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a(n) = -a(-4-n) = ((4n4*n+2)*F(2n2*n) + (7n7*n+5)*F(2n2*n+1))/5 with F(n) = A000045 (Fibonacci numbers).
a(n) = (Sum_{k=0..n} S(k, 3)*S(n-k, 3)) , where S(n, x) = U(n, x/2) is the n-th Chebyshev polynomials polynomial of the 2nd kind, A049310. - Paul Barry, Nov 14 2003
a(n) = Sum_{k=1..n+1} F(2k)*F(2(n-k+2)) , where F(k) is the k-th Fibonacci number. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008
a(n) = Sum_{k = 0..n} (n+2*k+1)*binomial(n+k, 2*k) ? - _From _Peter Bala_, Nov 02 05 2024: (Start)
a(n) = Sum_{k = 0..n} (n + 2*k + 1)*binomial(n+k, 2*k).
a(n) = (n+1) * hypergeom([-n, n+1, (n+3)/2], [1/2, (n+1)/2], -1/4).
Second-order recurrence: n*a(n) = 3*(n + 1)*a(n-1) - (n + 2)*a(n-2) with a(0) = 1 and a(1) = 6. (End)
a(n) = Sum_{k = 0..n} (n+2*k+1)*binomial(n+k, 2*k) ? - Peter Bala, Nov 02 2024
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Jean-Luc Baril, Toufik Mansour, José L. Ramírez, and Mark Shattuck. , <a href="http://jl.baril.u-bourgogne.fr/bmrs.pdf">Catalan words avoiding a pattern of length four</a>, Univ. de Bourgogne (France, 2024). See p. 5.
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