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A097811
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Numbers n such that n^3 is the sum of three or more consecutive positive cubes.
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4
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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20 is in this sequence because 11^3 + 12^3 + 13^3 + 14^3 = 20^3.
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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