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A208680
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Kashaev invariant for the (7,2)-torus knot.
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6
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1, 143, 58081, 48571823, 69471000001, 151763444497103, 470164385248041121, 1960764928973430783983, 10591336845363318048877441, 71933835058256664782546056463, 599982842750416411984319126244961
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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E.g.f.: 1/2*sin(2*x)/cos(7*x) = x + 143*x^3/3! + 58081*x^5/5! + ....
a(n) = (-1)^n/(4*n+4)*28^(2*n+1)*sum {k = 1..28} X(k)*B(2*n+2,k/28), where B(n,x) is a Bernoulli polynomial and X(n) is a periodic function modulo 28 given by X(n) = 0 except for X(28*n+5) = X(28*n+23) = 1 and X(28*n+9) = X(28*n+19) = -1.
a(n) = 1/2*(-1)^(n+1)*L(-2*n-1,X) in terms of the associated L-series attached to the periodic arithmetical function X.
O.g.f. as continued fraction: A(x) = 1/(1 + 25*x - 6*28*x/(1 - 8*28*x/(1 + 25*x -...- n*(7*n-1)*28*x/(1 - n*(7*n+1)*28*x/(1 + 25*x - ... ))))).
Also, A(x) = 1/(1 + 81*x - 8*28*x/(1 - 6*28*x/(1 + 81*x -...- n*(7*n+1)*28*x/(1 - n*(7*n-1)*28*x/(1 + 81*x - ... ))))). (End)
a(n) ~ sin(Pi/7) * 2^(4*n) * 7^(2*n-1) * n^(2*n-1/2) / (Pi^(2*n-1/2) * exp(2*n)). - Vaclav Kotesovec, May 18 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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