A208679 Kashaev's invariant for the (5,2)-torus knot (Solomon's seal knot).
1, 71, 14641, 6242711, 4555133281, 5076970085351, 8024733763147921, 17074591123571719991, 47056485265721520250561, 163059403058191163396938631, 693897612604719894794535433201
Offset: 1
Links
- Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
- K. Hikami, Volume Conjecture and Asymptotic Expansion of q-Series, Experimental Mathematics Vol. 12, Issue 3 (2003).
Crossrefs
Programs
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Maple
A208679 := proc(n) option remember; if n = 1 then 1; else (-4)^(n-1) - add((-25)^k*binomial(2*n-1,2*k)*procname(n-k),k=1..n) ; end if; end proc: seq(A208679(n),n = 1..20) # Peter Bala, Dec 20 2021 A208679 := 5*10^(2*n-2)*(-1)^n*euler(2*n-1,3/10): seq(A208679(n),n = 1..11); # Miles Wilson, Aug 05 2024
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Mathematica
nmax = 20; Table[(CoefficientList[Series[1/2*Sin[2*x]/Cos[5*x], {x, 0, 2*nmax}], x] * Range[0, 2*nmax - 1]!)[[j]], {j, 2, 2*nmax + 1, 2}] (* Vaclav Kotesovec, Aug 30 2015 *) -
PARI
my(x='x+O('x^30), v=Vec(serlaplace((1/2)*sin(2*x)/cos(5*x)))); vector(#v\2,n,v[2*n-1]) \\ Joerg Arndt, Aug 08 2024
Formula
E.g.f.: (1/2)*sin(2*x)/cos(5*x) = x + 71*x^3/3! + 14641*x^5/5! + ....
Define F(q) := Sum_{m,n >= 0} (q^(-m*n)*product {i = 1.. m+n} (1-q^i)). For the expansion of F(1-q) and F(exp(-t)) see A208733 and A208730 respectively. Kitami gives the conjectural e.g.f. exp(-9*t)*F(exp(-40*t)) = 1 + 71*t + 14641*t^2/2! + ....
a(n) = (-1)^n/(4*n+4)*20^(2*n+1)*Sum_{k = 1..20} X(k)*B(2*n+2,k/20), where B(n,x) is a Bernoulli polynomial and X(n) is a periodic function modulo 20 given by X(n) = 0 except for X(20*n+3) = X(20*n+17) = 1 and X(20*n+7) = X(20*n+13) = -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.
a(n) ~ (2*n-1)! * 2^(2*n-3/2) * 5^(2*n-1) * sqrt(5-sqrt(5)) / Pi^(2*n). - Vaclav Kotesovec, Aug 30 2015
From Peter Bala, May 11 2017: (Start)
Let X = 40*x. G.f. with offset 0: A(x) = 1 + 71*x + 14641*x^2 + ... = 1/(1 + 9*x - 2*X/(1 - 3*X/(1 + 9*x - 9*X/(1 - 11*X/(1 + 9*x - 21*X/(1 - 24*X/(1 + 9*x - ...))))))), where the sequence [2, 3, 9, 11, ..., n*(5*n - 1)/2, n*(5*n + 1)/2, ...] of unsigned coefficients in the partial numerators of the continued fraction is A057569.
A(x) = 1/(1 + 49*x - 3*X/(1 - 2*X/(1 + 49*x - 11*X/(1 - 9*X/(1 + 49*x - 24*X/(1 - 21*X/(1 + 49*x - 42*X/(1 - 38*X/(1 + 49*x - ...))))))))), where the sequence [3, 2, 11, 9, 24, 21, ...] of unsigned coefficients in the partial numerators of the continued fraction is obtained by swapping pairs of adjacent terms of A057569. Let B(x) = 1/(1 - 9*x)*A(x/(1 - 9*x)), that is, B(x) is the 9_th binomial transform of A(x). Then B(x/40) = 1 + 2*x + 10*x^2 + 104*x^3 + ... is the o.g.f. for A208730. (End)
From Peter Bala, Dec 20 2021: (Start)
a(1) = 1, a(n) = (-4)^(n-1) - Sum_{k = 1..n} (-25)^k*C(2*n-1,2*k)*a(n-k).
a(n) == 71^(n-1) ( mod (2^7)*3*(5^2) ). (End)
a(n) = 5*10^(2*n - 2)*(-1)^n*E(2*n - 1, 3/10), where E(n,x) is the n-th Euler polynomial in x (A060096/A060097). - Miles Wilson, Aug 05 2024
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