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%I #17 May 18 2017 11:49:38
%S 1,143,58081,48571823,69471000001,151763444497103,470164385248041121,
%T 1960764928973430783983,10591336845363318048877441,
%U 71933835058256664782546056463,599982842750416411984319126244961
%N Kashaev invariant for the (7,2)-torus knot.
%C Compare with A156370. For other Kashaev invariants see A002439 ((3,2)-torus knot), A208679 and A208681.
%H P. Bala, <a href="/A002439/a002439.pdf">Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)</a>
%H K. Hikami, <a href="http://www.emis.de/journals/EM/expmath/volumes/12/12.3/Hikami.pdf">Volume Conjecture and Asymptotic Expansion of q-Series</a>, Experimental Mathematics Vol. 12, Issue 3 (2003).
%F E.g.f.: 1/2*sin(2*x)/cos(7*x) = x + 143*x^3/3! + 58081*x^5/5! + ....
%F 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.
%F 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.
%F From _Peter Bala_, May 16 2017: (Start)
%F 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 - ... ))))).
%F 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)
%F 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
%Y Cf. A002439 ((3,2)-torus knot), A156370, A208679, A208681.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Mar 01 2012
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