

A248575


Rounded sums of the noninteger cube roots of n, as partitioned by the integer roots: round[sum(j from n^3+1 to (n+1)^31, j^(1/3))].


3



0, 10, 46, 128, 272, 498, 822, 1264, 1840, 2570, 3470, 4560, 5856, 7378, 9142, 11168, 13472, 16074, 18990, 22240, 25840, 29810, 34166, 38928, 44112, 49738, 55822, 62384, 69440, 77010, 85110, 93760, 102976, 112778, 123182, 134208, 145872, 158194, 171190, 184880, 199280, 214410, 230286, 246928, 264352, 282578, 301622
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

The fractional portions of each sum converge to 1/4 and 3/4, alternately.
The corresponding sums for square roots are given by A014105.
See A247112 for additional references to similar sequences and a conjecture.


LINKS



FORMULA

a(n) = round[sum(j from n^3+1 to (n+1)^31, j^(1/3))].
a(n) = ((n+1)*a(n2) + 3*a(n1)  30)/(n2)  20.
(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 nonrecursive formula, but I suspect one exists.)
a(n) = 3*a(n1)  2*a(n2)  2*a(n3) + 3*a(n4)  a(n5), 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
a(n) = (1(1)^n+8*n+18*n^2+12*n^3)/4.  Colin Barker, Dec 30 2014
G.f.: 2*x*(5*x^2+8*x+5) / ((x1)^4*(x+1)).  Colin Barker, Dec 30 2014


MATHEMATICA

Below are tables for both formulas, with the former showing the fractional portion:
Table[AccountingForm[N[Sum[j^(1/3), {j, n^3 + 1, (n + 1)^3  1}], 10]], {n, 0, 50}]
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}]


PROG

(PARI) a(n) = round(sum(j=n^3+1, (n+1)^31, j^(1/3))); \\ Michel Marcus, Dec 09 2014
(PARI) concat(0, Vec(2*x*(5*x^2+8*x+5)/((x1)^4*(x+1)) + O(x^100))) \\ Colin Barker, Dec 30 2014


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



