|
%I #57 Jun 26 2023 17:58:39
%S 0,1,1,1,3,5,4,13,21,17,55,89,72,233,377,305,987,1597,1292,4181,6765,
%T 5473,17711,28657,23184,75025,121393,98209,317811,514229,416020,
%U 1346269,2178309,1762289,5702887,9227465,7465176,24157817,39088169,31622993
%N Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.
%C Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - _Jonas Holmvall_, Jun 21 2023
%H Vincenzo Librandi, <a href="/A167808/b167808.txt">Table of n, a(n) for n = 0..1000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,4,0,0,1).
%F a(n) = (a(n-1)*A131534(n) + a(n-2)*A131534(n+2))/A131534(n+1) for n > 1.
%F a(3*n) = A001076(n) = (a(3*n-1) + a(3*n-2))/2;
%F a(3*n+1) = A033887(n) = 2*a(3*n-1) + a(3*n-2);
%F a(3*n+2) = A015448(n+1) = a(3*n-1) + 2*a(3*n-2).
%F From _Johannes W. Meijer_, Jul 01 2010: (Start)
%F a(2*n) = A001906(n)/A131534(n+1) for n >= 0 and a(2*n+1) = A179131(n)/5 for n >= 1.
%F a(6*n+2) - 2*a(6*n) = A134493(n);
%F 2*a(6*n+1) - a(6*n+3) = A023039(n);
%F 2*a(6*n+4) - a(6*n+2) = A134497(n);
%F a(6*n+5) - 2*a(6*n+3) = A103134(n);
%F 2*a(6*n+4) - a(6*n+6) = A075796(n).
%F (End)
%F From _Gary Detlefs_, Dec 11 2010: (Start)
%F a(n) = numerator(A000045(n)/A000032(n)).
%F If n mod 3 = 0 then a(n) = Fibonacci(n)/2, else a(n)= Fibonacci(n). (End)
%F G.f.: -x*(1 + x + x^2 - x^3 + x^4) / ( (x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1) ). - _R. J. Mathar_, Mar 08 2011
%F a(n) = 4*a(n-3) + a(n-6). - _Muniru A Asiru_, Oct 16 2018
%p nmax:=39; x(0):=0: x(1):=1/2:for n from 2 to nmax do x(n):=x(n-1)+x(n-2) od: for n from 0 to nmax do a(n):= numer(x(n)) od: seq(a(n),n=0..nmax); # _Johannes W. Meijer_, Jul 01 2010
%p with(combinat):f:=n->fibonacci(n):L:=n->f(n)+2*f(n-1):seq(numer(f(n)/L(n)), n=0..39); # _Gary Detlefs_, Dec 11 2010
%t f[n_]:=Numerator[Fibonacci[n]/Fibonacci[n+3]];Array[f,100,0] (* _Vladimir Joseph Stephan Orlovsky_, Feb 17 2011*)
%t Numerator[LinearRecurrence[{1,1},{0,1/2},40]] (* _Harvey P. Dale_, Aug 08 2014 *)
%t CoefficientList[Series[-x (1 + x + x^2 - x^3 + x^4)/((x^2 + x - 1) (x^4 - x^3 + 2 x^2 + x + 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Aug 08 2014 *)
%t LinearRecurrence[{0, 0, 4, 0, 0, 1},{0, 1, 1, 1, 3, 5},40] (* _Ray Chandler_, Aug 03 2015 *)
%t a[n_]:=If[Mod[n,3]==0, Fibonacci[n]/2, Fibonacci[n]]; Array[a, 40, 0] (* _Stefano Spezia_, Oct 16 2018 *)
%o (GAP) a:=[0,1,1,1,3,5];; for n in [7..40] do a[n]:=4*a[n-3]+a[n-6]; od; a; # _Muniru A Asiru_, Oct 16 2018
%Y Cf. A000045, A130196 (denominator).
%Y The a(2*n) appear in A179135. - _Johannes W. Meijer_, Jul 01 2010
%K nonn,frac
%O 0,5
%A _Reinhard Zumkeller_, Nov 12 2009
%E Typo in title corrected by _Johannes W. Meijer_, Jun 26 2010
|
