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A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits.

Edited 17.07.22, 15:00 CEST:

I did this verification with 100 digits precision, verifying $c_{5,6}$ and extending the series to 10th order. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} +\frac{32519 \pi ^7}{5160960 n^7}\\ +\frac{135001 \pi ^8}{20643840 n^8} +\frac{45750727 \pi ^9}{5945425920 n^9} +\frac{1198585643 \pi ^{10}}{118908518400 n^{10}} +\ldots. $$ Note that the denominator of $c_n$ is divisible by $n!$.

All constants were successively determined by a least-square fit of a 20th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a known function with this series expansion.

For reference, I'll append the 30 digits results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}

A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits.

Edited 17.07.22, 15:00 CEST:

I did this verification with 100 digits precision, verifying $c_{5,6}$ and extending the series to 10th order. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} +\frac{32519 \pi ^7}{5160960 N^7} +\frac{135001 \pi ^8}{20643840 N^8} +\frac{45750727 \pi ^9}{5945425920 N^9} +\frac{1198585643 \pi ^{10}}{118908518400 N^{10}} +\ldots. $$ Note that the denominators of $c_n$ are divisible by $n!$.

All constants were successively determined by a least-square fit of a 20th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a known function with this series expansion.

For reference, I'll append the 30 digits results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}

A quick numerical investigation for $N$ up to 512 leads to a conjectured expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} + \ldots. $$

All constants were determined by a least-square fit of a 10th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a function with this series expansion.

For reference, I'll append the high-precision results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}

A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits.

Edited 17.07.22, 15:00 CEST:

I did this verification with 100 digits precision, verifying $c_{5,6}$ and extending the series to 10th order. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} +\frac{32519 \pi ^7}{5160960 n^7}\\ +\frac{135001 \pi ^8}{20643840 n^8} +\frac{45750727 \pi ^9}{5945425920 n^9} +\frac{1198585643 \pi ^{10}}{118908518400 n^{10}} +\ldots. $$ Note that the denominator of $c_n$ is divisible by $n!$.

All constants were successively determined by a least-square fit of a 20th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a known function with this series expansion.

For reference, I'll append the 30 digits results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}

A quick numerical investigation for $N$ up to 2048 leads to a conjectured expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

For reference, I'll append some high-precision results: \begin{array}{lll} N & λ_0(N)& c_5^{-1}(N)\\ 128 & 3.95121320281088603898478521135 & -112.33736106154642063 \\ 256 & 3.97553152996970070083424756460 & -113.483268010223740 \\ 384 & 3.98367098180919839228844082991 & -113.86471637351927 \\ 512 & 3.98774696887736782094115794509 & -114.05534610371244 \\ \infty & 4 & -114.63(1) \end{array}

A quick numerical investigation for $N$ up to 512 leads to a conjectured expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} + \ldots. $$

All constants were determined by a least-square fit of a 10th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a function with this series expansion.

For reference, I'll append the high-precision results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}