A177237 Partial sums of round(n^2/19).

0, 0, 0, 0, 1, 2, 4, 7, 10, 14, 19, 25, 33, 42, 52, 64, 77, 92, 109, 128, 149, 172, 197, 225, 255, 288, 324, 362, 403, 447, 494, 545, 599, 656, 717, 781, 849, 921, 997, 1077, 1161, 1249, 1342, 1439, 1541, 1648, 1759, 1875, 1996, 2122, 2254

Offset: 0

Author

Mircea Merca, Dec 10 2010

Keywords

• nonn
• easy

Comments

The round 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(19) = 0 + 0 + 0 + 0 + 1 + 1 + 2 + 3 + 3 + 4 + 5 + 6 + 8 + 9 + 10 + 12 + 13 + 15 + 17 + 19 = 128.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..895
• Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

Crossrefs

Cf. A177100, A177116.

Programs

Magma

[Floor((2*n^3+3*n^2-11*n+42)/114): n in [0..50]]; // Vincenzo Librandi, Apr 29 2011

Maple

seq(round((2*n^3+3*n^2-11*n)/114),n=0..50)

Mathematica

Accumulate[Round[Range[0,50]^2/19]] (* Harvey P. Dale, Aug 15 2022 *)

SageMath

[(2*n^3 +3*n^2 -11*n +42)//114 for n in range(61)] # G. C. Greubel, Apr 27 2024

Formula

a(n) = round((n-2)*(n+3)*(2*n+1)/114).

a(n) = floor((2*n^3 + 3*n^2 - 11*n + 42)/114).

a(n) = ceiling((2*n^3 + 3*n^2 - 11*n - 54)/114).

a(n) = round((2*n^3 + 3*n^2 - 11*n)/114).

a(n) = a(n-19) + (n+1)*(n-19) + 128, n > 18.

From R. J. Mathar, Dec 13 2010: (Start)

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-19) - 3*a(n-20) + 3*a(n-21) - a(n-22).

G.f.: x^4*(1+x)*(1 - x + x^2 - x^3 + x^4)*(1 - x + x^2 - x^4 + x^6 - x^7 + x^8)/((1-x)^3 * (1 - x^19)). (End)