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A268484 a(n) = (n + 1)*(4*n^2 + 14*n + 9)/3.

%I A268484 #29 Nov 16 2024 21:30:34
%S A268484 3,18,53,116,215,358,553,808,1131,1530,2013,2588,3263,4046,4945,5968,
%T A268484 7123,8418,9861,11460,13223,15158,17273,19576,22075,24778,27693,30828,
%U A268484 34191,37790,41633,45728,50083,54706,59605,64788,70263,76038,82121,88520,95243
%N A268484 a(n) = (n + 1)*(4*n^2 + 14*n + 9)/3.
%H A268484 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A268484 G.f.: (3 + 6*x - x^2)/(x - 1)^4.
%F A268484 a(n) = Sum_{k = 0..n} (2*k + 1)*(2*k + 3) = Sum_{k = 0..n} A005408(k)*A005408(k + 1).
%F A268484 Sum_{n>=0} 1/a(n) = 0.4315109123788144393864...
%e A268484 a(0) = 1*3 = 3;
%e A268484 a(1) = 1*3 + 3*5 = 18;
%e A268484 a(2) = 1*3 + 3*5 + 5*7 = 53;
%e A268484 a(3) = 1*3 + 3*5 + 5*7 + 7*9 = 116, etc.
%t A268484 Table[(n + 1) ((4 n^2 + 14 n + 9)/3), {n, 0, 40}]
%t A268484 LinearRecurrence[{4, -6, 4, -1}, {3, 18, 53, 116}, 40]
%o A268484 (PARI) a(n)=(n+1)*(4*n^2+14*n+9)/3 \\ _Charles R Greathouse IV_, Jul 26 2016
%Y A268484 Cf. A000466, A005408, A007290, A135036.
%K A268484 nonn,easy
%O A268484 0,1
%A A268484 _Ilya Gutkovskiy_, Feb 12 2016