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A036659 Product of n with sum of next n consecutive integers.

%I A036659 #39 Jan 10 2023 01:50:05
%S A036659 2,14,45,104,200,342,539,800,1134,1550,2057,2664,3380,4214,5175,6272,
%T A036659 7514,8910,10469,12200,14112,16214,18515,21024,23750,26702,29889,
%U A036659 33320,37004,40950,45167,49664,54450,59534,64925,70632,76664,83030,89739,96800,104222,112014
%N A036659 Product of n with sum of next n consecutive integers.
%H A036659 Vincenzo Librandi, <a href="/A036659/b036659.txt">Table of n, a(n) for n = 1..1000</a>
%H A036659 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A036659 a(n) = 2 + (3*n*n + 10*n + 11)*n/2. - _Frank Ellermann_, Mar 16 2002
%F A036659 a(n) = n^2(3n + 1)/2. - Ronnie B Kon (ronniekon(AT)yahoo.com), Jun 09 2006
%F A036659 G.f.: x*(2 + 6*x + x^2)/(1 - x)^4. - _Colin Barker_, Mar 27 2012
%F A036659 From _Amiram Eldar_, Jan 10 2023: (Start)
%F A036659 Sum_{n>=1} 1/a(n) = Pi^2/3 + sqrt(3)*Pi + 9*log(3) - 18.
%F A036659 Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/6 - 2*sqrt(3)*Pi - 12*log(2) + 18. (End)
%e A036659 n=3 -> 3*(4+5+6)=45
%t A036659 CoefficientList[Series[(2+6*x+x^2)/(1-x)^4,{x,0,40}],x] (* _Vincenzo Librandi_, Apr 25 2012 *)
%t A036659 Table[(n^2*(1+3n))/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{2,14,45,104},40] (* _Harvey P. Dale_, Jan 04 2015 *)
%o A036659 (PARI) a(n)=n^2*(3*n+1)/2 \\ _Charles R Greathouse IV_, Dec 08 2011
%o A036659 (Magma) I:=[2, 14, 45, 104]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Apr 25 2012
%Y A036659 Cf. A036660.
%K A036659 nonn,easy
%O A036659 1,1
%A A036659 _G. L. Honaker, Jr._, Dec 15 1998