%I #36 Aug 12 2022 09:31:18
%S 1,4,18,75,320,1352,5733,24276,102850,435655,1845504,7817616,33116057,
%T 140281700,594243090,2517253683,10663258432,45170286424,191344405725,
%U 810547906740,3433536036866,14544692047439,61612304237568
%N a(n) = Fibonacci(n)*Fibonacci(n+1)^2.
%H Harry J. Smith, <a href="/A066259/b066259.txt">Table of n, a(n) for n = 1..200</a>
%H D. Zeitlin, <a href="http://dx.doi.org/10.1090/S0002-9947-1965-0185301-0">Generating Functions for Products of Recursive Sequences</a>, Transactions A.M.S., 116, Apr. 1965, p. 304.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).
%F 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)).
%F 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
%F a(n) = (1/5)*(F(3n+2) - (-1)^n*F(n-1)). - _Ralf Stephan_, Jul 26 2005
%F a(n) = (F(n+2)^3 - 2*F(n)^3 - F(n-1)^3)/6. - _Greg Dresden_, Aug 12 2022
%t First[#]Last[#]^2&/@Partition[Fibonacci[Range[30]],2,1] (* _Harvey P. Dale_, Mar 04 2011 *)
%o (PARI) { for (n=1, 200, a=fibonacci(n) * fibonacci(n+1)^2; write("b066259.txt", n, " ", a) ) } \\ _Harry J. Smith_, Feb 07 2010
%Y Cf. A000045, A065563, A066258, A000045, A056570.
%K nonn,easy
%O 1,2
%A _Len Smiley_, Dec 09 2001