A173722 Partial sums of round(n^2/8).

0, 0, 1, 2, 4, 7, 12, 18, 26, 36, 49, 64, 82, 103, 128, 156, 188, 224, 265, 310, 360, 415, 476, 542, 614, 692, 777, 868, 966, 1071, 1184, 1304, 1432, 1568, 1713, 1866, 2028, 2199, 2380, 2570, 2770, 2980, 3201, 3432, 3674, 3927, 4192, 4468, 4756, 5056, 5369

Offset: 0

Author

Mircea Merca, Nov 26 2010

Keywords

• nonn
• easy

Comments

Partial sums of A001971.

Examples

			a(5) = round(1/8) + round(4/8) + round(9/8) + round(16/8) + round(25/8) = 0 + 1 + 1 + 2 + 3 = 7.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..5000
• 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,-3,1,1,-3,3,-1).

Crossrefs

Cf. A001971.

Programs

Maple

A057077 := proc(n) op( 1+(n mod 4),[1,1,-1,-1]) ; end proc:
A173722 := proc(n) 3*(-1)^n/32+n^2/16+n/12+n^3/24+1/32-A057077(n)/8 ; end proc:
seq(A173722(n),n=0..80) ; # R. J. Mathar, Nov 26 2010

Mathematica

Table[Floor[(n + 2)*(2*n^2 - n + 6)/48], {n,0,50}] (* G. C. Greubel, Nov 29 2016 *)

PARI

a(n)=(n+2)*(2*n^2-n+6)\48 \\ Charles R Greathouse IV, Oct 19 2022

Formula

a(n) = Sum_{k=0..n} round(k^2/8).

a(n) = round((2*n^3+3*n^2+4*n)/48).

a(n) = round((2*n+1)*(2*n^2+2*n+3)/96).

a(n) = floor((n+2)*(2*n^2-n+6)/48).

a(n) = ceiling((2*n^3+3*n^2+4*n-9)/48).

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

G.f.: x^2*(1-x+x^2) / ( (1+x)*(x^2+1)*(x-1)^4 ). - R. J. Mathar, Nov 26 2010

a(n) = 3*(-1)^n/32+n^2/16+n/12+n^3/24+1/32-A057077(n)/8. - R. J. Mathar, Nov 26 2010