A238419 a(n) = the Wiener index of the Fibonacci cube G_n.

0, 1, 4, 16, 54, 176, 548, 1667, 4968, 14592, 42348, 121728, 347112, 983173, 2768812, 7758928, 21648546, 60172784, 166687436, 460356359, 1267964496, 3483818880, 9550754520, 26129950080, 71356349520, 194529354505, 529485228244, 1439096616976, 3906061968654, 10588691040176, 28670559059444

Offset: 0

Author

Emeric Deutsch, Mar 26 2014

Keywords

• nonn
• easy

Comments

The Fibonacci cube G_n is defined in the Klavzar and Mollard reference (as Gamma_n).

Examples

			a(2)=4 because the Fibonacci cube G_2 is the path P_3 having Wiener index 1 + 1 + 2 = 4.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• S. Klavzar, M. Mollard, Wiener index and Hosoya polynomial of Fibonacci and Lucas cubes, MATCH Commun. Math. Comput. Chem., 68, 2012, 311-324.
• Eric Weisstein's World of Mathematics, Fibonacci Cube Graph
• Eric Weisstein's World of Mathematics, Wiener Index
• Index entries for linear recurrences with constant coefficients, signature (4,0,-10,0,4,-1).

Crossrefs

Cf. A000045.

Programs

Magma

[(4*(n+1)*Fibonacci(n)^2 + (9*n+2)*Fibonacci(n)*Fibonacci(n+1) + 6*n*Fibonacci(n+1)^2)/25: n in [0..30]]; // Vincenzo Librandi, Mar 27 2014

Maple

with(combinat): F := proc (n) options operator, arrow: Fibonacci(n) end proc: a := proc (n) options operator, arrow: (1/25)*(4*n+4)*F(n)^2+(1/25)*(9*n+2)*F(n)*F(n+1)+(6/25)*n*F(n+1)^2 end proc: seq(a(n), n = 0 .. 30);

Mathematica

wifc[n_]:=Module[{f1=Fibonacci[n],f2=Fibonacci[n+1]},(4*(n+1)*f1^2+ (9*n+2)*f1*f2+ 6*n*f2^2)/25]; wifc[Range[0, 30]] (* Harvey P. Dale, Sep 22 2014 *)
LinearRecurrence[{4, 0, -10, 0, 4, -1}, {0, 1, 4, 16, 54, 176}, 100] (* G. C. Greubel, Mar 27 2016 *)
CoefficientList[Series[x/(1 - 2 x - 2 x^2 + x^3)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 07 2017 *)
Table[(n LucasL[2 (n + 2)] + 2 Fibonacci[2 n + 1] - (n + 2) (-1)^n)/25, {n, 0, 20}] (* Eric W. Weisstein, Sep 07 2017 *)

PARI

a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -1,4,0,-10,0,4]^n*[0;1;4;16;54;176])[1,1] \\ Charles R Greathouse IV, Sep 07 2017

Formula

a(n) = ( 4*(n + 1)*F(n)^2 + (9*n + 2)*F(n)*F(n+1) + 6*n*F(n+1)^2 )/25, where F = A000045 (Fibonacci numbers).

a(n) = sum( F(i)*F(i+1)*F(n-i+1)*F(n-i+2), i=1..n ), where F = A000045.

Empirical g.f.: x / ((x+1)^2*(x^2-3*x+1)^2). - Colin Barker, Mar 26 2014

The g.f. above is correct because a(n) is the convolution of F(n)F(n+1) by F(n+1)F(n+2) (see Klavzar and Mollard paper). - Michel Mollard, Aug 20 2014.

The g.f. above is correct because it is equal to (dG/dt)A246173.%20-%20_Emeric%20Deutsch">{t=1}, where G is the bivariate g.f. given in A246173. - _Emeric Deutsch, Oct 01 2014

a(n+3) = (2*(n+3)*a(n+2)+2*(n+4)*a(n+1)-(n+5)*a(n))/(n+2). - Robert Israel, Aug 29 2014

25*a(n) = n*Lucas(2*(n+2))+2*Fibonacci(2*n+1)-(n+2)*(-1)^n. - Ehren Metcalfe, Mar 26 2016

a(n) = 4*a(n-1) - 10*a(n-3) + 4*a(n-5) - a(n-6). - G. C. Greubel, Mar 27 2016

25*a(n) = -A001906(n+2) +5*A001871(n) -(n+2)*(-1)^n. - R. J. Mathar, Jul 24 2022