cp's OEIS Frontend

A175812 Partial sums of ceiling(n^2/6).

%I A175812 #40 Sep 08 2022 08:45:52
%S A175812 0,1,2,4,7,12,18,27,38,52,69,90,114,143,176,214,257,306,360,421,488,
%T A175812 562,643,732,828,933,1046,1168,1299,1440,1590,1751,1922,2104,2297,
%U A175812 2502,2718,2947,3188,3442,3709,3990,4284,4593,4916,5254,5607,5976,6360,6761,7178
%N A175812 Partial sums of ceiling(n^2/6).
%C A175812 Partial sums of A008747.
%C A175812 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 A175812 Vincenzo Librandi, <a href="/A175812/b175812.txt">Table of n, a(n) for n = 0..10000</a>
%H A175812 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1,-1,0,2,-1).
%H A175812 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.
%F A175812 a(n) = round((2*n+1)*(2*n^2 + 2*n + 17)/72).
%F A175812 a(n) = floor((n+1)*(2*n^2 + n + 17)/36).
%F A175812 a(n) = ceiling((2*n^3 + 3*n^2 + 18*n)/36).
%F A175812 a(n) = round((2*n^3 + 3*n^2 + 18*n)/36).
%F A175812 a(n) = a(n-6) + (n+1)*(n-6) + 18, n > 5.
%F A175812 From _Mircea Merca_, Jan 09 2011: (Start)
%F A175812 a(n) = 2*a(n-1) - a(n-3) - a(n-4) + 2*a(n-6) - a(n-7), n > 6.
%F A175812 G.f.: x*(x^4+1) / ( (x+1)*(x^2+x+1)*(x-1)^4 ). (End)
%e A175812 a(6) = 0 + 1 + 1 + 2 + 3 + 5 + 6 = 18.
%p A175812 seq(floor((n+1)*(2*n^2+n+17)/36),n=0..50)
%t A175812 Accumulate[Ceiling[Range[0,50]^2/6]] (* _Harvey P. Dale_, Jan 17 2016 *)
%o A175812 (Magma) [Round((2*n+1)*(2*n^2+2*n+17)/72): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011
%o A175812 (PARI) a(n) = (n+1)*(2*n^2+n+17)\36; \\ _Altug Alkan_, Sep 21 2018
%Y A175812 Cf. A008747.
%K A175812 nonn,easy
%O A175812 0,3
%A A175812 _Mircea Merca_, Dec 05 2010