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A061646
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a(-1) = 1, a(0) = 1, a(1) = 1; for n>1, a(n) = 2*a(n-1)+2*a(n-2)-a(n-3).
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11
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| -1,4
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COMMENTS
| 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 a (n+1)xn 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 a (n+2)x(n+2) square. - Roberto Tauraso, Jun 05 2004.
Let P = 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 (qntmpkt(AT)yahoo.com), 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 (baruchel(AT)users.sourceforge.net), 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). W. Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Jul 22 2005.
Contribution by L. Edson Jeffery, April 20, 2011. (Start):
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)
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REFERENCES
| J. H. Conway, The Sensual (Quadratic) Form, MAA.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, Wiley-Interscience, 2001; pages 383-384.
K. T. Atanassov, V. Atanassova, A. G. Shannon and J. C. Turner, New visual perspectives on Fibonacci numbers, World Scientific, 2002.
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LINKS
| Reinhard Zumkeller, Table of n, a(n) for n = -1..1000
Hwa, Theodore, Posting to sci.math.
L. E. Jeffery, Unit-primitive matrices.
Roberto Tauraso, A New Domino Tiling Sequence, Journal of Integer Sequences, Article 04.2.3 (2004).
Roberto Tauraso, A New Domino Tiling Sequence, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.3.
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FORMULA
| a(n) =Fib(n)^2+Fib(n)*Fib(n-1)+Fib(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 (se16(AT)btinternet.com), 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 (baruchel(AT)users.sourceforge.net), 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). - W. Lang, Jul 22 2005
G.f.: (1-x-x^2)/((1+x)*(1-3x+x^2)), for n>=0 . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 16 2008]
a(n) = F(n)*F(n+1) + F(n-1)^2, n>=-1
a(n) = sum(F(k)^2, k=0..n) + F(n-1)^2, n>=-1
a(n)=2*F(n)^2+(-1)^n, n>-1
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 offset for A061646 is -1. - L. Edson Jeffery, April 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.
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EXAMPLE
| a(7)=337 since 2*a(6)+2*a(5)-a(4)=2*129+2*49-19=337
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MAPLE
| with (combinat):
F:= n-> fibonacci(n):
seq (F(n)*F(n+1) +F(n-1)^2, n=-1..27);
seq (sum(F(k)^2, k=0..n) +F(n-1)^2, n=-1..27);
seq(2*F(n)^2+(-1)^n, n=0..27);
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MATHEMATICA
| LinearRecurrence[{2, 2, -1}, {1, 1, 1}, 100] (* From Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
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PROG
| (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
-- Reinhard Zumkeller, Dec 02 2011
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CROSSREFS
| Cf. A389316.
Sequence in context: A007288 A191824 A191757 * A017926 A017927 A116903
Adjacent sequences: A061643 A061644 A061645 * A061647 A061648 A061649
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KEYWORD
| nonn,changed
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AUTHOR
| Darrin Frey (freyd(AT)cedarville.edu), Jun 14 2001
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