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a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1).
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%I #4 Nov 30 2023 16:54:32

%S 0,1,21,751,38500,2617756,225629712,23924915568,3053853073152,

%T 461404969871616,81403191005875200,16580318776579814400,

%U 3861255442546368921600,1019529192596773592678400

%N a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1).

%C This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+1)*a(n) + n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} (-1)^(k+1)/k^2 for the constant 1/2*zeta(2). For other cases see A142999 (m=0), A143000 (m=1) and A143001 (m=2).

%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

%F a(n) = n!^2*p(n)*sum {k = 1..n} (-1)^(k+1)/(k^2*p(k-1)*p(k)), where p(n) = (n^6 + 3*n^5 + 22*n^4 + 39*n^3 + 85*n^2 + 66*n + 36)/36. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 7. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+ 1^4/(21+ 2^4/(35+ 3^4/(49+...+ (n-1)^4/(7*(2*n-1)))))), for n >=2. Lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(k^2*p(k-1)p(k)) = 1/(7+ 1^4/(21+ 2^4/(35+ 3^4/(49+...+ n^4/(7*(2*n+1)+...))))) = 1/2*(zeta(2)-(1+1/4+1/9)). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].

%p p := n -> (n^6+3*n^5+22*n^4+39*n^3+85*n^2+66*n+36)/36: a := n -> n!^2*p(n)*sum ((-1)^(k+1)/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20)

%t RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==7(2n+1)a[n]+n^4 a[n-1]},a,{n,15}] (* _Harvey P. Dale_, Nov 30 2023 *)

%Y Cf. A142999, A143000, A143001.

%K easy,nonn

%O 0,3

%A _Peter Bala_, Jul 18 2008