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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.
7
3, 33, 1035, 33273, 1070163, 34420113, 1107069147, 35607149289, 1145248319907, 36835122733569, 1184744167018155, 38105444942752473, 1225602095969542131, 39419576386041628017, 1267869080483024344443, 40779027899804588036553, 1311593714249667872790339
OFFSET
0,1
COMMENTS
The Berndt-type sequence number 12 for the argument 2*Pi/9 defined by the first trigonometric relations from the section "Formula" below (it is the complement of the sequence A215945). For more information see comments to A215945. We note that all a(n)/3 and 3^(-1 + floor((n+3)/3))*A(n) = A216034(n) are integers.
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.
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
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{33, -27, 3}, {3, 33, 1035}, 50]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Aug 28 2012
STATUS
approved