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%I #24 Mar 27 2016 07:47:54
%S 1,5,13,89,233,1597,4181,28657,75025,514229,1346269,9227465,24157817,
%T 165580141,433494437,2971215073,7778742049,53316291173,139583862445,
%U 956722026041,2504730781961,17167680177565,44945570212853,308061521170129,806515533049393
%N Odd Fibonacci numbers with odd index.
%C All prime Fibonacci numbers (A005478) except 2 and 3 are in this sequence. All terms equal 1 (mod 4). The indices of these Fibonacci numbers are 6k-1 or 6k+1.
%C This sequence can be thought of as two interlocking sequences, each of the form a(n) = 18a(n - 1) - a(n - 2).
%C Proof that all terms equal 1 (mod 4): From the Lucas 1876 identity Fib(2n+1) = Fib(n)^2 + Fib(n+1)^2 (see Weisstein, formula 60, or page 79 of Koshy), we see that odd-indexed Fibonacci numbers are the sum of two squares. Because a square is 0 or 1 (mod 4), the sum of two squares is 0, 1, or 2 (mod 4). All these terms are odd numbers. Hence, the only possibility is that they are 1 (mod 4). This can also be proved from the recursion formula.
%D Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.
%H T. D. Noe, <a href="/A190949/b190949.txt">Table of n, a(n) for n = 1..200</a>
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,18,0,-1).
%F a(n) = 18*a(n-2) - a(n-4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
%F G.f.: x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)). - _Colin Barker_, Jun 19 2012
%F a(n) = (-(2-sqrt(5))^n + (-2-sqrt(5))^n*(-2+sqrt(5)) + 2*(-2+sqrt(5))^n + sqrt(5)*(-2+sqrt(5))^n + (2+sqrt(5))^n)/(2*sqrt(5)) for n>0. - _Colin Barker_, Mar 27 2016
%t LinearRecurrence[{0,18,0,-1}, {1,5,13,89}, 50]
%o (PARI) a(n)=fibonacci(n\2*6+if(n%2,1,-1)) \\ _Charles R Greathouse IV_, Jun 08 2011
%o (PARI) Vec(x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)) + O(x^30)) \\ _Colin Barker_, Mar 27 2016
%Y Cf. A000045 (Fibonacci numbers).
%K nonn,easy
%O 1,2
%A _T. D. Noe_, May 24 2011
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