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
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Dec 05 2010
STATUS
approved