cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A190949 Odd Fibonacci numbers with odd index.

Original entry on oeis.org

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
Offset: 1

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Author

T. D. Noe, May 24 2011

Keywords

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 6k-1 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 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.

References

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

Crossrefs

Intersection of A001519 and A014437.
Cf. A000045 (Fibonacci numbers).

Programs

  • Mathematica
    LinearRecurrence[{0,18,0,-1}, {1,5,13,89}, 50]
  • PARI
    a(n)=fibonacci(n\2*6+if(n%2,1,-1)) \\ Charles R Greathouse IV, Jun 08 2011
    
  • 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

Formula

a(n) = 18*a(n-2) - a(n-4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
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
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
a(n) = A000045(A007310(n)). - Amiram Eldar, Jul 25 2024