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A166473
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a(n) = 2^L(n+1) * 3^L(n)/12, where L(n) is the n-th Lucas number (A000032(n)).
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3
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OFFSET
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1,1
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COMMENTS
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LINKS
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FORMULA
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For n>1, a(n) = 12*a(n-1) * a(n-2).
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MATHEMATICA
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Table[(2^LucasL[n+1] 3^LucasL[n])/12, {n, 10}] (* Harvey P. Dale, Aug 17 2011 *)
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PROG
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(PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1);
vector(10, n, 2^(lucas(n+1)-2)*3^(lucas(n)-1) ) \\ G. C. Greubel, Jul 22 2019
(Magma) [2^(Lucas(n+1)-2)*3^(Lucas(n)-1): n in [1..10]]; // G. C. Greubel, Jul 22 2019
(Sage) [2^(lucas_number2(n+1, 1, -1)-2)*3^(lucas_number2(n, 1, -1)-1) for n in (1..10)] # G. C. Greubel, Jul 22 2019
(GAP) List([1..10], n-> 2^(Lucas(1, -1, n+1)[2]-2)*3^(Lucas(1, -1, n)[2]-1)); # G. C. Greubel, Jul 22 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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