OFFSET
1,1
COMMENTS
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.
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
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
Note that by Fermat's theorem no cube is the sum of two positive cubes.
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
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..68 (terms n = 1..55 from Donovan Johnson)
Michael Bennett, Vandita Patel, and Samir Siksek, Perfect powers that are sums of consecutive cubes, arXiv:1603.08901 [math.NT], 2016. [But typo on last line of article where 1115 should be 1155]
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
FORMULA
a(n) = A131643(n)^(1/3). - Jonathan Sondow, Jan 10 2016
EXAMPLE
20 is in this sequence because 11^3 + 12^3 + 13^3 + 14^3 = 20^3.
MATHEMATICA
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Aug 25 2004; Sep 07 2004
EXTENSIONS
Name edited by Altug Alkan, Dec 07 2015
STATUS
approved