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)).
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
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).
Programs
-
Mathematica
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}] -
PARI
a(n) = round(sum(j=n^3+1,(n+1)^3-1, j^(1/3))); \\ Michel Marcus, Dec 09 2014
-
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
-
PARI
a(n)=n*(6*n^2+9*n+4)\/2 \\ Charles R Greathouse IV, Jul 16 2024
Formula
a(n) = round(Sum_{j=n^3+1..(n+1)^3-1} j^(1/3)).
a(n) = ((n+1)*a(n-2) + 3*a(n-1) - 30)/(n-2) - 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 non-recursive formula, but I suspect one exists.)
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
From Colin Barker, Dec 30 2014: (Start)
a(n) = (1-(-1)^n + 8*n + 18*n^2 + 12*n^3)/4.
G.f.: 2*x*(5*x^2 + 8*x + 5) / ((x-1)^4*(x+1)). (End)
From Amrit Awasthi, Jul 08 2024: (Start)
a(2n) = 24*n^3 + 18*n^2 + 4*n.
a(2n-1) = 24*n^3 - 18*n^2 + 4*n, for n >= 1. (End)
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
Comments