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A099839
a(n) = -10*a(n-1) - 5*a(n-2) with a(0) = 1, a(1) = -11.
3
1, -11, 105, -995, 9425, -89275, 845625, -8009875, 75870625, -718656875, 6807215625, -64478871875, 610752640625, -5785132046875, 54797557265625, -519049912421875, 4916511337890625, -46569863816796875, 441116081478515625, -4178311495701171875
OFFSET
0,2
COMMENTS
A transform of the Fibonacci numbers: The g.f. is the image of the g.f. x/(1-x-x^2) of A000045 under the transformation G(x) -> 1/(2*(1+x)) * G((1-x)/(2*(1+x)).
FORMULA
G.f.: (1-x)/(1 + 10*x + 5*x^2).
a(n) = (-5 - 2*sqrt(5))^n*(1/2 + 3*sqrt(5)/10) + (-5 + 2*sqrt(5))^n*(1/2 - 3*sqrt(5)/10).
a(n) = (1/2)*5^((n-1)/2)*(sqrt(5)*(1+(-1)^n)*Fibonacci(3*n+2) - (1-(-1)^n)*Lucas(3*n+2)). - G. C. Greubel, Apr 21 2023
MATHEMATICA
LinearRecurrence[{-10, -5}, {1, -11}, 41] (* G. C. Greubel, Apr 21 2023 *)
PROG
(Magma) [n le 2 select 13-12*n else -10*Self(n-1) -5*Self(n-2): n in [1..41]]; // G. C. Greubel, Apr 21 2023
(SageMath)
def A099839(n): return 5^((n-1)/2)*(sqrt(5)*((n+1)%2)*fibonacci(3*n+2) - (n%2)*lucas_number2(3*n+2, 1, -1))
[A099839(n) for n in range(41)] # G. C. Greubel, Apr 21 2023
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Oct 27 2004
STATUS
approved