A242300 a(n) = Sum_{0<=i<j<=n}L(i)*L(j), where L(k)=A000032(k) is the k-th Lucas number.

0, 2, 11, 35, 105, 292, 796, 2130, 5655, 14927, 39281, 103160, 270600, 709282, 1858291, 4867275, 12746265, 33375932, 87388676, 228801650, 599034975, 1568333527, 4106014561, 10749789360, 28143481680, 73680863042, 192899442971, 505018008755, 1322155461705

Offset

0,2

Comments

This sequence does for Lucas numbers what A190173 does for Fibonacci numbers.

Links

Paolo Xausa, Table of n, a(n) for n = 0..2000

Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).

Formula

The sums are (1) for L(2*k): (L(2*k+1)-1)^2 + L(2*k-1) + 1 and (2) for L(2*k+1): (L(2*k+2)-1)^2 + L(2*k) - 4.

G.f.: -x*(x^3+5*x^2-3*x-2) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). - Colin Barker, May 12 2014

a(n) = (L(n+1)-1)^2 + L(n-1) + (5*(-1)^n-3)/2. - Colin Barker, May 13 2014

Limit_{n->oo} a(n)/a(n-1) = A104457. - Bruce Nye, Jan 16 2026

Example

For L(12) = a(13) the sum is (L(13)-1)^2 + L(11) + 1 = 520^2 + 200 = 270600 and for L(13) = a(14) the sum is (L(14)-1)^2 + l(12) - 4 = 842^2 + 322 - 4 = 709282.

Mathematica

A242300[n_] := (LucasL[n+1] - 1)^2 + LucasL[n-1] + (5*(-1)^n - 3)/2; Array[A242300, 30, 0] (* or *)
LinearRecurrence[{4, -2, -6, 4, 2, -1}, {0, 2, 11, 35, 105, 292}, 30] (* Paolo Xausa, Jan 16 2026 *)

Prog

(PARI) concat(0, Vec(-x*(x^3+5*x^2-3*x-2)/((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)) + O(x^100))) \\ Colin Barker, May 13 2014
(SageMath)
[(lucas_number2(i+1, 1, -1)-1)^2+lucas_number2(i-1, 1, -1)+(5*(-1)^i-3)/2 for i in [0..50]] # Tom Edgar, May 13 2014

Crossrefs

Cf. A000032, A190173, A104457.

Sequence in context: A000914 A256317 A086735 * A078982 A154416 A184538

Adjacent sequences: A242297 A242298 A242299 * A242301 A242302 A242303

Keyword

nonn,easy

Author

J. M. Bergot, May 10 2014

Extensions

Typo in a(18) fixed by Colin Barker, May 12 2014

More terms from Colin Barker, May 12 2014

Status

approved