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A273623
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a(n) = Fibonacci(3*n) - (2 + (-1)^n)*Fibonacci(n).
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4
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1, 5, 32, 135, 605, 2560, 10933, 46305, 196384, 831875, 3524489, 14929920, 63245753, 267913165, 1134902560, 4807524015, 20365009477, 86267563520, 365435291981, 1548008735625, 6557470308896, 27777889982155, 117669030432337, 498454011740160, 2111485077903025
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OFFSET
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1,2
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COMMENTS
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This is a divisibility sequence: if n divides m then a(n) divides a(m). The sequence satisfies a linear recurrence of order 6. In general, for integers r and s, the sequence Fibonacci(r*n) - 2*Fibonacci((r - 2*s)*n) + Fibonacci((r - 4*s)*n) is a divisibility sequence of the sixth order. This is the case r = 3, s = 1. See A127595 (case r = 4, s = 1).
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LINKS
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FORMULA
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a(n) = Fibonacci(3*n) - 2*Fibonacci(n) + Fibonacci(-n).
a(2*n) = 5*Fibonacci(2*n)^3;
a(2n+1) = Fibonacci(2*n+1)*(5*Fibonacci(2*n+1)^2 - 4) = Fibonacci(2*n+1)*Lucas(2*n+1)^2.
O.g.f. x*(x^4 - x^3 + 8*x^2 + x + 1)/( (1 + x - x^2 )*(1 - x - x^2)*(1 - 4*x - x^2 ) ).
a(n) = 4*a(n-1) + 4*a(n-2) - 12*a(n-3) - 4*a(n-4) + 4*a(n-5) + a(n-6). - G. C. Greubel, Jun 02 2016
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MAPLE
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with(combinat):
seq(fibonacci(3*n) - (2 + (-1)^n)*fibonacci(n), n = 1..25);
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MATHEMATICA
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LinearRecurrence[{4, 4, -12, -4, 4, 1}, {1, 5, 32, 135, 605, 2560}, 100] (* G. C. Greubel, Jun 02 2016 *)
Table[Fibonacci[3 n] - (2 + (-1)^n) Fibonacci[n], {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)
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PROG
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(Magma) [Fibonacci(3*n)-(2+(-1)^n)*Fibonacci(n): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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