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

6, 20, 40, 60, 70, 180, 330, 540, 1155, 1581, 2805, 2856, 3876, 5544, 16830, 27060, 62244, 82680, 90090, 175440, 237456, 249424, 273819, 413820, 431548, 534660, 860706, 1074744, 1205750, 1306620, 1630200, 1764070, 1962820, 1983150

Offset: 1

Author

T. D. Noe, Aug 25 2004; Sep 07 2004

Keywords

• nonn

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

Examples

			20 is in this sequence because 11^3 + 12^3 + 13^3 + 14^3 = 20^3.
		

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

Crossrefs

Cf. A097812 (n^2 is the sum of consecutive squares), A265845.

See A131643 for the actual cubes.

Programs

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]

Formula

a(n) = A131643(n)^(1/3). - Jonathan Sondow, Jan 10 2016

Extensions

Name edited by Altug Alkan, Dec 07 2015