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The Berndt-type sequence number 13 for the argument 2Pi2*Pi/9 defined by the relation a(n)*sqrt(3) = s(1)^(2*n-1) - s(2)^(2*n-1) + s(4)^(2*n-1), where s(j) := 2*sin(2*Pi*j/9). For the respective sums with the even powers of sines - see A215634.
We note that X(n) = s(1)^n + (-s(2))^n + s(4)^n - - see Witula's book for details. Moreover the numbers of the form a(n)*3^(-1-floor((n-1)/3)) are integers.
The following summation formulas holdshold: sumSum_{k=3..n} a(k) = 3*(2*a(n-1) - a(n-2) + 1), X(2*n+1) - X(1)*3^n = X(2*n+1) = -sqrt(3)*Sum_{k=1..n} X(2*(n-k))*3^(k-1), and X(2*n) - X(0)*3^n = X(2*n) - 3^(n+1) = -sqrt(3,)*Sum{k=1..,n} X(2*(n-k))*3^(k-1).
a(k) = 3*(2*a(n-1) - a(n-2) + 1), X(2*n+1) - X(1)*3^n = X(2*n+1) = -sqrt(3)*sum{k=1,..,n} X(2*(n-k))*3^(k-1), and X(2*n) - X(0)*3^n = X(2*n) - 3^(n+1) = -sqrt(3)*sum{k=1,..,n} X(2*(n-k))*3^(k-1).
We have s(1)^5 - s(2)^5 + s(4)^5 = 5*(s(1)^3 - s(2)^3 + s(4)^3) = -15*sqrt(3), s(1)^9 - s(2)^9 + s(4)^9 = 4*(s(1)^7 - s(2)^7 + s(4)^7) = -252*sqrt(3),
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a(n) = Sum_{k=0..n} 3*(-1)^k*(binomial(2*n-1, n+9*k+7) - binomial(2*n-1, n+9*k+1)). - Greg Dresden, Jan 28 2023
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G. C. Greubel, <a href="/A216757/b216757.txt">Table of n, a(n) for n = 1..1000</a>
CoefficientList[Series[-3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3), {x, 0, 5 0}], x] (* G. C. Greubel, Apr 17 2017 *)
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