OFFSET
0,3
COMMENTS
(a(n) + a(n+1))/2 = F(2n+2).
a(n) = -a(-n-1), using the negative Fibonacci values.
First differences equal 2*A059929.
Partial sums equal A192873.
Unlike Fibonacci, the divisibility of a(n) by the primes is quite limited, specifically to p = 5, 11, 19, 31, 59, 71, 79, 109, ... where those after 5 are only a subset of primes congruent to {1,4} mod 5.
Values of a(n) mod p, for all primes p exhibit repeating pattern cycles of length k = (p-1)/m or (p+1)/m (except p = 5), based on whether p is congruent to {1,4} mod 5 or {2,3} mod 5. For p = 5, k = 2p = 10. Only the slightest similarity exists here with Fibonacci: there are formulas like this for a cycle length k, but for Fibonacci those are "divisibility cycles" for prime p, not the "pattern cycles" on mod p, and the m values differ for many primes, creating different cycle lengths for the same p.
a(n) has the property: a(k/2 + i) mod p + a(k/2 - 1 - i) mod p = p or 0, for all primes p, and all i 0 <= i <= k/2, in every cycle of length k. Thus, when plotted, the lower and upper halves of a every cycle have an inverted (i.e., flipped) symmetry.
For some primes (e.g., 13, 17, 37, 53, 61, 89, 97) each half-cycle (of length k/2) is internally symmetric (i.e., the second quarter-cycle is a mirror image of the first quarter cycle, and the fourth is a mirror image of the third, on each side of some value at k/4), while the flipped symmetry still holds for the upper and lower halves. See example for p = 61, with k = 30 in pdf file below.
No such symmetries on mod p, of either type, exist for Fibonacci.
a(n) is also (apart from sign) the determinant of a 2 X 2 matrix of squares of successive Fibonacci numbers: a(n) = (-1)^(n)*(F(n+2)^2*F(n-1)^2 -F(n)^2*F(n+1)^2). - R. M. Welukar, Aug 30 2014
For n>1 a(n) is the ceiling of the maximum area of a quadrilateral having sides of length in increasing order F(n), F(n+1), L(n), and L(n+1) with L(n)=A000032(n). - J. M. Bergot, Jan 19 2016
For n>1 a(n) is the numerator of the continued fraction [1, 1, ... 1, 2, 1, 1, ... 1, 2] with n-2 1's before each 2. - Greg Dresden and Kevin Zhanming Zheng, Aug 16 2020
a(n) is the number of edge covers in the rocket graph R_{3,n+1,n}. A rocket graph R_{m,i,j} is a m-cycle with two paths attached to adjacent vertices of the cycle, which have lengths i and j respectively. This is similar to a tadpole graph but with two tails. - Bridget Rozema, Oct 09 2024
REFERENCES
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Richard R. Forberg, Plot of a(n) mod 61
Bridget Rozema and Maisie Smith, Edge Covers of Unions of Path and Cycle Graphs, 2024.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
FORMULA
G.f.: (x^2 - x + 1)/((x + 1)*(x^2 - 3*x + 1)). - Joerg Arndt, Feb 23 2014
a(n) = (2*Lucas(2*n+1) + 3*(-1)^n)/5. - Ralf Stephan, Feb 27 2014
a(n) = 2*a(n-1) + 2*a(n-2)-a(n-3). - Colin Barker, Dec 20 2014
a(n) = F(n-1)*F(n+2) + F(n)*F(n+1). - J. M. Bergot, Dec 20 2014
a(n) = 2*F(n)*F(n+1) + (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = F(2*n+1) - F(n-1)^2 +(-1)^n for n>0. - J. M. Bergot, Jan 19 2016
a(n) = (2^(-n)*(3*(-2)^n-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
a(n) = F(n+1)^2 + F(n)*F(n-1). See also A099016, tenth formula. - Bruno Berselli, Feb 15 2017
2*a(n) = L(n)*L(n+1) - F(n)*F(n+1), where L = A000032. - Bruno Berselli, Sep 27 2017
MATHEMATICA
a[n_] := Fibonacci[n+1]^2 + Fibonacci[n+1]*Fibonacci[n] - Fibonacci[n]^2; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Feb 27 2014 *)
LinearRecurrence[{2, 2, -1}, {1, 1, 5}, 40] (* Vincenzo Librandi, Jan 20 2016 *)
PROG
(PARI) F=fibonacci;
a(n)=F(n+1)^2 + F(n+1)*F(n) - F(n)^2;
vector(33, n, a(n-1)) \\ Joerg Arndt, Feb 23 2014
(PARI) Vec((x^2-x+1)/((x+1)*(x^2-3*x+1)) + O(x^100)) \\ Colin Barker, Dec 20 2014
(PARI) a(n) = round((2^(-n)*(3*(-2)^n-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 28 2016
(Magma) [Fibonacci(n+1)^2+Fibonacci(n+1)*Fibonacci(n)- Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Jan 20 2016
(Magma) F:=Fibonacci; [F(n+1)^2+F(n)*F(n-1): n in [0..30]]; // Bruno Berselli, Feb 15 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard R. Forberg, Jan 25 2014
STATUS
approved