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The Berndt-type sequence number 8 for the argument 2Pi2*Pi/9 defined by the trigonometric relations from the Formula section below.
From the general recurrence relation: b(n) = -3*b(n-1) + 9*b(n-2) + 3*b(n-3), i.e. , b(n) - b(n-2) = 8*b(n-2) + 3(b(n-3) - b(n-1)) it can be easily deduced the following summation formulas can be easily deduced: b(2*n+1) + 3*b(2*n) - 3*b(0) - b(1) = 8*sumSum_{k=1,..,n} b(2*k-1) and b(2*n+2) + 3*b(2*n+1) - b(2) - 3*b(1) = 8*sumSum_{k=1,..n} b(2*k). Hence it follows that (a(2*n+1) + 3*a(2*n))/2 are all integers congruent to 3 modulo 4, and (a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.
(a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.
The following decomposition hold holds true: (X - k(1)^n)*(X - (-k(2))^n)*(X - k(3)^n) = X^3 - sqrt(3)^(-n)*a(n)*X^2 + sqrt(3)^(-n)*T(n) - sqrt(3)^(-n), where T(2*n+1) = sqrt(3)*A215945(n) and T(2*n) = A215948(n). [Roman Witula, Aug 30 2012]
G.f.: (3 + 6*x - 9*x^2)/(1 + 3*x - 9*x^2 - 3*x^3). [corrected by Georg Fischer, May 10 2019]
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G.f.: (3+6*x-9*x^2)/(1+3*x-9*x^2-3*x^3). - Corrected [corrected by Georg Fischer, May 10 2019]
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G.f.: (3+6*x-9*x^2)/(1-+3*x+3-9*x^2-273*x^3). - Corrected by _Georg Fischer_, May 10 2019
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