A177205 Partial sums of round(n^2/17).

0, 0, 0, 1, 2, 3, 5, 8, 12, 17, 23, 30, 38, 48, 60, 73, 88, 105, 124, 145, 169, 195, 223, 254, 288, 325, 365, 408, 454, 503, 556, 613, 673, 737, 805, 877, 953, 1034, 1119, 1208, 1302, 1401, 1505, 1614, 1728, 1847, 1971, 2101, 2237, 2378, 2525

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(17) = 0 + 0 + 0 + 1 + 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 10 + 12 + 13 + 15 + 17 = 105.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..890
• 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+n+36)/102): n in [0..50]]; // Vincenzo Librandi, Apr 29 2011

Maple

seq(round(n*(n+1)*(2*n+1)/102),n=0..50)

Mathematica

Accumulate[Round[Range[0,50]^2/17]] (* Harvey P. Dale, Jul 04 2022 *)

Formula

a(n) = round(n*(n+1)*(2*n+1)/102).

a(n) = floor((2*n^3 + 3*n^2 + n + 36)/102).

a(n) = ceiling((2*n^3 + 3*n^2 + n - 36)/102).

a(n) = a(n-17) + (n+1)*(n-17) + 105, n > 16.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-17) - 3*a(n-18) + 3*a(n-19) - a(n-20) with g.f. x^3 *(1+x) *(x^12 - 2*x^11 + 2*x^10 - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + 2*x^2 - 2*x + 1) / ( (x^16 + x^15 + x^14 + x^13 + x^12 + x^11 + x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) *(x-1)^4 ). - R. J. Mathar, Dec 13 2010