A172046 Partial sums of floor(n^2/7) (A056834).

0, 0, 0, 1, 3, 6, 11, 18, 27, 38, 52, 69, 89, 113, 141, 173, 209, 250, 296, 347, 404, 467, 536, 611, 693, 782, 878, 982, 1094, 1214, 1342, 1479, 1625, 1780, 1945, 2120, 2305, 2500, 2706, 2923, 3151, 3391, 3643, 3907, 4183, 4472, 4774, 5089, 5418, 5761, 6118

Offset: 0

Author

Mircea Merca, Nov 19 2010

Keywords

• nonn
• easy

Examples

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

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

Crossrefs

Cf. A056834.

Programs

Magma

[Round((2*n^3+3*n^2-11*n)/42): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

Maple

a:= n-> round((2*n^3+3*n^2-11*n)/42): seq (a(n), n=0..50);

Mathematica

LinearRecurrence[{3,-3,1,0,0,0,1,-3,3,-1},{0,0,0,1,3,6,11,18,27,38},60] (* Harvey P. Dale, Apr 12 2017 *)

Formula

a(n) = Sum_{k=0..n} floor(k^2/7).

a(n) = round((2*n^3 + 3*n^2 - 11*n)/42).

a(n) = round((2*n^3 + 3*n^2 - 11*n - 6)/42).

a(n) = floor((2*n^3 + 3*n^2 - 11*n + 6)/42).

a(n) = ceiling((2*n^3 + 3*n^2 - 11*n - 18)/42).

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

G.f.: x^3*(x+1)*(x^2 - x + 1)/((x-1)^4*(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)). [Colin Barker, Oct 26 2012]