A066259 a(n) = Fibonacci(n)*Fibonacci(n+1)^2.
1, 4, 18, 75, 320, 1352, 5733, 24276, 102850, 435655, 1845504, 7817616, 33116057, 140281700, 594243090, 2517253683, 10663258432, 45170286424, 191344405725, 810547906740, 3433536036866, 14544692047439, 61612304237568, 260993908980000, 1105587940186225, 4683345669678532
Offset: 1
Links
- Harry J. Smith, Table of n, a(n) for n = 1..200
- D. Zeitlin, Generating Functions for Products of Recursive Sequences, Transactions A.M.S., 116, Apr. 1965, p. 304.
- Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).
Programs
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Mathematica
First[#]Last[#]^2&/@Partition[Fibonacci[Range[30]],2,1] (* Harvey P. Dale, Mar 04 2011 *)
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PARI
a(n) = { fibonacci(n) * fibonacci(n+1)^2 } \\ Harry J. Smith, Feb 07 2010
Formula
O.g.f.: (x+x^2)/(1-3x-6x^2+3x^3+x^4) = x(1+x)/((1+x-x^2)(1-4x-x^2)).
a(n) = second term from left in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., a(4) = 75 since M^4 * [1 0 0 0] = [125 75 45 27] = [A056570(5) a(4) A066258(3) A056570(4)]. - Gary W. Adamson, Oct 31 2004
a(n) = (1/5)*(F(3n+2) - (-1)^n*F(n-1)). - Ralf Stephan, Jul 26 2005
a(n) = (F(n+2)^3 - 2*F(n)^3 - F(n-1)^3)/6. - Greg Dresden, Aug 12 2022