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A079472
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Number of perfect matchings on an n X n L-shaped graph.
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15
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0, 2, 4, 12, 30, 80, 208, 546, 1428, 3740, 9790, 25632, 67104, 175682, 459940, 1204140, 3152478, 8253296, 21607408, 56568930, 148099380, 387729212, 1015088254, 2657535552, 6957518400, 18215019650, 47687540548, 124847601996, 326855265438, 855718194320
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OFFSET
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1,2
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COMMENTS
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a(n+1) = 2*F(n)*F(n+1) appears as the second component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, with F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), a(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
a(n+1) is the numerator 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 denominators, see A061646. - Greg Dresden and Max Liu, Jun 25 2023
For n >= 3, a(n) equals the sum of the sides of the right triangle with side lengths [(F(n)*F(n-3), 2*F(n-1)*F(n-2), F(2*n-3)] (n = 4 corresponds to the 3-4-5 right triangle). - Peter Bala, Nov 03 2023
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REFERENCES
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Daniele Corradetti, La Metafisica del Numero, 2008
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 178, 255.
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LINKS
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FORMULA
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a(n) = 2*F(n)*F(n-1) where F(n) are the Fibonacci numbers (A000045).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003: (Start)
a(n+1) = a(n) + 2*F(n)^2.
G.f.: 2*x^2/((1+x)*(1-3*x+x^2)). (End)
a(n) = Im( (F(n) + i*F(n+1))^2 ) (cf. A121646). - Daniele Corradetti (d.corradetti(AT)gmail.com), May 02 2008
a(n) = F(n+1)^2 - F(n)^2 - F(n-1)^2.
a(1 - n) = -a(n). (End)
a(n) = ( 2*(-1)^n - (1+sqrt(5))*((3-sqrt(5))/2)^n - (1-sqrt(5))*((3+sqrt(5))/2)^n )/5. - Colin Barker, Sep 27 2016
a(n) = F(2n-2) + F(n-1)^2, where F(n) is the n-th Fibonacci number.
a(n) = M^(n+1)[2,1], for n>0 where M=[0,0,1;0,1,2;1,1,1]. (End)
a(n) = F(n)^2 + F(n-1)^2 - F(n-2)^2. - Michael Somos, Mar 02 2023
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EXAMPLE
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a(7) = 2*13*8 = 208 = number of matchings. F(7) = 13 F(6) = 8
a(3) = 4 because in the graph with vertex set {(0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2)} and edge set {h(0,0), h(1,0), h(0,1), h(1,1), h(0,2), v(0,0), v(0,1), v(1,0), v(1,1), v(2,0)}, where h(i,j) (v(i,j)) is a horizontal (vertical) edge of unit length starting from vertex (i,j), we have the following four perfect matchings: {h(0,0), h(0,1), h(0,2), v(2,0)}, {h(0,0), v(0,1), v(1,1), v(2,0)}, {v(0,0), v(1,0), v(2,0), h(0,2)} and {v(0,0), h(1,0), h(1,1), h(0,2)}. - Emeric Deutsch, Dec 30 2004
G.f. = 2*x^2 + 4*x^3 + 12*x^4 + 30*x^5 + 80*x^6 + 208*x^7 + 546*x^8 + ...
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MAPLE
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with(combinat, fibonacci):seq(2*fibonacci(n)*fibonacci(n-1), n=1..30);
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MATHEMATICA
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PROG
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(PARI) {a(n) = 2 * fibonacci(n) * fibonacci(n-1)}; \\ Michael Somos, Jun 28 2014
(PARI) concat(0, Vec(2*x^2/((x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Sep 27 2016
(Magma) [2*Fibonacci(n)*Fibonacci(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 29 2014
(Sage) [2*fibonacci(n-1)*fibonacci(n) for n in (1..30)] # G. C. Greubel, Jan 07 2019
(GAP) List([1..30], n -> 2*Fibonacci(n-1)*Fibonacci(n)); # G. C. Greubel, Jan 07 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Helen King (h.king(AT)uea.ac.uk), Jan 15 2003
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EXTENSIONS
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More terms from Benoit Cloitre and Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003
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STATUS
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approved
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