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a(n) = 2 * T(n,(n+1)/2) where T(n,x) is a Chebyshev polynomial of the first kind.
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%I #22 Mar 15 2024 12:50:18

%S 2,2,7,52,527,6726,103682,1874888,38925119,912670090,23855111399,

%T 687808321212,21687295069442,742397047217294,27420344506901023,

%U 1086932029484351248,46027034321342899967,2073668380220713167378,99042070146811639444802

%N a(n) = 2 * T(n,(n+1)/2) where T(n,x) is a Chebyshev polynomial of the first kind.

%H Seiichi Manyama, <a href="/A343260/b343260.txt">Table of n, a(n) for n = 0..386</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Chebyshev_polynomials">Chebyshev polynomials</a>.

%F a(n) = 2 * cos(n*arccos((n+1)/2)).

%F a(n) = 2 * n * Sum_{k=0..n} (n-1)^k * binomial(n+k,2*k)/(n+k) for n > 0.

%F a(n) ~ exp(1) * n^n. - _Vaclav Kotesovec_, Apr 09 2021

%F Conjecture: a(p^r) == 1 (mod p^(2*r)) for positive integer r and all primes p >= 5. - _Peter Bala_, Mar 11 2024

%t Table[2*ChebyshevT[n, (n+1)/2], {n, 0, 18}] (* _Amiram Eldar_, Apr 09 2021 *)

%o (PARI) a(n) = 2*polchebyshev(n, 1, (n+1)/2);

%o (PARI) a(n) = round(2*cos(n*acos((n+1)/2)));

%o (PARI) a(n) = if(n==0, 2, 2*n*sum(k=0, n, (n-1)^k*binomial(n+k, 2*k)/(n+k)));

%Y Cf. A097691, A115066, A342205, A343259, A343261.

%K nonn,easy

%O 0,1

%A _Seiichi Manyama_, Apr 09 2021