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A100673
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A Graham-Pollak-like sequence with cube root instead of square root.
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1
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1, 2, 3, 4, 6, 8, 11, 15, 20, 26, 34, 44, 56, 71, 90, 114, 144, 182, 230, 291, 367, 463, 584, 737, 929, 1171, 1476, 1860, 2344, 2954, 3723, 4691, 5911, 7448, 9385, 11825, 14899, 18772, 23652, 29800, 37546, 47306, 59603, 75096, 94616, 119209, 150195, 189235
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OFFSET
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0,2
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COMMENTS
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When the multiplier in the recurrence is 2 and the recurrence has two terms inside a square root, we have the Graham-Pollak sequence, where there is a remarkable exact explicit formula for a(n) in terms of the union of the set of integers and the set of integer multiples of Sqrt(2). As Weisstein summarizes Borwein & Bailey: "It is not known if sequences such as ... a(n) = a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3)) have corresponding properties." This sequence is the given one, having the multiplier in the recurrence as 2 and three terms inside a cube root and with a(0) = 1. Through n=50, the primes are when n = 1, 2, 3, 6, 13, 20, 21, 24, 25, 31. Through n=50, the semiprimes are when n = 4, 7, 9, 10, 19, 23, 32, 34, 36, 40, 42, 47, 49.
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REFERENCES
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Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003.
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LINKS
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FORMULA
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a(0) = 1, a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3))
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EXAMPLE
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a(6) = 11 because a(5) = 8; so a(6) = Floor((2*8*(8+1)*(8+2))^(1/3))
= floor(1440^(1/3)) = 11, which happens to be prime.
a(45) = 119209 because a(44) = 94616, so a(45) = Floor((2*94616*(94616+1)*(94616+2))^(1/3)) = floor(1694094050176992^(1/3)) = 119209 = 23 * 71 * 73, which happens to be a 3-brilliant number.
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MATHEMATICA
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NestList[Floor[Surd[2#(#+1)(#+2), 3]]&, 1, 50] (* Harvey P. Dale, Feb 24 2016 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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