A248621 Floor of sums of the squares of the non-integer cube roots of n, as partitioned by the integer roots: floor(Sum_{j=n^3+1..(n+1)^3-1} j^(2/3)).

0, 16, 120, 456, 1240, 2760, 5376, 9520, 15696, 24480, 36520, 52536, 73320, 99736, 132720, 173280, 222496, 281520, 351576, 433960, 530040, 641256, 769120, 915216, 1081200, 1268800, 1479816, 1716120, 1979656, 2272440, 2596560, 2954176, 3347520, 3778896, 4250680

Offset: 0

Author

Richard R. Forberg, Dec 02 2014

Keywords

• nonn
• easy

Comments

The fractional portion of each sum converges to 1/10.

See A248575 for the corresponding sums of the cube root.

See A247112 for the cube of the square roots, other references and a conjecture.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A247112, A248575, A248698.

Programs

Mathematica

Table[2 n + 5 n^2 + 6 n^3 + 3 n^4, {n, 0, 50}]

PARI

a(n) = floor(sum(j=n^3+1, (n+1)^3-1, j^(2/3))); \\ Michel Marcus, Dec 22 2014

PARI

concat(0, Vec(-8*x*(x+2)*(2*x+1)/(x-1)^5 + O(x^100))) \\ Colin Barker, Dec 30 2014

Formula

a(n) = floor(Sum_{j=n^3+1..(n+1)^3-1} j^(2/3)).

a(n) = 2*n + 5*n^2 + 6*n^3 + 3*n^4.

G.f.: -8*x*(x+2)*(2*x+1) / (x-1)^5. - Colin Barker, Dec 30 2014

E.g.f.: exp(x)*x*(16 + 44*x + 24*x^2 + 3*x^3). - Stefano Spezia, Jul 13 2024