A292696 a(n) = L(n)^2 - 5*(-1)^n = L(n+1)*L(n-1), where L = A000032.

-1, 6, 4, 21, 44, 126, 319, 846, 2204, 5781, 15124, 39606, 103679, 271446, 710644, 1860501, 4870844, 12752046, 33385279, 87403806, 228826124, 599074581, 1568397604, 4106118246, 10749957119, 28143753126, 73681302244, 192900153621, 505019158604, 1322157322206, 3461452807999

Offset: 0

Author

Bruno Berselli, Sep 21 2017

Keywords

• sign
• easy

Comments

This is the case k=1 of the identity L(n)^2 + 5*F(k)^2*(-1)^(n+k) = L(n+k)*L(n-k), where F = A000045. See also the comment in A292612.

References

• Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover Publications (2008), page 29 (the formula 20b, for h=-k, gives the identity in Comments section).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Ron Knott, Fibonacci and Lucas Numbers (see 31st formula: the identity in the Comments section is the case i=-k).
• Eric Weisstein's World of Mathematics, Lucas Number (see the formula n. 5).
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A000032, A001622, A001654, A292612.

Cf. A059929: Fibonacci(n+2)*Fibonacci(n).

Programs

GAP

a:=[-1,6,4];; for n in [4..10^3] do a[n]:= 2*a[n-1]+2*a[n-2]-a[n-3]; od; A292696:=a; # Muniru A Asiru, Oct 03 2017

Magma

[Lucas(n)^2-5*(-1)^n: n in [0..40]];

Maple

A292696:=proc(n) option remember:
if n=0 then -1 elif n=1 then 6 elif n=2 then 4 elif  n>=3 then 2*procname(n-1)+2*procname(n-2)-procname(n-3) fi; end:
seq(A292696(n),n=0..10^2); # Muniru A Asiru, Oct 03 2017

Mathematica

Table[LucasL[n]^2 - 5 (-1)^n, {n, 0, 40}]
LinearRecurrence[{2,2,-1},{-1,6,4},40] (* Harvey P. Dale, Oct 02 2018 *)

PARI

Vec(-(1 - 8*x + 6*x^2) / ((1 + x)*(1 - 3*x + x^2)) + O(x^30)) \\ Colin Barker, Sep 21 2017

Sage

[lucas(n)^2-5*(-1)^n for n in range(40)]

Formula

G.f.: (-1 + 8*x - 6*x^2)/((1 + x)*(1 - 3*x + x^2)).

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

a(n) = -A001654(n+1) + 8*A001654(n) - 6*A001654(n-1) with A001654(-1)=0.

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

Sum_{i=0..n} a(i) = L(2*n+1) - (3*(-1)^n + 1)/2.

a(n) = 2^(-n)*(-3*(-2)^n + (3-sqrt(5))^n + (3+sqrt(5))^n). - Colin Barker, Sep 21 2017

From Peter Bala, Oct 14 2019: (Start)

Sum_{n >= 1} L(n)/a(n) = 3/2.

Sum_{n >= 1} (-1)^n*L(n)/a(n) = 1/2. (End)

From Amiram Eldar, Oct 06 2020: (Start)

Sum_{n>=1} 1/a(n) = 1/2.

Sum_{n>=1} (-1)^n/a(n) = 7/10 - 2*phi/5, where phi is the golden ratio (A001622). (End)