A177239 Partial sums of round(n^2/20).

0, 0, 0, 0, 1, 2, 4, 6, 9, 13, 18, 24, 31, 39, 49, 60, 73, 87, 103, 121, 141, 163, 187, 213, 242, 273, 307, 343, 382, 424, 469, 517, 568, 622, 680, 741, 806, 874, 946, 1022, 1102, 1186, 1274, 1366, 1463, 1564, 1670, 1780, 1895, 2015, 2140

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(20) = 0 + 0 + 0 + 0 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 18 + 20 = 141.
		

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.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1,1,-2,0,2,-1).

Crossrefs

Cf. A001304, A177100, A177116.

Programs

Magma

[Floor((n+4)*(2*n^2-5*n+6)/120): n in [0..50]]; // Vincenzo Librandi, Apr 29 2011

Maple

seq(round(n*(n-2)*(2*n+7)/120),n=0..50)

Mathematica

f[n_] := Round[n^2/20]; Accumulate@ Array[f, 51, 0] (* Robert G. Wilson v, Dec 20 2010 *)

SageMath

[(n+4)*(2*n^2 -5*n +6)//120 for n in range(56)] # G. C. Greubel, Apr 27 2024

Formula

a(n) = A001304(n-4).

a(n) = round((2*n+1)*(2*n^2 + 2*n - 15)/240).

a(n) = floor((n+4)*(2*n^2 - 5*n + 6)/120).

a(n) = ceiling((n-3)*(2*n^2 + 9*n + 13)120).

a(n) = round(n*(n-2)*(2*n+7)/120).

a(n) = a(n-20) + (n+1)*(n-20) + 141, n > 19.

From R. J. Mathar, Dec 12 2010: (Start)

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) + a(n-5) - 2*a(n-6) + 2*a(n-8) - a(n-9).

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