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%I #34 Jun 13 2015 00:55:16
%S 0,16,120,456,1240,2760,5376,9520,15696,24480,36520,52536,73320,99736,
%T 132720,173280,222496,281520,351576,433960,530040,641256,769120,
%U 915216,1081200,1268800,1479816,1716120,1979656,2272440,2596560,2954176,3347520,3778896,4250680
%N Floor of sums of the squares of the non-integer cube roots of n, as partitioned by the integer roots: floor[sum(j from n^3+1 to (n+1)^3-1, j^(2/3))].
%C The fractional portion of each sum converges to 1/10.
%C See A248575 for the corresponding sums of the cube root.
%C See A247112 for the cube of the square roots, other references and a conjecture.
%H Colin Barker, <a href="/A248621/b248621.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = floor[sum(j from n^3+1 to (n+1)^3-1, j^(2/3))].
%F a(n) = 2*n + 5*n^2 + 6*n^3 + 3*n^4.
%F G.f.: -8*x*(x+2)*(2*x+1) / (x-1)^5. - _Colin Barker_, Dec 30 2014
%t Table[AccountingForm[N[Sum[j^(2/3), {j, n^3 + 1, (n + 1)^3 - 1}], 50]], {n, 0, 50}]
%t Table[2 n + 5 n^2 + 6 n^3 + 3 n^4, {n, 0, 50}]
%o (PARI) a(n) = floor(sum(j=n^3+1, (n+1)^3-1, j^(2/3))); \\ _Michel Marcus_, Dec 22 2014
%o (PARI) concat(0, Vec(-8*x*(x+2)*(2*x+1)/(x-1)^5 + O(x^100))) \\ _Colin Barker_, Dec 30 2014
%Y Cf. A248575, A247112.
%K nonn,easy
%O 0,2
%A _Richard R. Forberg_, Dec 02 2014
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