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A242300 a(n) = Sum_{0<=i<j<=n}L(i)*L(j), where L(k)=A000032(k) is the k-th Lucas number. 3

%I #41 Feb 09 2024 11:16:44

%S 0,2,11,35,105,292,796,2130,5655,14927,39281,103160,270600,709282,

%T 1858291,4867275,12746265,33375932,87388676,228801650,599034975,

%U 1568333527,4106014561,10749789360,28143481680,73680863042,192899442971,505018008755,1322155461705

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

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

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-6,4,2,-1).

%F 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.

%F 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

%F a(n) = (L(n+1)-1)^2 + L(n-1) + (5*(-1)^n-3)/2. - _Colin Barker_, May 13 2014

%e 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.

%o (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

%o (Sage)

%o [(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

%Y Cf. A000032, A190173.

%K nonn,easy

%O 0,2

%A _J. M. Bergot_, May 10 2014

%E Typo in a(18) fixed by _Colin Barker_, May 12 2014

%E More terms from _Colin Barker_, May 12 2014

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Last modified April 25 09:14 EDT 2024. Contains 371967 sequences. (Running on oeis4.)