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A242300
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a(n) = Sum_{0<=i<j<=n}L(i)*L(j), where L(k)=A000032(k) is the k-th Lucas number.
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3
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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
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OFFSET
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0,2
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COMMENTS
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This sequence does for Lucas numbers what A190173 does for Fibonacci numbers.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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PROG
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(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
(Sage)
[(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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