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,-3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-3,3,-1).
FORMULA
a(n) = round((2*n+1)*(n^2 + n + 30)/114).
a(n) = floor((2*n^3 + 3*n^2 + 61*n + 78)/114).
a(n) = ceiling((2*n^3 + 3*n^2 + 61*n - 18)/114).
a(n) = round((2*n^3 + 3*n^2 + 61*n)/114).
a(n) = a(n-19) + (n+1)*(n-19) + 140, n > 18.
G.f.: x*(x+1)*(x^2 - x + 1)*(x^4 - x^2 + 1)*(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*(x^6 - x^3 + 1)/((x-1)^4*(x^18 + x^17 + 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)). - Colin Barker, Oct 26 2012
EXAMPLE
a(19) = 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 12 + 14 + 16 + 18 + 19 = 140.
MAPLE
seq(round((2*n^3+3*n^2+61*n)/114), n=0..50)
MATHEMATICA
Accumulate[Ceiling[Range[0, 50]^2/19]] (* Harvey P. Dale, Mar 19 2018 *)
PROG
(Magma) [Round((2*n+1)*(n^2+n+30)/114): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
(PARI) a(n)=(2*n^3+3*n^2+61*n+78)\114 \\ Charles R Greathouse IV, Jul 06 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Dec 05 2010
STATUS
approved