A171965 Partial sums of floor(n^2/6) (A056827).

0, 0, 0, 1, 3, 7, 13, 21, 31, 44, 60, 80, 104, 132, 164, 201, 243, 291, 345, 405, 471, 544, 624, 712, 808, 912, 1024, 1145, 1275, 1415, 1565, 1725, 1895, 2076, 2268, 2472, 2688, 2916, 3156, 3409, 3675, 3955, 4249, 4557, 4879, 5216, 5568, 5936, 6320, 6720, 7136

Offset: 0

Author

Mircea Merca, Nov 19 2010

Keywords

• nonn
• easy

Comments

Quasipolynomial.

Examples

			a(5) = 7 = 0 + 0 + 0 + 1 + 2 + 4.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,1,-3,3,-1).

Crossrefs

Cf. A056827.

Programs

Magma

[Round((2*n^3+3*n^2-12*n-6)/36): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

Maple

a(n):=round((2*n^3 +3*n^2 -12*n-6)/36)

Mathematica

Accumulate[Floor[Range[0,50]^2/6]] (* Harvey P. Dale, Jul 30 2020 *)

PARI

a(n)=(2*n^3+3*n^2-12*n+7)\36 \\ Charles R Greathouse IV, Jan 29 2012

Python

a171965 = [0]
for n in range(1, 60): a171965.append(a171965[-1] + n*n//6)
print(a171965) # Gennady Eremin, Mar 13 2022

Formula

a(n) = Sum_{k=0..n} floor(k^2/6).

a(n) = round((2*n^3 + 3*n^2 - 12*n - 6)/36).

a(n) = round((4*n^3 + 6*n^2 - 24*n - 13)/72).

a(n) = floor((2*n^3 + 3*n^2 - 12*n + 7)/36).

a(n) = ceiling((2*n^3 + 3*n^2 - 12*n - 20)/36).

a(n) = a(n-6) + n^2 - 5*n + 7, n > 5.

G.f.: x^3*(1+x^2) / ( (1+x)*(1+x+x^2)*(x^2-x+1)*(x-1)^4 ). - R. J. Mathar, Jan 28 2012