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A011199 a(n) = (n+1)*(2*n+1)*(3*n+1).

%I A011199 #25 Oct 18 2022 15:01:34
%S A011199 1,24,105,280,585,1056,1729,2640,3825,5320,7161,9384,12025,15120,
%T A011199 18705,22816,27489,32760,38665,45240,52521,60544,69345,78960,89425,
%U A011199 100776,113049,126280,140505,155760,172081,189504,208065,227800,248745,270936,294409,319200
%N A011199 a(n) = (n+1)*(2*n+1)*(3*n+1).
%H A011199 Ivan Panchenko, <a href="/A011199/b011199.txt">Table of n, a(n) for n = 0..1000</a>
%H A011199 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A011199 G.f.: (1 + 20*x + 15*x^2)/(x-1)^4. - _Alois P. Heinz_, Sep 04 2014
%F A011199 a(n) = 6*n^3 + 11*n^2 + 6*n + 1. - _Reinhard Zumkeller_, Jun 08 2015
%F A011199 E.g.f.: (1 + 23*x + 29*x^2 + 6*x^3)*exp(x). - _G. C. Greubel_, Mar 03 2020
%F A011199 From _Amiram Eldar_, Jan 13 2021: (Start)
%F A011199 Sum_{n>=0} 1/a(n) = sqrt(3)*Pi/4 - 4*log(2) + 9*log(3)/4.
%F A011199 Sum_{n>=0} (-1)^n/a(n) = 2*log(2) - (1 - sqrt(3)/2)*Pi. (End)
%p A011199 seq(mul(j*n+1, j=1..3), n = 0..40); # _G. C. Greubel_, Mar 03 2020
%t A011199 Product[j*Range[0,40] +1, {j,3}] (* _G. C. Greubel_, Mar 03 2020 *)
%t A011199 LinearRecurrence[{4,-6,4,-1},{1,24,105,280},40] (* _Harvey P. Dale_, Apr 21 2020 *)
%o A011199 (Haskell)
%o A011199 a011199 n = product $ map ((+ 1) . (* n)) [1, 2, 3]
%o A011199 -- _Reinhard Zumkeller_, Jun 08 2015
%o A011199 (PARI) vector(41, n, my(m=n-1); prod(j=1,3, j*m+1)) \\ _G. C. Greubel_, Mar 03 2020
%o A011199 (Magma) [&*[j*n+1:j in [1..3]]: n in [0..40]]; // _G. C. Greubel_, Mar 03 2020
%o A011199 (Sage) [product(j*n+1 for j in (1..3)) for n in (0..40)] # _G. C. Greubel_, Mar 03 2020
%o A011199 (GAP) List([0..40], n-> (n+1)*(2*n+1)*(3*n+1) ); # _G. C. Greubel_, Mar 03 2020
%Y A011199 Cf. A079588.
%K A011199 nonn,easy
%O A011199 0,2
%A A011199 _N. J. A. Sloane_