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a(n) = n*(2*n+5)*(2*n+7).
0

%I #18 Oct 18 2022 15:29:52

%S 0,63,198,429,780,1275,1938,2793,3864,5175,6750,8613,10788,13299,

%T 16170,19425,23088,27183,31734,36765,42300,48363,54978,62169,69960,

%U 78375,87438,97173,107604,118755,130650,143313,156768,171039,186150,202125,218988,236763

%N a(n) = n*(2*n+5)*(2*n+7).

%D Eric Harold Neville, Jacobian Elliptic Functions, 2nd ed., p. 38.

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

%F From _Wesley Ivan Hurt_, Oct 05 2020: (Start)

%F a(n) = 4*n^3 + 24*n^2 + 35*n.

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

%F G.f.: 3*x*(21-18*x+5*x^2)/(1-x)^4. (End)

%t Table[n*(2*n + 5)*(2*n + 7), {n, 0, 60}] (* _Wesley Ivan Hurt_, Oct 05 2020 *)

%o (Magma) [n*(2*n+5)*(2*n+7) : n in [0..60]]; // _Wesley Ivan Hurt_, Oct 05 2020

%o (PARI) a(n)=n*(2*n+5)*(2*n+7) \\ _Charles R Greathouse IV_, Oct 18 2022

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Sean A. Irvine_, Oct 05 2020