n-1$ and each such algebra is a quotient algebra of the one with kupisch series $[n,2n-1,2n-2,...,n+1]$, which is also the unique such algebra with largest Loewy length $2n-1$).

Now what is left to do is calculate the permanent of the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$.

The Cartan matrix in this cases is the matrix with rows $[1,1,....,1], [1,2,2,....,2,2], [1,1,2,2,...,2,2]....,[1,1,1,....,1,2]$.

What is left to do is to show that the permanent of this matrix is given by $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$, see https://oeis.org/A000670.

I try my luck on proving that now, but feel free to post a proof if you have a quick argument (I have no experience with calculating permanents)!

I found a proof I guess: First note that a Nakayama algebra with a circle as a quiver and $n \geq 2$ simple modules has Kupisch series of the form $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=c_0+1 \geq 3$ and $c_i-1 \leq c_{i+1}$ else. Now in case one has $c_0 \geq n+1$ then the simple modules $S_0$ should have infinite projective dimension and thus the algebra has infinite global dimension. But the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$ has finite global dimension. (this should also prove a slight generalisation of a result of gustafson in http://ac.els-cdn.com/0021869385900699/1-s2.0-0021869385900699-main.pdf?_tid=cdb3970c-8c15-11e7-be52-00000aab0f27&acdnat=1503941273_ffdc739fc29a56625414330e0bcbe36f, namely a Nakayama algebra of finite global dimension with $n$ simple modules has Loewy length at most $2n-1$ and each such algebra is a quotient algebra of the one with kupisch series $[n,2n-1,2n-2,...,n+1]$, which is also the unique such algebra with largest Loewy length $2n-1$).

Now what is left to do is calculate the permanent of the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$.

The Cartan matrix in this cases is the matrix with rows $[1,1,....,1], [1,2,2,....,2,2], [1,1,2,2,...,2,2]....,[1,1,1,....,1,2]$.

What is left to do is to show that the permanent of this matrix is given by $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$, see https://oeis.org/A000670.

I try my luck on proving that now, but feel free to post a proof if you have a quick argument (I have no experience with calculating permanents)!

The sequence is $\frac{1}{2}$ of the sequence https://oeis.org/A000629 , which also has to do with circles (necklaces). Is there a good reason why?

I found a proof I guess: First note that a Nakayama algebra with a circle as a quiver and $n \geq 2$ simple modules has Kupisch series of the form $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=c_0+1 \geq 3$ and $c_i-1 \leq c_{i+1}$ else. Now in case one has $c_0 \geq n+1$ then the simple modules $S_0$ should have infinite projective dimension and thus the algebra has infinite global dimension. But the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$ has finite global dimension. (this should also prove a slight generalisation of a result of gustafson in http://ac.els-cdn.com/0021869385900699/1-s2.0-0021869385900699-main.pdf?_tid=cdb3970c-8c15-11e7-be52-00000aab0f27&acdnat=1503941273_ffdc739fc29a56625414330e0bcbe36f, namely a Nakayama algebra of finite global dimension with $n$ simple modules has Loewy length at most $2n-1$ and each such algebra is a quotient algebra of the one with kupisch series $[n,2n-1,2n-2,...,n+1]$, which is also the unique such algebra with largest Loewy length $2n-1$).

Now what is left to do is calculate the permanent of the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$.

The Cartan matrix in this cases is the matrix with rows $[1,1,....,1], [1,2,2,....,2,2], [1,1,2,2,...,2,2]....,[1,1,1,....,1,2]$.

What is left to do is to show that the permanent of this matrix is given by $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$, see https://oeis.org/A000670.

Does anyone has a direct proof of that? Im not good with calculating permanents.

I found a proof I guess: First note that a Nakayama algebra with a circle as a quiver and $n \geq 2$ simple modules has Kupisch series of the form $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=c_0+1 \geq 3$ and $c_i-1 \leq c_{i+1}$ else. Now in case one has $c_0 \geq n+1$ then the simple modules $S_0$ should have infinite projective dimension and thus the algebra has infinite global dimension. But the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$ has finite global dimension. (this should also prove a slight generalisation of a result of gustafson in http://ac.els-cdn.com/0021869385900699/1-s2.0-0021869385900699-main.pdf?_tid=cdb3970c-8c15-11e7-be52-00000aab0f27&acdnat=1503941273_ffdc739fc29a56625414330e0bcbe36f, namely a Nakayama algebra of finite global dimension with $n$ simple modules has Loewy length at most $2n-1$ and each such algebra is a quotient algebra of the one with kupisch series $[n,2n-1,2n-2,...,n+1]$, which is also the unique such algebra with largest Loewy length $2n-1$).

Now what is left to do is calculate the permanent of the algebra with Kupisch series $[n,2n-1,2n-2,...,n+1]$.

The Cartan matrix in this cases is the matrix with rows $[1,1,....,1], [1,2,2,....,2,2], [1,1,2,2,...,2,2]....,[1,1,1,....,1,2]$.

What is left to do is to show that the permanent of this matrix is given by $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$, see https://oeis.org/A000670.

I try my luck on proving that now, but feel free to post a proof if you have a quick argument (I have no experience with calculating permanents)!