%I #30 Sep 08 2022 08:45:52
%S 0,1,2,3,4,6,8,11,15,20,26,33,41,50,61,73,87,103,121,140,162,186,212,
%T 240,271,304,340,379,421,466,514,565,619,677,738,803,872,945,1021,
%U 1102,1187,1276,1369,1467,1569,1676,1788,1905,2027,2154,2286
%N Partial sums of ceiling(n^2/19).
%C 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).
%H Vincenzo Librandi, <a href="/A175864/b175864.txt">Table of n, a(n) for n = 0..10000</a>
%H Mircea Merca, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Merca/merca3.html">Inequalities and Identities Involving Sums of Integer Functions</a> J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-3,3,-1).
%F a(n) = round((2*n+1)*(n^2 + n + 30)/114).
%F a(n) = floor((2*n^3 + 3*n^2 + 61*n + 78)/114).
%F a(n) = ceiling((2*n^3 + 3*n^2 + 61*n - 18)/114).
%F a(n) = round((2*n^3 + 3*n^2 + 61*n)/114).
%F a(n) = a(n-19) + (n+1)*(n-19) + 140, n > 18.
%F 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
%e a(19) = 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 12 + 14 + 16 + 18 + 19 = 140.
%p seq(round((2*n^3+3*n^2+61*n)/114),n=0..50)
%t Accumulate[Ceiling[Range[0,50]^2/19]] (* _Harvey P. Dale_, Mar 19 2018 *)
%o (Magma) [Round((2*n+1)*(n^2+n+30)/114): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011
%o (PARI) a(n)=(2*n^3+3*n^2+61*n+78)\114 \\ _Charles R Greathouse IV_, Jul 06 2017
%K nonn,easy
%O 0,3
%A _Mircea Merca_, Dec 05 2010