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%I #49 Jun 13 2015 00:55:16
%S 0,10,46,128,272,498,822,1264,1840,2570,3470,4560,5856,7378,9142,
%T 11168,13472,16074,18990,22240,25840,29810,34166,38928,44112,49738,
%U 55822,62384,69440,77010,85110,93760,102976,112778,123182,134208,145872,158194,171190,184880,199280,214410,230286,246928,264352,282578,301622
%N Rounded sums of the non-integer cube roots of n, as partitioned by the integer roots: round[sum(j from n^3+1 to (n+1)^3-1, j^(1/3))].
%C The fractional portions of each sum converge to 1/4 and 3/4, alternately.
%C The corresponding sums for square roots are given by A014105.
%C See A247112 for additional references to similar sequences and a conjecture.
%H Colin Barker, <a href="/A248575/b248575.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 (3,-2,-2,3,-1).
%F a(n) = round[sum(j from n^3+1 to (n+1)^3-1, j^(1/3))].
%F a(n) = ((n+1)*a(n-2) + 3*a(n-1) - 30)/(n-2) - 20.
%F (Thanks to Mathematica for finding the recursive formula from the first 12 terms, as a DifferenceRoot, reformatted here for OEIS format and verified to n = 100. I could not "coax" Mathematica to produce a simple non-recursive formula, but I suspect one exists.)
%F a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5), for n > 5. This implies that the digital roots (A010888) of the terms are cyclic with a period of eighteen. - _Ivan N. Ianakiev_, Dec 13 2014
%F a(n) = (1-(-1)^n+8*n+18*n^2+12*n^3)/4. - _Colin Barker_, Dec 30 2014
%F G.f.: 2*x*(5*x^2+8*x+5) / ((x-1)^4*(x+1)). - _Colin Barker_, Dec 30 2014
%t Below are tables for both formulas, with the former showing the fractional portion:
%t Table[AccountingForm[N[Sum[j^(1/3), {j, n^3 + 1, (n + 1)^3 - 1}], 10]], {n, 0, 50}]
%t RecurrenceTable [{a[n] == ((n + 1)*a[n - 2] + 3* a[n - 1] - 30)/(n - 2) - 20,
%t a[1] == 10, a[2] == 46}, a, {n, 1, 50}]
%o (PARI) a(n) = round(sum(j=n^3+1,(n+1)^3-1, j^(1/3))); \\ _Michel Marcus_, Dec 09 2014
%o (PARI) concat(0, Vec(2*x*(5*x^2+8*x+5)/((x-1)^4*(x+1)) + O(x^100))) \\ _Colin Barker_, Dec 30 2014
%Y Cf. A247112, A014105.
%K nonn,easy
%O 0,2
%A _Richard R. Forberg_, Dec 02 2014
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