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%I #55 Sep 04 2022 04:04:36
%S 0,12,60,180,420,840,1512,2520,3960,5940,8580,12012,16380,21840,28560,
%T 36720,46512,58140,71820,87780,106260,127512,151800,179400,210600,
%U 245700,285012,328860,377580,431520,491040,556512,628320,706860,792540,885780,987012
%N a(n) = n*(n + 1)*(n + 2)*(n + 3)/2.
%C a(n) is the area of an irregular quadrilateral with vertices at (1,1), (n+1, n+2), ((n+1)^2, (n+2)^2) and ((n+1)^3, (n+2)^3). - _Art Baker_, Dec 08 2018
%H G. C. Greubel, <a href="/A033486/b033486.txt">Table of n, a(n) for n = 0..5000</a> (terms 0..680 from Vincenzo Librandi)
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = 6*A034827(n+3) = 12*A000332(n+3).
%F G.f.: 12*x/(1 - x)^5. - _Colin Barker_, Mar 01 2012
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) with a(0) = 0, a(1) = 12, a(2) = 60, a(3) = 180, a(4) = 420. - _Harvey P. Dale_, Feb 04 2015
%F E.g.f.: (24*x + 36*x^2 + 12*x^3 + x^4)*exp(x)/2. - _Franck Maminirina Ramaharo_, Dec 08 2018
%F From _Amiram Eldar_, Sep 04 2022: (Start)
%F Sum_{n>=1} 1/a(n) = 1/9.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 8*(3*log(2)-2)/9. (End)
%p [seq(12*binomial(n+3,4),n=0..32)]; # _Zerinvary Lajos_, Nov 24 2006
%t Table[n*(n + 1)*(n + 2)*(n + 3)/2, {n, 0, 50}] (* _David Nacin_, Mar 01 2012 *)
%t LinearRecurrence[{5,-10,10,-5,1},{0,12,60,180,420},40] (* _Harvey P. Dale_, Feb 04 2015 *)
%o (Magma) [n*(n+1)*(n+2)*(n+3)/2: n in [0..40]]; // _Vincenzo Librandi_, Apr 28 2011
%o (PARI) a(n)=n*(n+1)*(n+2)*(n+3)/2 \\ _Charles R Greathouse IV_, Oct 07 2015
%o (GAP) List([0..40],n->n*(n+1)*(n+2)*(n+3)/2); # _Muniru A Asiru_, Dec 08 2018
%o (Sage) [12*binomial(n+3,4) for n in range(40)] # _G. C. Greubel_, Dec 08 2018
%Y Cf. A000332, A050534, A033488, A033487, A060008.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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