A177189 Partial sums of round(n^2/16).

0, 0, 0, 1, 2, 4, 6, 9, 13, 18, 24, 32, 41, 52, 64, 78, 94, 112, 132, 155, 180, 208, 238, 271, 307, 346, 388, 434, 483, 536, 592, 652, 716, 784, 856, 933, 1014, 1100, 1190, 1285, 1385, 1490, 1600, 1716, 1837, 1964, 2096, 2234, 2378, 2528, 2684

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(16) = 0 + 0 + 0 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 8 + 9 + 11 + 12 + 14 + 16 = 94.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..885
• 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((n+3)*(2*n^2-3*n+13)/96): n in [0..50]]; // Vincenzo Librandi, Apr 29 2011

Maple

seq(round((2*n^3+3*n^2+4*n)/96),n=0..50)

Mathematica

Accumulate[Round[Range[0,50]^2/16]]  (* Harvey P. Dale, Mar 16 2011 *)

Formula

a(n) = round((2*n+1)*(2*n^2 + 2*n + 3)/192).

a(n) = floor((n+3)*(2*n^2 - 3*n + 13)/96).

a(n) = ceiling((n-2)*(2*n^2 + 7*n + 18)/96).

a(n) = round((2*n^3 + 3*n^2 + 4*n)/96).

a(n) = a(n-16) + (n+1)*(n-16) + 94, n > 15.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-8) - 3*a(n-9) + 3*a(n-10) - a(n-11) with g.f. x^3*(1 - x + x^2 + x^4 - x^3) / ( (1+x)*(1+x^2)*(1+x^4)*(x-1)^4 ). - R. J. Mathar, Dec 13 2010