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A product of consecutive doubled Fibonacci numbers.
3

%I #18 Sep 08 2022 08:45:48

%S 1,3,6,16,40,105,273,715,1870,4896,12816,33553,87841,229971,602070,

%T 1576240,4126648,10803705,28284465,74049691,193864606,507544128,

%U 1328767776,3478759201,9107509825,23843770275,62423800998,163427632720

%N A product of consecutive doubled Fibonacci numbers.

%H G. C. Greubel, <a href="/A166536/b166536.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-3,1).

%F G.f.: (1 - 3*x^2 + x^3)/(1 - 3*x + 3*x^3 - x^4).

%F a(n) = F(n+1)*F(n+2) + (1 - (-1)^n)/2, where F = A000045.

%F a(n) = (F(n+2)*(1 + (-1)^n)/2 + F(n)*(1 - (-1)^n)/2)*(F(n+3)*(1 - (-1)^n)/2 + F(n+1)*(1 + (-1)^n)/2).

%F a(n)*a(n+2) - a(n+1)^2 = (-1)^n*(F(2*n+4) - 1).

%F a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - _G. C. Greubel_, May 16 2016

%F a(n) = Sum_{i=0..n+1} F(i+1)*F(i-1), where F(-1) = 1. - _Bruno Berselli_, Feb 16 2017

%t LinearRecurrence[{3, 0, -3, 1}, {1, 3, 6, 16}, 30] (* _G. C. Greubel_, May 16 2016 *)

%o (Magma) /* From the sixth formula: */ F:=Fibonacci; [&+[F(i+1)*F(i-1): i in [0..n+1]]: n in [0..30]]; // _Bruno Berselli_, Feb 15 2017

%o (PARI) my(x='x+O('x^30)); Vec((1-3*x^2+x^3)/(1-3*x+3*x^3-x^4)) \\ _G. C. Greubel_, Jan 09 2019

%o (Sage) ((1-3*x^2+x^3)/(1-3*x+3*x^3-x^4)).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Jan 09 2019

%o (GAP) a:=[1,3,6,16];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # _G. C. Greubel_, Jan 09 2019

%Y Cf. A000045, A166516.

%K nonn,easy

%O 0,2

%A _Paul Barry_, Oct 16 2009