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A143002
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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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3
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0, 1, 21, 751, 38500, 2617756, 225629712, 23924915568, 3053853073152, 461404969871616, 81403191005875200, 16580318776579814400, 3861255442546368921600, 1019529192596773592678400
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OFFSET
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0,3
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COMMENTS
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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).
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REFERENCES
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Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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LINKS
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FORMULA
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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].
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MAPLE
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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)
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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