A100673 A Graham-Pollak-like sequence with cube root instead of square root.

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

Offset: 0

Author

Jonathan Vos Post, Dec 06 2004

Keywords

• easy
• nonn

Comments

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.

Examples

			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.
		

References

• Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.
• Eric Weisstein's World of Mathematics, Graham-Pollak sequence

Crossrefs

Cf. A001521, A091522, A091523.

Programs

Mathematica

NestList[Floor[Surd[2#(#+1)(#+2),3]]&,1,50] (* Harvey P. Dale, Feb 24 2016 *)

Formula

a(0) = 1, a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3))