A175868 Partial sums of ceiling(n^2/20).

0, 1, 2, 3, 4, 6, 8, 11, 15, 20, 25, 32, 40, 49, 59, 71, 84, 99, 116, 135, 155, 178, 203, 230, 259, 291, 325, 362, 402, 445, 490, 539, 591, 646, 704, 766, 831, 900, 973, 1050, 1130, 1215, 1304, 1397, 1494, 1596, 1702, 1813, 1929, 2050, 2175

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..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 (3,-3,1,0,0,0,0,0,0,1,-3,3,-1).

Formula

a(n) = round((2*n+1)*(2*n^2 + 2*n + 69)/240).

a(n) = floor((2*n^3 + 3*n^2 + 70*n + 72)/120).

a(n) = ceiling((2*n^3 + 3*n^2 + 70*n - 3)/120).

a(n) = a(n-20) + (n+1)*(n-20) + 155.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-10) - 3*a(n-11) + 3*a(n-12) - a(n-13); a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=4, a(5)=6, a(6)=8, a(7)=11, a(8)=15, a(9)=20, a(10)=25, a(11)=32, a(12)=40. - Harvey P. Dale, Jun 10 2011

G.f.: x*(x^4 - x^2 + 1)*(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)/((x-1)^4*(x+1)*(x^4 - x^3 + x^2 - x + 1)*(x^4 + x^3 + x^2 + x + 1)). - Colin Barker, Oct 26 2012

Example

a(20) = 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 5 + 7 + 8 + 9 + 10 + 12 + 13 + 15 + 17 + 19 + 20 = 155.

Maple

seq(ceil((2*n^3+3*n^2+70*n-3)/120), n=0..50)

Mathematica

Accumulate[Ceiling[Range[0, 60]^2/20]] (* Harvey P. Dale, Jun 10 2011 *)

Prog

(PARI) a(n)=(2*n^3+3*n^2+70*n+72)\120 \\ Charles R Greathouse IV, Jun 10 2011
(Magma) [Round((2*n+1)*(2*n^2+2*n+69)/240): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Crossrefs

Sequence in context: A265254 A303944 A039855 * A035950 A175864 A133153

Adjacent sequences: A175865 A175866 A175867 * A175869 A175870 A175871

Keyword

nonn,easy

Author

Mircea Merca, Dec 05 2010

Status

approved