This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A100673 #12 Feb 16 2025 08:32:55 %S A100673 1,2,3,4,6,8,11,15,20,26,34,44,56,71,90,114,144,182,230,291,367,463, %T A100673 584,737,929,1171,1476,1860,2344,2954,3723,4691,5911,7448,9385,11825, %U A100673 14899,18772,23652,29800,37546,47306,59603,75096,94616,119209,150195,189235 %N A100673 A Graham-Pollak-like sequence with cube root instead of square root. %C A100673 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 A100673 Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003. %H A100673 Harvey P. Dale, <a href="/A100673/b100673.txt">Table of n, a(n) for n = 0..1000</a> %H A100673 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 A100673 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Graham-PollakSequence.html">Graham-Pollak sequence</a> %F A100673 a(0) = 1, a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3)) %e A100673 a(6) = 11 because a(5) = 8; so a(6) = Floor((2*8*(8+1)*(8+2))^(1/3)) %e A100673 = floor(1440^(1/3)) = 11, which happens to be prime. %e A100673 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 A100673 NestList[Floor[Surd[2#(#+1)(#+2),3]]&,1,50] (* _Harvey P. Dale_, Feb 24 2016 *) %Y A100673 Cf. A001521, A091522, A091523. %K A100673 easy,nonn %O A100673 0,2 %A A100673 _Jonathan Vos Post_, Dec 06 2004