A175812 Partial sums of ceiling(n^2/6).

0, 1, 2, 4, 7, 12, 18, 27, 38, 52, 69, 90, 114, 143, 176, 214, 257, 306, 360, 421, 488, 562, 643, 732, 828, 933, 1046, 1168, 1299, 1440, 1590, 1751, 1922, 2104, 2297, 2502, 2718, 2947, 3188, 3442, 3709, 3990, 4284, 4593, 4916, 5254, 5607, 5976, 6360, 6761, 7178

Offset: 0

Author

Mircea Merca, Dec 05 2010

Keywords

• nonn
• easy

Comments

Partial sums of A008747.

There are several sequences of integers of the form ceiling(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,8,10,11,12, 14,15,16,19,20,23,24).

Examples

			a(6) = 0 + 1 + 1 + 2 + 3 + 5 + 6 = 18.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,0,2,-1).
• Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

Crossrefs

Cf. A008747.

Programs

Magma

[Round((2*n+1)*(2*n^2+2*n+17)/72): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple

seq(floor((n+1)*(2*n^2+n+17)/36),n=0..50)

Mathematica

Accumulate[Ceiling[Range[0,50]^2/6]] (* Harvey P. Dale, Jan 17 2016 *)

PARI

a(n) = (n+1)*(2*n^2+n+17)\36; \\ Altug Alkan, Sep 21 2018

Formula

a(n) = round((2*n+1)*(2*n^2 + 2*n + 17)/72).

a(n) = floor((n+1)*(2*n^2 + n + 17)/36).

a(n) = ceiling((2*n^3 + 3*n^2 + 18*n)/36).

a(n) = round((2*n^3 + 3*n^2 + 18*n)/36).

a(n) = a(n-6) + (n+1)*(n-6) + 18, n > 5.

From Mircea Merca, Jan 09 2011: (Start)

a(n) = 2*a(n-1) - a(n-3) - a(n-4) + 2*a(n-6) - a(n-7), n > 6.

G.f.: x*(x^4+1) / ( (x+1)*(x^2+x+1)*(x-1)^4 ). (End)