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3, 7, 5, 9, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 281, 311, 347, 419, 431, 461, 521, 569, 599, 617, 641, 659, 809, 821, 827, 857, 881, 1019, 1031, 1049, 1061, 1091, 1151, 1229, 1277, 1289, 1301, 1319, 1427, 1451, 1481, 1487
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refs;
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history;
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OFFSET
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1,1
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COMMENTS
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Let p and p+2 be twin primes. Then Lucas(p) = 1 + p*A006206(p) and Lucas(p+2) = 1 + (p+2)*A006206(p+2). It follows from Lucas(n) + Lucas(n+1) = Lucas(n+2) that p = (Lucas(p+1) - 2*A006206(p+2))/(A006206(p+2) - A006206(p))
For i = 3, 4, 5, 6, 7, 8, 9, 10, 11: ((Lucas(i+1) - 2*A006206(i+2))/(A006206(i+2) - A006206(i))) = (3, 7, 5, 19/3, 31/4, 9, 87/10, 149/14, 11, 135/11, 663/50, 1094/77, 1787/120, 2939/181, 17, 7849/434, 12799/672, 20894/1041, 34031/1622, 55469/2514, 45131/1962, 146921/6115, 238915/9554, 194252/7465, 631347/23386, 1025917/36617, 29, 2706059/90178, 4393211/141710, 3565643/111405, 11573003/350702). - Creighton Dement, Jan 31 2006
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LINKS
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FORMULA
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It is conjectured that a(n+4) = A001359(n+2) for all n.
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MAPLE
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# First 63 Terms with(combinat): with(numtheory): A006206 := proc(n) local sum; sum := 0; for d in divisors(n) do sum := sum + mobius(n/d)*(fibonacci(d+1)+fibonacci(d-1)) od; RETURN(sum/n); end; A000204 := n->fibonacci(n+1)+fibonacci(n-1); T := n -> (A000204(n+1) - 2*A006206(n+2))/(A006206(n+2)-A006206(n)); A113910 := []: for i from 3 by 1 to 2000 do if is(T(i) = floor(T(i))) then A113910 := [op(A113910), T(i)]; fi: od: A113910; # Creighton Dement, Jan 15 2009
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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