A215948 a(n) = 3^n*A(2*n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.
3, 33, 1035, 33273, 1070163, 34420113, 1107069147, 35607149289, 1145248319907, 36835122733569, 1184744167018155, 38105444942752473, 1225602095969542131, 39419576386041628017, 1267869080483024344443, 40779027899804588036553, 1311593714249667872790339
Offset: 0
Examples
We have t(1)^4 + t(2)^4 + t(4)^4 = 1035 = (345/11)*(t(1)^2 + t(2)^2 + t(4)^2) and (1 - 4*s(1)/sqrt(3))^4 + (1 + 4*s(2)/sqrt(3))^4 + (1 - 4*s(4)/sqrt(3))^4 = 115. Moreover we get a(2)/a(1) = 31,(36), a(3)/a(1) = 1008,(27), a(4)/a(1) = 32429,(18).
References
- 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).
Links
- Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6.
- Index entries for linear recurrences with constant coefficients, signature (33,-27,3).
Programs
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Mathematica
LinearRecurrence[{33,-27,3}, {3,33,1035}, 50]
Formula
a(n) = t(1)^(2*n) + t(2)^(2*n) + t(4)^(2*n) = (-sqrt(3) + 4*s(1))^(2*n) + (sqrt(3) + 4*s(2))^(2*n) + (-sqrt(3) + 4*s(4))^(2*n), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9). For the respective sums of odd powers - see A215945.
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
G.f.: 3*(1-22*x+9*x^2)/(1-33*x+27*x^2-3*x^3).
a(n) = cot(Pi/18)^(2*n) + cot(5*Pi/18)^(2*n) + cot(7*Pi/18)^(2*n). - Greg Dresden, Oct 01 2020
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