1,2

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.

I. Gutman and S. J. Cyvin, A result on 1-factors related to Fibonacci numbers, The Fibonacci Quarterly, 28 (1990), pp. 81-84.

a(n) = 2F(n)F(n-1) where F(n) are the Fibonacci numbers (A000045).

a(n) = 2F(n)F(n-1) = 2*A001654(n) = F(2n)-F(n)^2 = A001906(n)-A007598(n) = (F(n+1)^2-F(n-2)^2)/2 = (A007598(n+1)-A007598(n-2))/2 = 2(L(2*n-1)+(-1)^n)/5 = (2/5)*(A002878(n-1)+ A033999(n)), where L(n) = A000032(n); Recurrence: a(n) = a(n-1)+2*F(n)^2; G.f.: 2*x/((x+1)*(x^2-3*x+1)) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003

a(n) = Im( (F(n)+i*F(n+1))^2 ) (cf. A121646). - Daniele Corradetti (d.corradetti(AT)gmail.com), May 02 2008

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 (deutsch(AT)duke.poly.edu), Dec 30 2004

with(combinat, fibonacci):seq(2*fibonacci(n)*fibonacci(n-1), n=1..30);

Cf. A001654, A121646.

Sequence in context: A048618 A083554 A059412 this_sequence A006948 A148186 A148187

Adjacent sequences: A079469 A079470 A079471 this_sequence A079473 A079474 A079475

easy,nonn

Helen King (h.king(AT)uea.ac.uk), Jan 15 2003

More terms from Benoit Cloitre (benoit7848c(AT)orange.fr) and Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003