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A143413
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Apery-like numbers for the constant e: a(n) = 1/(n-1)!*sum {k = 0..n+1} (-1)^k*C(n+1,k)*(2*n-k)! for n >= 1.
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5
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-1, 1, 11, 181, 3539, 81901, 2203319, 67741129, 2346167879, 90449857081, 3843107102339, 178468044946621, 8994348275804891, 488964835817842021, 28523735794360301039, 1777328098986754744081, 117817961601577138782479, 8279178465722546926265329
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OFFSET
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0,3
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COMMENTS
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This sequence satisfies the recursion (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), which leads to a rapidly converging series for Napier's constant: e = 2 * sum {n = 1..inf} (-1)^n * n^2/(a(n)*a(n-1)). Notice the striking parallels with the theory of the Apery numbers A(n) = A005258(n), which satisfy a similar recurrence relation n^2*A(n) - (n-1)^2*A(n-2) = (11*n^2-11*n+3)*A(n-1) and which appear in the series acceleration formula zeta(2) = 5*sum {n = 1..inf} 1/(n^2*A(n)*A(n-1)) = 5*[1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...].
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LINKS
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Table of n, a(n) for n=0..17.
A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report. Math. Intelligencer 1 (1978/79), no 4, 195-203.
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FORMULA
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a(0):= -1, a(n) = 1/(n-1)!*sum {k = 0..n+1} (-1)^k*C(n+1,k)*(2*n-k)! for n >= 1. Apart from the initial term, this sequence is the second superdiagonal of the square array A060475; equivalently, the second subdiagonal of the square array A086764. Recurrence relation: a(0) = -1, a(1) = 1, (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), n >= 2. Let b(n) denote the solution to this recurrence with initial conditions b(0) = 0, b(1) = 2. Then b(n) = A143414(n) = 1/(n-1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!. The rational number b(n)/a(n) is equal to the Pade approximation to exp(x) of degree (n-1,n+1) evaluated at x = 1 and b(n)/a(n) -> e very rapidly. For example, b(100)/a(100) - e is approximately 1.934 * 10^(-436). The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^n *2*n^2 leads to rapidly converging series for e and 1/e: e = 2 * sum {n = 1..inf} (-1)^n * n^2/(a(n)*a(n-1)) = 2*[1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...]; 1/e = 1/2 - 2*sum {n = 2..inf} (-1)^n * n^2/(b(n)*b(n-1)) = 1/2 - 2*[2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...]. Conjectural congruences: for r >= 0 and odd prime p, calculation suggests that a(p^r*(p+1)) + a(p^r) == 0 (mod p^(r+1)).
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MAPLE
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with(combinat): a := n -> 1/(n-1)!*add ((-1)^k*binomial(n+1, k)*(2*n-k)!, k = 0..n+1): seq(a(n), n = 1..19);
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CROSSREFS
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Cf. A005258, A060475, A086764, A143414, A143415.
Sequence in context: A020456 A036935 A205088 * A009118 A112943 A057618
Adjacent sequences: A143410 A143411 A143412 * A143414 A143415 A143416
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KEYWORD
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easy,sign
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AUTHOR
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Peter Bala, Aug 14 2008
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STATUS
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approved
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