This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A248575 #73 Jul 17 2024 09:58:27
%S A248575 0,10,46,128,272,498,822,1264,1840,2570,3470,4560,5856,7378,9142,
%T A248575 11168,13472,16074,18990,22240,25840,29810,34166,38928,44112,49738,
%U A248575 55822,62384,69440,77010,85110,93760,102976,112778,123182,134208,145872,158194,171190,184880,199280,214410,230286,246928,264352,282578,301622
%N A248575 Rounded sums of the non-integer cube roots of n, as partitioned by the integer roots: round(Sum_{j=n^3+1..(n+1)^3-1} j^(1/3)).
%C A248575 The fractional portions of each sum converge to 1/4 and 3/4, alternately.
%C A248575 The corresponding sums for square roots are given by A014105.
%C A248575 See A247112 for additional references to similar sequences and a conjecture.
%H A248575 Colin Barker, <a href="/A248575/b248575.txt">Table of n, a(n) for n = 0..1000</a>
%H A248575 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-2,3,-1).
%F A248575 a(n) = round(Sum_{j=n^3+1..(n+1)^3-1} j^(1/3)).
%F A248575 a(n) = ((n+1)*a(n-2) + 3*a(n-1) - 30)/(n-2) - 20.
%F A248575 (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 A248575 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 A248575 From _Colin Barker_, Dec 30 2014: (Start)
%F A248575 a(n) = (1-(-1)^n + 8*n + 18*n^2 + 12*n^3)/4.
%F A248575 G.f.: 2*x*(5*x^2 + 8*x + 5) / ((x-1)^4*(x+1)). (End)
%F A248575 From _Amrit Awasthi_, Jul 08 2024: (Start)
%F A248575 a(2n) = 24*n^3 + 18*n^2 + 4*n.
%F A248575 a(2n-1) = 24*n^3 - 18*n^2 + 4*n, for n >= 1. (End)
%F A248575 E.g.f.: (x*(19 +27*x + 6*x^2)*cosh(x) + (1 + 19*x + 27*x^2 + 6*x^3)*sinh(x))/2. - _Stefano Spezia_, Jul 17 2024
%t A248575 RecurrenceTable [{a[n] == ((n + 1)*a[n - 2] + 3* a[n - 1] - 30)/(n - 2) - 20, a[1] == 10, a[2] == 46}, a, {n, 1, 50}]
%o A248575 (PARI) a(n) = round(sum(j=n^3+1,(n+1)^3-1, j^(1/3))); \\ _Michel Marcus_, Dec 09 2014
%o A248575 (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
%o A248575 (PARI) a(n)=n*(6*n^2+9*n+4)\/2 \\ _Charles R Greathouse IV_, Jul 16 2024
%Y A248575 Cf. A010888, A247112, A014105, A374384.
%K A248575 nonn,easy
%O A248575 0,2
%A A248575 _Richard R. Forberg_, Dec 02 2014