A175831 Partial sums of ceiling(n^2/12).

0, 1, 2, 3, 5, 8, 11, 16, 22, 29, 38, 49, 61, 76, 93, 112, 134, 159, 186, 217, 251, 288, 329, 374, 422, 475, 532, 593, 659, 730, 805, 886, 972, 1063, 1160, 1263, 1371, 1486, 1607, 1734, 1868, 2009, 2156, 2311, 2473, 2642, 2819, 3004, 3196, 3397, 3606

Offset: 0

Author

Mircea Merca, Dec 05 2010

Keywords

• nonn
• easy

Comments

Partial sums of A036410.

There are several sequences of integers of the form ceiling(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,8,10,11,12, 14,15,16,19,20,23,24).

Examples

			a(12) = 0 + 1 + 1 + 1 + 2 + 3 + 3 + 5 + 6 + 7 + 9 + 11 + 12 = 61.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• 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,-1,-1,0,2,-1).

Crossrefs

Cf. A036410.

Programs

Magma

[Round((2*n+1)*(2*n^2+2*n+41)/144): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple

seq(floor((n+1)*(2*n^2+n+41)/72),n=0..50)

Mathematica

Accumulate[Ceiling[Range[0,50]^2/12]] (* or *) LinearRecurrence[{2,0,-1,-1,0,2,-1},{0,1,2,3,5,8,11},60] (* Harvey P. Dale, Apr 16 2023 *)

PARI

a(n)=(n+1)*(2*n^2+n+41)\72 \\ Charles R Greathouse IV, Jul 06 2017

Formula

a(n) = round((2*n+1)*(2*n^2 + 2*n + 41)/144).

a(n) = floor((n+1)*(2*n^2 + n + 41)/72).

a(n) = ceiling((2*n^3 + 3*n^2 + 42*n)/72).

a(n) = a(n-12) + (n+1)*(n-12) + 61.

G.f.: x*(1-x^2+x^4) / ( (1+x)*(1+x+x^2)*(x-1)^4 ). - R. J. Mathar, Jun 22 2011