A037451 a(n) = Fibonacci(n) * Fibonacci(2*n).

0, 1, 3, 16, 63, 275, 1152, 4901, 20727, 87856, 372075, 1576279, 6676992, 28284569, 119814747, 507544400, 2149990983, 9107510539, 38580029568, 163427634589, 692290558575, 2932589884016, 12422650070163, 52623190204271, 222915410823168, 944284833600625, 4000054745057907, 16944503814103696, 71778070001033487

Offset: 0

Author

Gary W. Adamson, Feb 01 2000

Keywords

• nonn
• easy

Comments

Let F(n) = Fibonacci(n), then abs(det([F(n), F(n+k); F(n+2k), F(n+3k)])) = a(k), independent of n. - R. M. Welukar, Aug 26 2014

From Joerg Arndt, Aug 26 2014: (Start)

This is a special case of Johnson's identity (relation 32 in the Mathworld link).

F(a)*F(b) - F(c)*F(d) = (-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)), where a+b = c+d and r arbitrary.

Here a = n, b = n+3*k, c = n+k, d = n+2*k, and r = c, so that

(-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)) =

(-1)^c*(F(a-c)*F(b-c) - F(c-c)*F(d-c)) =

(-1)^c*(F(a-c)*F(b-c) - 0) =

(-1)^c*(F(-k)*F(-2*k)), taking the absolute value gives a(k).

(End)

Let L(n) = A000032(n), then abs(det([L(n), L(n+k); L(n+2k), L(n+3k)])) = 5*a(k), independent of n. - M. N. Deshpande and R. M. Welukar, Aug 30 2014

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Eric Weisstein, Fibonacci Number (MathWorld).
• Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Crossrefs

Cf. A000032, A000045, A098317, A273622.

Programs

Magma

[Fibonacci(n)*Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

Maple

seq((fibonacci(2*n)*fibonacci(n)), n=0..25); # Zerinvary Lajos, Jun 24 2006

Mathematica

Table[Fibonacci[n]Fibonacci[2n],{n,0,40}] (* Harvey P. Dale, Mar 13 2011 *)

PARI

concat([0], Vec( x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)) + O(x^66) ) ) \\ Joerg Arndt, Aug 26 2014

Formula

From Emanuele Munarini, Jul 18 2003: (Start)

G.f.: ( x + x^3 )/( 1 - 3 x - 6 x^2 + 3 x^3 + x^4 ).

a(n+4) = 3*a(n+3) + 6*a(n+2) - 3*a(n+1) - a(n).

(End)

G.f.: x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)). - Joerg Arndt, Aug 26 2014

a(n) = (1/5)*(Lucas(3*n) - (-1)^n*Lucas(n)) = (1/5)*(Lucas(3*n) - Lucas(-n)). In general, for r = s (mod 2) the sequence Lucas(r*n) - Lucas(s*n) is a divisibility sequence. Cf. A273622. - Peter Bala, May 27 2016

Lim_{n->infinity} a(n+1)/a(n) = 2 + sqrt(5) = A098317. - Ilya Gutkovskiy, Jun 01 2016

a(n) = (-(1/2*(-1-sqrt(5)))^n+(2-sqrt(5))^n-(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)/5. - Colin Barker, Jun 03 2016