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a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n-2).
2

%I #24 Jul 13 2021 19:21:04

%S 1,7,28,123,515,2192,9269,39291,166396,704935,2986039,12649248,

%T 53582777,226980767,961505180,4073002563,17253513691,73087060144,

%U 309601749709,1311494066355,5555578003196,23533806098447,99690802365743,422297015611968,1788878864731825

%N a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n-2).

%C An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons. Type A have sides A066259(n+1), A220360(n+1), A066259(n+1), A220360(n+1), A066259(n+1), and opposite diagonals A056570(n+2), A056570(n+2), A220361(n+2), A056570(n+2), A056570(n+2), for n=1,2,...

%D R. K. Guy, Unsolved Problems in Number Theory, D20.

%H Indranil Ghosh, <a href="/A220361/b220361.txt">Table of n, a(n) for n = 2..1593</a>

%H J. H. Jordan, B. E. Peterson, <a href="http://projecteuclid.org/euclid.rmjm/1181072619">Almost regular integer Fibonacci pentagons</a>, Rocky Mountain J. Math. Volume 23, Number 1 (1993), 243-247.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).

%F a(n) = 3*a(n-1)+6*a(n-2)-3*a(n-3)-a(n-4). G.f.: x^2*(x^2+4*x+1) / ((x^2-x-1)*(x^2+4*x-1)). - _Colin Barker_, Sep 23 2014

%p with(combinat): A220361:=n->fibonacci(n)^3+(-1)^n*fibonacci(n-2): seq(A220361(n), n=2..30); # _Wesley Ivan Hurt_, Apr 26 2017

%t Table[Fibonacci[n]^3 + (-1)^n * Fibonacci[n - 2], {n, 2, 30}] (* _T. D. Noe_, Dec 13 2012 *)

%t LinearRecurrence[{3,6,-3,-1},{1,7,28,123},30] (* _Harvey P. Dale_, Jul 13 2021 *)

%o (PARI) Vec(x^2*(x^2+4*x+1)/((x^2-x-1)*(x^2+4*x-1)) + O(x^100)) \\ _Colin Barker_, Sep 23 2014

%o (PARI) a(n) = fibonacci(n)^3 + (-1)^n*fibonacci(n-2) \\ _Charles R Greathouse IV_, Feb 14 2017

%K nonn,easy

%O 2,2

%A _Michel Marcus_, Dec 12 2012