login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A Graham-Pollak-like sequence with cube root instead of square root.
1

%I #11 Sep 21 2017 03:46:50

%S 1,2,3,4,6,8,11,15,20,26,34,44,56,71,90,114,144,182,230,291,367,463,

%T 584,737,929,1171,1476,1860,2344,2954,3723,4691,5911,7448,9385,11825,

%U 14899,18772,23652,29800,37546,47306,59603,75096,94616,119209,150195,189235

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

%C 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.

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

%H Harvey P. Dale, <a href="/A100673/b100673.txt">Table of n, a(n) for n = 0..1000</a>

%H R. L. Graham and H. O. Pollak, <a href="http://www.jstor.org/stable/2688390">Note on a nonlinear recurrence related to sqrt(2)</a>, Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Graham-PollakSequence.html">Graham-Pollak sequence</a>

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

%e a(6) = 11 because a(5) = 8; so a(6) = Floor((2*8*(8+1)*(8+2))^(1/3))

%e = floor(1440^(1/3)) = 11, which happens to be prime.

%e 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.

%t NestList[Floor[Surd[2#(#+1)(#+2),3]]&,1,50] (* _Harvey P. Dale_, Feb 24 2016 *)

%Y Cf. A001521, A091522, A091523.

%K easy,nonn

%O 0,2

%A _Jonathan Vos Post_, Dec 06 2004