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A175831 Partial sums of ceiling(n^2/12).

%I A175831 #32 Apr 16 2023 18:43:51
%S A175831 0,1,2,3,5,8,11,16,22,29,38,49,61,76,93,112,134,159,186,217,251,288,
%T A175831 329,374,422,475,532,593,659,730,805,886,972,1063,1160,1263,1371,1486,
%U A175831 1607,1734,1868,2009,2156,2311,2473,2642,2819,3004,3196,3397,3606
%N A175831 Partial sums of ceiling(n^2/12).
%C A175831 Partial sums of A036410.
%C A175831 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 A175831 Vincenzo Librandi, <a href="/A175831/b175831.txt">Table of n, a(n) for n = 0..10000</a>
%H A175831 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 A175831 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1,-1,0,2,-1).
%F A175831 a(n) = round((2*n+1)*(2*n^2 + 2*n + 41)/144).
%F A175831 a(n) = floor((n+1)*(2*n^2 + n + 41)/72).
%F A175831 a(n) = ceiling((2*n^3 + 3*n^2 + 42*n)/72).
%F A175831 a(n) = a(n-12) + (n+1)*(n-12) + 61.
%F A175831 G.f.: x*(1-x^2+x^4) / ( (1+x)*(1+x+x^2)*(x-1)^4 ). - _R. J. Mathar_, Jun 22 2011
%e A175831 a(12) = 0 + 1 + 1 + 1 + 2 + 3 + 3 + 5 + 6 + 7 + 9 + 11 + 12 = 61.
%p A175831 seq(floor((n+1)*(2*n^2+n+41)/72),n=0..50)
%t A175831 Accumulate[Ceiling[Range[0,50]^2/12]] (* or *) LinearRecurrence[{2,0,-1,-1,0,2,-1},{0,1,2,3,5,8,11},60] (* _Harvey P. Dale_, Apr 16 2023 *)
%o A175831 (Magma) [Round((2*n+1)*(2*n^2+2*n+41)/144): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011
%o A175831 (PARI) a(n)=(n+1)*(2*n^2+n+41)\72 \\ _Charles R Greathouse IV_, Jul 06 2017
%Y A175831 Cf. A036410.
%K A175831 nonn,easy
%O A175831 0,3
%A A175831 _Mircea Merca_, Dec 05 2010