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A061646
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a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) with a(-1) = 1, a(0) = 1, a(1) = 1.
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20
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1, 1, 1, 3, 7, 19, 49, 129, 337, 883, 2311, 6051, 15841, 41473, 108577, 284259, 744199, 1948339, 5100817, 13354113, 34961521, 91530451, 239629831, 627359043, 1642447297, 4299982849, 11257501249, 29472520899, 77160061447, 202007663443, 528862928881
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OFFSET
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-1,4
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COMMENTS
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Beginning at the well for the topograph of a positive definite quadratic form with values 1, 1, 1 at a superbase (i.e., 1, 1 and 1 are the vonorms of the superbase), these numbers indicate the values of the quadratic form at vectors adjacent to the path in the topograph of greatest rate of ascent of labels of the edges of the topograph.
For n > 1, a_n is the number of domino tilings of the L-grid obtained by removing the upper-right (n-1) X (n-2) rectangle from an (n+1) X n rectangle; also, for n > 1, (2*a_n)^2 is the number of domino tilings of the holey square obtained by removing the centered (n-2) X (n-2) square from an (n+2) X (n+2) square. - Roberto Tauraso, Jun 05 2004
Let P denote the 3 X 3 Fibonacci matrix [ 0 0 1 / 0 1 2 / 1 1 1 ]. Then a(n) is the central term of P^n. - Gary W. Adamson, May 13 2003
Coefficient of 1 when looking for the simplest linear dependence between (phi^(n-1) + Fibonacci(n-1)) / (phi^n + Fibonacci(n)) - 1/phi, 1 and phi. Thus a(n) is given by lindep([phi^(n-1) + Fibonacci(n-1))/(phi^n + Fibonacci(n))-1/phi,1,phi],80)[2] when using PARI/GP with enough digits of precision. - Thomas Baruchel, Nov 19 2004
a(n), n >= 2, is twice the area of the plane triangle in three-dimensional space with vertices (F(n-1),0,0), (0,F(n),0) and (0,0,F(n+1)). See the Atanassov et al. reference p. 88 (misprint in eq. (1.3): it should read F_{2n-1} not (F_{2n-1})^2). - Wolfdieter Lang, Jul 22 2005
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(10,2) =
(0 0 1 0 0)
(0 1 0 1 0)
(1 0 1 0 1)
(0 1 0 2 0)
(0 0 2 0 1).
Then a(n) = (Trace(U^n))/5 = (U^n)_(5,5) = ((U^n)_(1,1)+(U^n)_(4,4))/2 = ((U^n)_(2,2)+(U^n)_(3,3))/2, n>=0, recalling that the offset for A061646 is -1. (See also A189316.) (End)
a(n+1) is the denominator of the continued fraction [1,...,1,2,1,...,1] with n 1's to the left of the central 2, and n 1's to the right of the central 2. For the numerators, see A079472. - Greg Dresden and Max Liu, Jun 25 2023
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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.
J. H. Conway, The Sensual (Quadratic) Form, MAA.
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LINKS
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I. Amburg, K. Dasaratha, L. Flapan, T. Garrity, C. Lee, C. Mihailak, N. Neumann-Chun, S. Peluse, M. Stoffregen, Stern Sequences for a Family of Multidimensional Continued Fractions: TRIP-Stern Sequences, arXiv:1509.05239 [math.CO], 2015-2017.
I. Amburg, K. Dasaratha, L. Flapan, T. Garrity, C. Lee, C. Mihailak, N. Neumann-Chun, S. Peluse, M. Stoffregen, Stern Sequences for a Family of Multidimensional Continued Fractions: TRIP-Stern Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.1.7.
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FORMULA
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a(n) = Fibonacci(n)^2 + Fibonacci(n)*Fibonacci(n-1) + Fibonacci(n-1)^2 = A007598(n+1) - A001654(n-1). Area of triangle with sides sqrt(a(n)), sqrt(a(n-1)) and sqrt(a(n-2)) is sqrt(3)/4, i.e., 2*(a(n)*a(n-1) + a(n)*a(n-2) + a(n-1)*a(n-2)) - (a(n)^2 + a(n-1)^2 + a(n-2)^2) = 3. - Henry Bottomley, Jan 09 2003
a(n) = (2*Fibonacci(n)*Fibonacci(n+1)*(Fibonacci(n+2) + phi*Fibonacci(n+1)) + (1/phi)^n)/(Fibonacci(n)*phi + Fibonacci(n+1)). - Thomas Baruchel, Nov 19 2004
a(n) = F(2*n-1) + F(n-1)*F(n), with F(-3):=2, F(-2):=-1 and F(-1):=1 (corrected eq. (1.3) of the Atanassov et al. reference) with F(n):=A000045 (Fibonacci). - Wolfdieter Lang, Jul 22 2005
a(n) = (1/5)*Sum_{k=1..5) ((w_k)^2-1)^n, w_k = 2*cos((2*k-1)*Pi/10), n >= 0, recalling that the sequence offset is -1. - L. Edson Jeffery, Apr 20 2011
a(n) = ((-2)^n + 2*(3-sqrt(5))^n + 2*(3+sqrt(5))^n)/(5*2^n). - L. Edson Jeffery, Apr 21 2011
a(n) = L(2*n-2) + F(n-3)*F(n-2) for n >= 3, where L=A000032 are Lucas numbers. - J. M. Bergot, Aug 08 2012
a(n) = (F(n-1)^2 + F(n)^2 + F(n+1)^2)/2.
a(n) = F(n+1)*F(n-1) + F(n)^2.
a(n) = F(n+1)^2 - F(n)*F(n-1). (In addition, see Nov 20 2010 Formula entry from Gary Detlefs.) (End)
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EXAMPLE
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a(7)=337 since 2*a(6) + 2*a(5) - a(4) = 2*129 + 2*49 - 19 = 337.
a(7)=337 since (F(9)^2 + 3* F(6)^2)/4 = (34^2 + 3*8^2) = 1348/4 = 337. - Philippe Deléham, Oct 17 2020
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MAPLE
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with (combinat):
F:= n-> fibonacci(n):
seq (F(n)*F(n+1) +F(n-1)^2, n=-1..27);
seq(2*F(n)^2+(-1)^n, n=0..27);
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MATHEMATICA
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#[[1]]^2+#[[2]]^2+#[[1]]#[[2]]&/@Partition[Fibonacci[Range[-2, 30]], 2, 1] (* Harvey P. Dale, Feb 11 2022 *)
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PROG
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(Haskell)
a061646 n = a061646_list !! (n + 1)
a061646_list = 1 : 1 : 1 : zipWith (-) (map (* 2)
(zipWith (+) (drop 2 a061646_list) (tail a061646_list))) a061646_list
(Magma) I:=[1, 1, 1]; [n le 3 select I[n] else 2*Self(n-1) +2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 07 2019
(Sage) [2*fibonacci(n)^2 + (-1)^n for n in (-1..30)] # G. C. Greubel, Jan 07 2019
(GAP) a:=[1, 1, 1];; for n in [4..30] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 07 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Darrin Frey (freyd(AT)cedarville.edu), Jun 14 2001
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STATUS
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approved
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