A175870 Partial sums of ceiling(n^2/24).

0, 1, 2, 3, 4, 6, 8, 11, 14, 18, 23, 29, 35, 43, 52, 62, 73, 86, 100, 116, 133, 152, 173, 196, 220, 247, 276, 307, 340, 376, 414, 455, 498, 544, 593, 645, 699, 757, 818, 882, 949, 1020, 1094, 1172, 1253, 1338, 1427, 1520, 1616, 1717, 1822

Offset

0,3

Comments

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).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..900

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,-1,1,-1,-1,1,-1,2,-1).

Formula

a(n) = round((2*n+1)*(2*n^2 + 2*n + 95)/288).

a(n) = floor((n+1)*(2*n^2 + n + 95)/144).

a(n) = ceiling((2*n^3 + 3*n^2 + 96*n)/144).

a(n) = a(n-24) + (n+1)*(n-24) + 220.

G.f.: x*(x^6 - x^3 + 1)/((x-1)^4*(x+1)*(x^2+1)*(x^2+x+1)). - Colin Barker, Oct 26 2012

Example

a(24) = 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 5 + 6 + 6 + 8 + 9 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 21 + 23 + 24 = 220.

Maple

seq(floor((n+1)*(2*n^2+n+95)/144), n=0..50)

Prog

(Magma) [Floor((n+1)*(2*n^2+n+95)/144): n in [0..50]]; // Vincenzo Librandi, Apr 29 2011
(PARI) a(n)=(n+1)*(2*n^2+n+95)\144 \\ Charles R Greathouse IV, Jul 06 2017

Crossrefs

Sequence in context: A075535 A238383 A134953 * A114829 A175869 A007279

Adjacent sequences: A175867 A175868 A175869 * A175871 A175872 A175873

Keyword

nonn,easy

Author

Mircea Merca, Dec 05 2010

Status

approved