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A215945
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a(n) = - 3^n*A(2*n+1), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3, with A(0)=A(1)=3, A(2)=11.
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7
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-3, -105, -3387, -108945, -3504051, -112702329, -3624894315, -116589061665, -3749904995427, -120609834867081, -3879226882922139, -124769271310005681, -4013008656895890963, -129072153032843014809, -4151404124161560449739
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OFFSET
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0,1
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COMMENTS
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The Berndt-type sequence number 11 for the argument 2Pi/9 connecting with the following sequence T(n) := t(1)^n + (-t(2))^n + t(4)^n, where t(j) := tan(2*Pi*j/9). More precisely we have sqrt(3)*a(n) = T(2*n+1) = (-sqrt(3))^n*W(n;2*i/sqrt(3)), where W(n;d) := (1 + 2*i*d*s(1))^n
+ (1 - 2*i*d*s(2))^n + (1 + 2*i*d*s(4))^n, n=0,1,..., d in C. For example we have W(0;d)=W(1;d)=3, W(2;d)=3-6*d^2. The characteristic polynomial of W(n;d) has the form ((x-1)-2*i*d*s(1))*((x-1)-2*i*d*s(2))*((x-1)-2*i*d*s(4)) = (x-1)^3 + 3*d^2*(x-1) - sqrt(3)*i*d^3 = x^3 - 3*x^2 + 3*(1+d^2)*x - 1 - 3*d^2 - sqrt(3)*i*d^3, which implies the recurrence form of W(n;d) = 3*W(n-1;d) - 3*(1+d^2)*W(n-2;d) + (1+3*d^2+sqrt(3)*i*d^3)*W(n-3;d). In consequence we obtain the title recurrence relation for
A(n) := W(n;2*i/sqrt(3)). The polynomials W(n;d) are equivalent of the big omega function with index n of argument d defined and discussed in Witula-Slota's paper (Section 6) and in comments to A215794.
We note that all numbers a(n)/3 and 3^(-1+floor((n+1)/3))*A(n) = A216034(n) are integers.
The following decomposition hold true: (X - t(1)^n)*(X - (-t(2))^n)*(X - t(4)^n) = X^3 - (-sqrt(3))^n*A(n)*X^2 + A215829(n)*X - sqrt(3)^n. Moreover we have A215829(n) = (T(n)^2 - T(2*n))/2, which implies A215829(2*n+1) = (3*a(n)^2 - A215948(2*n+1))/2 and A215829(2*n) = (A215948(n)^2 - A215948(2*n))/2.
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REFERENCES
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D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).
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LINKS
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FORMULA
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sqrt(3)*a(n) = t(1)^(2*n+1) - t(2)^(2*n+1) + t(4)^(2*n+1) = (-sqrt(3) + 4*s(1))^(2*n+1) + (-sqrt(3) - 4*s(2))^(2*n+1) + (-sqrt(3) + 4*s(4))^(2*n+1), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9).
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
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EXAMPLE
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We have 35*(t(1) - t(2) + t(3)) = t(1)^3 - t(2)^3 + t(4)^3, t(1)^7 - t(2)^7 + t(4)^7 = -5*81*269*sqrt(3) and t(1)^9 - t(2)^9 + t(4)^9 = -9*389339*sqrt(3).
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MATHEMATICA
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LinearRecurrence[{33, -27, 3}, {-3, -105, -3387}, 17] (* Bruno Berselli, Aug 29 2012 *)
CoefficientList[Series[-3 (1 + x)^2/(1 - 33 x + 27 x^2 - 3 x^3), {x, 0, 20}], x] (* Vincenzo Librandi, Mar 19 2013 *)
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PROG
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(Magma) m:=17; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(1+x)^2/(1-33*x+27*x^2-3*x^3))); // Bruno Berselli, Aug 29 2012
(Magma) I:=[-3, -105, -3387]; [n le 3 select I[n] else 33*Self(n-1)-27*Self(n-2)+3*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Mar 19 2013
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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