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%I #55 Oct 16 2023 23:32:46
%S 0,0,5,21,54,110,195,315,476,684,945,1265,1650,2106,2639,3255,3960,
%T 4760,5661,6669,7790,9030,10395,11891,13524,15300,17225,19305,21546,
%U 23954,26535,29295,32240,35376,38709,42245,45990,49950,54131,58539
%N a(n) = n^3 - n*(n+1)/2.
%C n-th cube (A000578(n)) minus n-th triangular number (A000217(n)).
%C Partial sums of A045944. - _Vladimir Joseph Stephan Orlovsky_, Jun 25 2009
%C The sum of the n-1 numbers between n^2 and n*(n+1) = a(n). - _J. M. Bergot_, Apr 15 2013
%C Use the terms in A061885 to form the antidiagonals for an array. The antidiagonals begin: 0;2,3;6,7,8;12,13,14,15;20,21,22,23,24,25. The sum of the terms in these antidiagonals = a(n)for n > 0. - _J. M. Bergot_, Jul 08 2013
%C a(n) is the sum of the n numbers strictly between n^2-n-1 and n^2. - _Charlie Marion_, Feb 21 2020
%H G. C. Greubel, <a href="/A160378/b160378.txt">Table of n, a(n) for n = 0..1000</a>
%H Milan Janjic and B. Petkovic, <a href="http://arxiv.org/abs/1301.4550">A counting function</a>, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = (2*n^3 - n^2 - n)/2. - _Vincenzo Librandi_, Dec 12 2010; edited by _Klaus Brockhaus_, Dec 12 2010
%F From _Chai Wah Wu_, Aug 03 2022: (Start)
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.
%F G.f.: x^2*(5 + x)/(1 - x)^4. (End)
%F E.g.f.: (x^2/2)*(5 + 2*x)*exp(x). - _G. C. Greubel_, Oct 14 2023
%e a(4) = 4^3 - 4*5/2 = 64 - 10 = 54.
%t Table[n^3 - n*(n+1)/2, {n,0,40}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 25 2009 *)
%o (Magma) [ n^3-n*(n+1)/2: n in [0..50] ];
%o (PARI) a(n)=n^3-n*(n+1)/2 \\ _Charles R Greathouse IV_, Oct 18 2022
%o (SageMath) [n^3 -binomial(n+1,2) for n in range(41)] # _G. C. Greubel_, Oct 14 2023
%Y Cf. A000217, A000578, A045944.
%K nonn,easy
%O 0,3
%A _Gil Broussard_, May 11 2009
%E Definition clarified and offset changed from 1 to 0 by _Klaus Brockhaus_, Dec 12 2010
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