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a(n) = (n+1)*(2*n+1)*(3*n+1).
7

%I #25 Oct 18 2022 15:01:34

%S 1,24,105,280,585,1056,1729,2640,3825,5320,7161,9384,12025,15120,

%T 18705,22816,27489,32760,38665,45240,52521,60544,69345,78960,89425,

%U 100776,113049,126280,140505,155760,172081,189504,208065,227800,248745,270936,294409,319200

%N a(n) = (n+1)*(2*n+1)*(3*n+1).

%H Ivan Panchenko, <a href="/A011199/b011199.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: (1 + 20*x + 15*x^2)/(x-1)^4. - _Alois P. Heinz_, Sep 04 2014

%F a(n) = 6*n^3 + 11*n^2 + 6*n + 1. - _Reinhard Zumkeller_, Jun 08 2015

%F E.g.f.: (1 + 23*x + 29*x^2 + 6*x^3)*exp(x). - _G. C. Greubel_, Mar 03 2020

%F From _Amiram Eldar_, Jan 13 2021: (Start)

%F Sum_{n>=0} 1/a(n) = sqrt(3)*Pi/4 - 4*log(2) + 9*log(3)/4.

%F Sum_{n>=0} (-1)^n/a(n) = 2*log(2) - (1 - sqrt(3)/2)*Pi. (End)

%p seq(mul(j*n+1, j=1..3), n = 0..40); # _G. C. Greubel_, Mar 03 2020

%t Product[j*Range[0,40] +1, {j,3}] (* _G. C. Greubel_, Mar 03 2020 *)

%t LinearRecurrence[{4,-6,4,-1},{1,24,105,280},40] (* _Harvey P. Dale_, Apr 21 2020 *)

%o (Haskell)

%o a011199 n = product $ map ((+ 1) . (* n)) [1, 2, 3]

%o -- _Reinhard Zumkeller_, Jun 08 2015

%o (PARI) vector(41, n, my(m=n-1); prod(j=1,3, j*m+1)) \\ _G. C. Greubel_, Mar 03 2020

%o (Magma) [&*[j*n+1:j in [1..3]]: n in [0..40]]; // _G. C. Greubel_, Mar 03 2020

%o (Sage) [product(j*n+1 for j in (1..3)) for n in (0..40)] # _G. C. Greubel_, Mar 03 2020

%o (GAP) List([0..40], n-> (n+1)*(2*n+1)*(3*n+1) ); # _G. C. Greubel_, Mar 03 2020

%Y Cf. A079588.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_