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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
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
This sequence does for Lucas numbers what A190173 does for Fibonacci numbers.
LINKS
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
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.
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
(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
CROSSREFS
Sequence in context: A000914 A256317 A086735 * A078982 A154416 A184538
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

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Last modified June 23 13:32 EDT 2024. Contains 373648 sequences. (Running on oeis4.)