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Partial sums of round(n^2/6).
2

%I #41 Sep 08 2022 08:45:50

%S 0,0,1,3,6,10,16,24,35,49,66,86,110,138,171,209,252,300,354,414,481,

%T 555,636,724,820,924,1037,1159,1290,1430,1580,1740,1911,2093,2286,

%U 2490,2706,2934,3175,3429,3696,3976,4270,4578,4901,5239,5592,5960,6344,6744,7161

%N Partial sums of round(n^2/6).

%C Partial sums of A056829.

%H Vincenzo Librandi, <a href="/A173691/b173691.txt">Table of n, a(n) for n = 0..5000</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_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,1,-3,3,-1).

%F a(n) = Sum_{k=0..n} round(k^2/6).

%F a(n) = round((2*n^3 + 3*n^2 + 6*n)/36).

%F a(n) = round((4*n^3 + 6*n^2 + 12*n + 5)/72).

%F a(n) = floor((2*n^3 + 3*n^2 + 6*n + 16)/36).

%F a(n) = ceiling((2*n^3 + 3*n^2 + 6*n - 11)/36).

%F a(n) = a(n-6) + n^2 - 5*n + 10, n > 5.

%F G.f.: x^2*(1+x^4)/((1+x)*(1-x+x^2)*(1+x+x^2)*(1-x)^4). - _Bruno Berselli_, Jan 12 2011

%e a(5) = round(1/6) + round(4/6) + round(9/6) + round(16/6) + round(25/6) = 0 + 1 + 2 + 3 + 4.

%e Note that 9/6 = 1.5 is rounded up.

%p a(n):=round((2*n^(3)+3*n^(2)+6*n)/(36))

%t Accumulate[Round[Range[0, 50]^2/6]] (* or *) LinearRecurrence[{3,-3,1,0, 0,1,-3,3,-1}, {0,0,1,3,6,10,16,24,35}, 60] (* _Harvey P. Dale_, Jan 08 2014 *)

%t CoefficientList[Series[x^2(1+x^4)/((1+x)(1-x+x^2)(1+x+x^2)(1-x)^4), {x, 0, 60}], x] (* _Vincenzo Librandi_, Mar 26 2014 *)

%o (Magma) [Floor((2*n^3+3*n^2+6*n+16)/36): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011

%o (PARI) vector(60, n, n--; (16+6*n+3*n^2+2*n^3)\36) \\ _G. C. Greubel_, Jul 02 2019

%o (Sage) [floor((16+6*n+3*n^2+2*n^3)/36) for n in (0..60)] # _G. C. Greubel_, Jul 02 2019

%Y Cf. A056829.

%K nonn,easy

%O 0,4

%A _Mircea Merca_, Nov 25 2010