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From Vaclav Kotesovec, Apr 21 2024: (Start)
Recurrence: 2*(n+1)*(2*n + 1)*(13*n^3 - 17*n^2 - 12*n + 10)*a(n) = (143*n^5 - 44*n^4 - 525*n^3 + 320*n^2 + 94*n - 60)*a(n-1) + 4*(n-1)*(26*n^4 + 5*n^3 - 108*n^2 + 7*n + 22)*a(n-2) + 16*(n-2)*(n-1)*(13*n^3 + 22*n^2 - 7*n - 6)*a(n-3).
a(n) ~ sqrt((247 - 9131*13^(2/3)/(1788163 + 409728*sqrt(78))^(1/3) + (13*(1788163 + 409728*sqrt(78)))^(1/3))/13) * ((11 + (1/3)*(172017 - 16848*sqrt(78))^(1/3) + (6371 + 624*sqrt(78))^(1/3))^n / (sqrt(Pi*n) * 2^(2*n + 3) * 3^(n + 1/2))). (End)
Table[Sum[(1 + (-1)^(n+j))/2 * Binomial[n, j] * Binomial[n+1, (n-j)/2] * (j+1)/(n+1), {j, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 21 2024 *)
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a(n) = Sum_{k=0..n} binomial(n, k)*CT(n, k) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*CT(n, k), where CT(n, k) is the Catalan triangle A053121.
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a(n) = Sum_{j=0..n} even(n + j)*binomial(n, j)*binomial(n + 1, (n - j)/2)*(j + 1)/(n + 1), where even(k) = 1 if k is even and otherwise 0.