

A190949


Odd Fibonacci numbers with odd index.


2



1, 5, 13, 89, 233, 1597, 4181, 28657, 75025, 514229, 1346269, 9227465, 24157817, 165580141, 433494437, 2971215073, 7778742049, 53316291173, 139583862445, 956722026041, 2504730781961, 17167680177565, 44945570212853, 308061521170129, 806515533049393
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OFFSET

1,2


COMMENTS

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 6k1 or 6k+1.
This sequence can be thought of as two interlocking sequences, each of the form a(n) = 18a(n  1)  a(n  2).
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 oddindexed 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.


REFERENCES

Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.


LINKS



FORMULA

a(n) = 18*a(n2)  a(n4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
G.f.: x*(1x)*(1+6*x+x^2)/((1+4*xx^2)*(14*xx^2)).  Colin Barker, Jun 19 2012
a(n) = ((2sqrt(5))^n + (2sqrt(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


MATHEMATICA

LinearRecurrence[{0, 18, 0, 1}, {1, 5, 13, 89}, 50]


PROG

(PARI) Vec(x*(1x)*(1+6*x+x^2)/((1+4*xx^2)*(14*xx^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



