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A236428
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a(n) = F(n+1)^2 + F(n+1)*F(n) - F(n)^2, where F = A000045.
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7
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1, 1, 5, 11, 31, 79, 209, 545, 1429, 3739, 9791, 25631, 67105, 175681, 459941, 1204139, 3152479, 8253295, 21607409, 56568929, 148099381, 387729211, 1015088255, 2657535551, 6957518401, 18215019649, 47687540549, 124847601995, 326855265439, 855718194319
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OFFSET
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0,3
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COMMENTS
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(a(n) + a(n+1))/2 = F(2n+2).
a(n) = -a(-n-1), using the negative Fibonacci values.
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
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REFERENCES
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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.
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LINKS
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FORMULA
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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) = 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
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MATHEMATICA
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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 *)
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PROG
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(PARI) F=fibonacci;
a(n)=F(n+1)^2 + F(n+1)*F(n) - F(n)^2;
(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
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CROSSREFS
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Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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