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a(n) = Fibonacci(n) * Fibonacci(2*n).
3

%I #52 Sep 08 2022 08:44:52

%S 0,1,3,16,63,275,1152,4901,20727,87856,372075,1576279,6676992,

%T 28284569,119814747,507544400,2149990983,9107510539,38580029568,

%U 163427634589,692290558575,2932589884016,12422650070163,52623190204271,222915410823168,944284833600625,4000054745057907,16944503814103696,71778070001033487

%N a(n) = Fibonacci(n) * Fibonacci(2*n).

%C Let F(n) = Fibonacci(n), then abs(det([F(n), F(n+k); F(n+2k), F(n+3k)])) = a(k), independent of n. - _R. M. Welukar_, Aug 26 2014

%C From _Joerg Arndt_, Aug 26 2014: (Start)

%C This is a special case of Johnson's identity (relation 32 in the Mathworld link).

%C F(a)*F(b) - F(c)*F(d) = (-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)), where a+b = c+d and r arbitrary.

%C Here a = n, b = n+3*k, c = n+k, d = n+2*k, and r = c, so that

%C (-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)) =

%C (-1)^c*(F(a-c)*F(b-c) - F(c-c)*F(d-c)) =

%C (-1)^c*(F(a-c)*F(b-c) - 0) =

%C (-1)^c*(F(-k)*F(-2*k)), taking the absolute value gives a(k).

%C (End)

%C Let L(n) = A000032(n), then abs(det([L(n), L(n+k); L(n+2k), L(n+3k)])) = 5*a(k), independent of n. - M. N. Deshpande and _R. M. Welukar_, Aug 30 2014

%H Colin Barker, <a href="/A037451/b037451.txt">Table of n, a(n) for n = 0..1000</a>

%H Eric Weisstein, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a> (MathWorld).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).

%F From _Emanuele Munarini_, Jul 18 2003: (Start)

%F G.f.: ( x + x^3 )/( 1 - 3 x - 6 x^2 + 3 x^3 + x^4 ).

%F a(n+4) = 3*a(n+3) + 6*a(n+2) - 3*a(n+1) - a(n).

%F (End)

%F G.f.: x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)). - _Joerg Arndt_, Aug 26 2014

%F a(n) = (1/5)*(Lucas(3*n) - (-1)^n*Lucas(n)) = (1/5)*(Lucas(3*n) - Lucas(-n)). In general, for r = s (mod 2) the sequence Lucas(r*n) - Lucas(s*n) is a divisibility sequence. Cf. A273622. - _Peter Bala_, May 27 2016

%F Lim_{n->infinity} a(n+1)/a(n) = 2 + sqrt(5) = A098317. - _Ilya Gutkovskiy_, Jun 01 2016

%F a(n) = (-(1/2*(-1-sqrt(5)))^n+(2-sqrt(5))^n-(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)/5. - _Colin Barker_, Jun 03 2016

%p seq((fibonacci(2*n)*fibonacci(n)), n=0..25); # _Zerinvary Lajos_, Jun 24 2006

%t Table[Fibonacci[n]Fibonacci[2n],{n,0,40}] (* _Harvey P. Dale_, Mar 13 2011 *)

%o (Magma) [Fibonacci(n)*Fibonacci(2*n): n in [0..30]]; // _Vincenzo Librandi_, Apr 18 2011

%o (PARI) concat([0], Vec( x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)) + O(x^66) ) ) \\ _Joerg Arndt_, Aug 26 2014

%Y Cf. A000032, A000045, A098317, A273622.

%K nonn,easy

%O 0,3

%A _Gary W. Adamson_, Feb 01 2000