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a(n) = (2n-2)^3 + (2n-2) - 1.
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%I #33 Jan 13 2022 02:26:41

%S -1,9,67,221,519,1009,1739,2757,4111,5849,8019,10669,13847,17601,

%T 21979,27029,32799,39337,46691,54909,64039,74129,85227,97381,110639,

%U 125049,140659,157517,175671,195169,216059,238389,262207,287561,314499,343069,373319

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

%C a(n)/a(n-1) tends to 1 as n becomes very large (of order 10^3 or more).

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

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

%F G.f.: x*(-1 + 13*x + 25*x^2 + 11*x^3)/(1-x)^4. - _Vincenzo Librandi_, Mar 17 2015

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4. - _Vincenzo Librandi_, Mar 17 2015

%e a(3) = (2*3-2)^3 + (2*3-2) - 1 = 67.

%t Table[(2 n - 2)^3 + (2 n - 2) - 1, {n, 30}] (* _Michael De Vlieger_, Mar 17 2015 *)

%o (PARI) a(n)=8*n^3 - 24*n^2 + 26*n - 11 \\ _Charles R Greathouse IV_, Mar 17 2015

%K sign,easy

%O 1,2

%A _Arka Mal_, Mar 08 2015

%E More terms from _Vincenzo Librandi_, Mar 17 2015