A177100 Partial sums of round(n^2/9).

0, 0, 0, 1, 3, 6, 10, 15, 22, 31, 42, 55, 71, 90, 112, 137, 165, 197, 233, 273, 317, 366, 420, 479, 543, 612, 687, 768, 855, 948, 1048, 1155, 1269, 1390, 1518, 1654, 1798, 1950, 2110, 2279, 2457, 2644, 2840, 3045, 3260, 3485, 3720, 3965, 4221, 4488, 4766

Offset: 0

Author

Mircea Merca, Dec 09 2010

Keywords

• nonn
• easy

Comments

Partial sums of A008740(n-3). [Corrected by Gerald Hillier, Dec 24 2017]

The round function, also called the nearest integer function, is defined here by round(x)=floor(x+1/2).

There are several sequences of integers of the form round(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,9,11,12,13,16,17,19,20, 28,29,36,44).

Examples

			a(9) = 0+0+0+1+2+3+4+5+7+9 = 31.
		

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,-3,1,0,0,0,0,0,1,-3,3,-1).

Crossrefs

Cf. A008740.

Programs

Maple

seq(round((2*n^3+3*n^2-3*n)/54),n=0..50)

Mathematica

Accumulate[Round[Range[0,80]^2/9]] (* Harvey P. Dale, Apr 06 2017 *)

PARI

a(n)=(n+3)*(2*n^2-3*n+6)\54 \\ Charles R Greathouse IV, Sep 28 2015

Formula

a(n) = round((n-1)*(n+2)*(2*n+1)/54);

a(n) = floor((n+3)*(2*n^2-3*n+6)/54);

a(n) = ceiling((n-2)*(2*n^2+7*n+11)/54);

a(n) = round((2*n^3+3*n^2-3*n)/54);

a(n) = a(n-9) + (n+1)*(n-9) + 31, n > 8.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-9) - 3*a(n-10) + 3*a(n-11) - a(n-12). - R. J. Mathar, Mar 11 2012

G.f.: x^3*(x+1)*(x^4-x^3+x^2-x+1)/((x-1)^4*(x^2+x+1)*(x^6+x^3+1)). - Colin Barker, Oct 10 2012