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A240137
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Sum of n consecutive cubes starting from n^3.
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9
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0, 1, 35, 216, 748, 1925, 4131, 7840, 13616, 22113, 34075, 50336, 71820, 99541, 134603, 178200, 231616, 296225, 373491, 464968, 572300, 697221, 841555, 1007216, 1196208, 1410625, 1652651, 1924560, 2228716, 2567573, 2943675, 3359656, 3818240, 4322241, 4874563
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OFFSET
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0,3
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COMMENTS
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Sum_{i>=1} 1/a(i) = 1.0356568858420883122567711052556541...
Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of cubes with side length q. - Wesley Ivan Hurt, Apr 15 2018
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LINKS
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FORMULA
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G.f.: x*(1 + 30*x + 51*x^2 + 8*x^3)/(1 - x)^5.
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EXAMPLE
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a(3) = 216 because 216 = 3^3 + 4^3 + 5^3.
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MAPLE
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MATHEMATICA
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Table[n^2 (3 n - 1) (5 n - 3)/4, {n, 0, 40}]
CoefficientList[Series[x (1 + 30 x + 51 x^2 + 8 x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
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PROG
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(Sage) [n^2*(3*n-1)*(5*n-3)/4 for n in [0..40]]
(Magma) [n^2*(3*n-1)*(5*n-3)/4: n in [0..40]];
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CROSSREFS
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Cf. A116149: sum of n consecutive cubes after n^3.
Cf. A050410: sum of n consecutive squares starting from n^2.
Cf. A000326 (pentagonal numbers): sum of n consecutive integers starting from n.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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