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A215877
a(n) = (A(n) - A215817(n))/sqrt(7), where A(n) = (6-sqrt(7))A(n-1) - (12-4*sqrt(7))A(n-2) + (8-3*sqrt(7))A(n-3), with A(0)=3, A(1)=6-sqrt(7), and A(2)=19-4*sqrt(7).
7
0, -1, -4, -16, -64, -254, -1000, -3913, -15248, -59263, -229996, -892033, -3459544, -13421784, -52104416, -202436819, -787231328, -3064347392, -11940020992, -46569416006, -181808493296, -710442293743, -2778591945620, -10876271461745, -42606078512048
OFFSET
0,3
COMMENTS
The Berndt-type sequence number 15 for the argument 2Pi/7 defined by requiring sqrt(7)*a(n) to be the irrational part of the trigonometric sum A(n) := c(1)^(2*n) + c(2)^(2*n) + c(4)^(2*n), where c(j) := 2*cos(Pi/4 + 2*Pi*j/7) = 2*cos((7+8*j)*Pi/28).
We note that A(n)-sqrt(7)*a(n)= A215817(n). For more facts on A(n) - see comments to A215817.
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
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5
FORMULA
sqrt(7)*a(n) = to the irrational part of c(1)^(2*n) + c(2)^(2*n) + c(4)^(2*n) = (1-s(1))^n + (1-s(2))^n + (1-s(4))^n, where c(j) = 2*cos((7+8*j)*Pi/28) and s(j) := sin(2*Pi*j/7).
Empirical g.f.: -x * (2*x-1)^2 * (x^2-4*x+1) / (x^6 -24*x^5 +86*x^4 -104*x^3 +53*x^2 -12*x +1). - Colin Barker, Jun 01 2013
EXAMPLE
We have a(2)/a(1) = a(3)/a(2) = a(4)/a(3) = 4, but a(5)-4*a(4)=2 and a(6)=4*(a(5)-a(2)). Moreover it follows
the relations: 4*A(1)-A(2) = 5 = (3+s(1))*(1-s(1)) + (3+s(2))*(1-s(2)) + (3+s(4))*(1-s(4)), 4*A(2)-A(3) = 10 = (3+s(1))*(1-s(1))^2 + (3+s(2))*(1-s(2))^2 + (3+s(4))*(1-s(4))^2, 4*A(3)-A(4) = 27 = (3+s(1))*(1-s(1))^3 + (3+s(2))*(1-s(2))^3 + (3+s(4))*(1-s(4))^3, whereas 4*A(4)-a(5) = 82-2*sqrt(7) = (3+s(1))*(1-s(1))^4 + (3+s(2))*(1-s(2))^4 + (3+s(4))*(1-s(4))^4.
CROSSREFS
KEYWORD
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AUTHOR
Roman Witula, Aug 25 2012
STATUS
approved