cp's OEIS Frontend

A097811 Numbers n such that n^3 is the sum of three or more consecutive positive cubes.

%I A097811 #37 Jun 19 2020 04:01:26
%S A097811 6,20,40,60,70,180,330,540,1155,1581,2805,2856,3876,5544,16830,27060,
%T A097811 62244,82680,90090,175440,237456,249424,273819,413820,431548,534660,
%U A097811 860706,1074744,1205750,1306620,1630200,1764070,1962820,1983150
%N A097811 Numbers n such that n^3 is the sum of three or more consecutive positive cubes.
%C A097811 These numbers were found by exhaustive search. The sums are not unique; for n=2856, there are two representations. The Mathematica code prints n, the range of cubes in the sum and the number of cubes in the sum. For instance, 82680^3 equals the sum of 6591 cubes! A faster program was used to check all sums s of consecutive cubes such that s < 2000000^3.
%C A097811 2856^3 is the only cube < 2*10^23 that is a sum in two different ways. 2856^3 = 213^3 +...+ 555^3 = 273^3 +...+ 560^3. - _Donovan Johnson_, Feb 22 2011
%C A097811 The terms of this sequence tend to contain only small primes. Is a(n)^(1/3) an upper bound for the largest prime factor of a(n)? - _Ralf Stephan_, May 22 2013
%C A097811 Note that by Fermat's theorem no cube is the sum of two positive cubes.
%C A097811 The cubes of the terms form a subsequence of A265845 (numbers that are sums of consecutive positive cubes in more than one way) which is sparse: among the first 1000 terms of A265845, only 17 are cubes. - _Jonathan Sondow_, Jan 10 2016
%H A097811 Chai Wah Wu, <a href="/A097811/b097811.txt">Table of n, a(n) for n = 1..68</a> (terms n = 1..55 from Donovan Johnson)
%H A097811 Michael Bennett, Vandita Patel, and Samir Siksek, <a href="https://arxiv.org/abs/1603.08901">Perfect powers that are sums of consecutive cubes</a>, arXiv:1603.08901 [math.NT], 2016. [But typo on last line of article where 1115 should be 1155]
%H A097811 K. S. Brown, <a href="http://www.mathpages.com/home/kmath147.htm">Sum of Consecutive Nth Powers Equals an Nth Power</a>
%F A097811 a(n) = A131643(n)^(1/3). - _Jonathan Sondow_, Jan 10 2016
%e A097811 20 is in this sequence because 11^3 + 12^3 + 13^3 + 14^3 = 20^3.
%t A097811 g[m0_, m1_] := (m1-m0+1)(m0+m1)(m0^2+m1^2+m1-m0)/4; lst={}; Do[n=g[m0, m1]^(1/3); If[IntegerQ[n], Print[{n, m0, m1, m1-m0+1}]; AppendTo[lst, n]], {m1, 2, 14000}, {m0, m1-1, 1, -1}]; Union[lst]
%Y A097811 Cf. A097812 (n^2 is the sum of consecutive squares), A265845.
%Y A097811 See A131643 for the actual cubes.
%K A097811 nonn
%O A097811 1,1
%A A097811 _T. D. Noe_, Aug 25 2004; Sep 07 2004
%E A097811 Name edited by _Altug Alkan_, Dec 07 2015