已知數列a1=a2=1,且a(n+2)-a(n)=1/a(n+1),求a(2020)?

a(n+1)a(n+2)  - a(n+1)a(n) = 1, for all n in N
∴ a(n+2)a(n+1) = a(n+1)a(n) + 1 = a(2)a(1) + n, for all n in N
∴ a(n+2) = (a(2)a(1) + n)/a(n+1) 
               =   (n+1)/a(n+1)
               = [(n+1)/(n)]a(n), for all n in N
∴ a(n) = [(n-1)/(n-2)] a(n-2)
           = [(n-1)/(n-2)](n-3)/a(n-3)
           = [(n-1)(n-3)]/[(n-2)(n-4)] a(n-4)

n = 2k-1, k in N, 則
    a(n) = [(n-1)(n-3)…2]/[(n-2)(n-4)…1] a(1)
           = [(2k-2)(2k-4)...2]/[(2k-3)(2k-5)...1]
           = [(2k-2)(2k-4)...2]^2/(2k-2)!
           = [2^(k-1) (k-1)!]^2/(2k-2)!
           = 2^(n-1){[(n-1)/2]!}^2/(n-1)!

n = 2k, k in N, 則
    a(n) = [(n-1)(n-3)…3]/[(n-2)(n-4)…2] a(2)
           = [(2k-1)(2k-3)...3]/[(2k-2)(2k-4)...2]
           = (2k-1)!/[(2k-2)(2k-4)...2]^2
           = (n-1)!/{2^(n-2)[(n/2-1)!]^2}

∴ a(2020) = 2019!/{2^2018(1009!)^2}