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A143414
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Apery-like numbers for the constant 1/e: a(n) = 1/(n-1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!.
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2
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0, 2, 30, 492, 9620, 222630, 5989242, 184139480, 6377545512, 245868202890, 10446648201110, 485126443539012, 24449173476952380, 1329144227959100462, 77535552689576436210, 4831278674685354629040, 320262424087652686405712
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OFFSET
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0,2
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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 the constant 1/e: 1/e = 1/2 - 2 * sum {n = 2..inf} (-1)^n * n^2/(a(n)*a(n-1)). Notice the striking resemblance to 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)). Compare with A143413 and A143415.
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LINKS
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Table of n, a(n) for n=0..16.
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(n) = 1/(n-1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!. Recurrence relation: a(0) = 0, a(1) = 2, (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) = -1, b(1) = 1. Then b(n) = A143413(n) = 1/(n-1)!*sum {k = 0..n+1} (-1)^k*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) -> 1/e very rapidly. For example, |b(100)/a(100) - 1/e| is approximately 2.177 * 10^(-437). The identity a(n)*b(n-1) - a(n-1)*b(n) = (-1)^n *2*n^2 leads to rapidly converging series for the constants 1/e and e: 1/e = 1/2 - 2*sum {n = 2..inf} (-1)^n * n^2/(a(n)*a(n-1)) = 1/2 - 2*[2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...]; e = 2 * sum {n = 1..inf} (-1)^n * n^2/(b(n)*b(n-1)) = 2*[1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...].
a(n) = (BesselK(n-1/2,1/2)-(1-2*n)*BesselK(n+1/2,1/2)) * exp(1/2)/(2*Pi^(1/2)) [From Mark van Hoeij, Nov 12 2009]
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MAPLE
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with(combinat): a := n -> 1/(n-1)!*add (binomial(n-1, k)*(2*n-k)!, k = 0..n-1): seq(a(n), n = 0..19);
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CROSSREFS
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Cf. A143413, A143415.
Sequence in context: A219706 A219869 A072976 * A099046 A020547 A208881
Adjacent sequences: A143411 A143412 A143413 * A143415 A143416 A143417
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KEYWORD
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easy,nonn
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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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