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a(n) = round((n+1/n)^3).
1

%I #25 Feb 05 2024 10:56:24

%S 8,16,37,77,141,235,364,536,756,1030,1364,1764,2236,2786,3420,4144,

%T 4964,5886,6916,8060,9324,10714,12236,13896,15700,17654,19764,22036,

%U 24476,27090,29884,32864,36036,39406,42980,46764,50764

%N a(n) = round((n+1/n)^3).

%H Vincenzo Librandi, <a href="/A197986/b197986.txt">Table of n, a(n) for n = 1..10000</a>

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

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(1)= 8, a{2}=16, a(3)=37, a(4)=77, a(5)=141, a(6)=235, a(7)=364, a(8)=536, a(9)=756, a(10)=1030. - _Harvey P. Dale_, Apr 05 2012

%F From _G. C. Greubel_, Feb 04 2024: (Start)

%F a(n) = n*(n^2+3) for n > 6, with a(1)=8, a(2)=16, a(3)=37, a(4)=77, a(5)=141, a(6)=235.

%F G.f.: x*(8 - 16*x + 21*x^2 - 7*x^3 - x^4 + x^5 - x^6 + 3*x^7 - 3*x^8 + x^9)/(1-x)^4.

%F E.g.f.: 4*x + x^2 + x^3/3! + x^4/4! + x^5/5! + x^6/6! + x*(4 + 3*x + x^2)*exp(x). (End)

%t Table[Round[(n+1/n)^3],{n,40}] (* or *) Join[{8,16,37,77,141,235}, LinearRecurrence[ {4,-6,4,-1},{364,536,756,1030},40]] (* _Harvey P. Dale_, Apr 05 2012 *)

%o (Magma) [Round((n+1/n)^3): n in [1..60]]

%o (SageMath) [8,16,37,77,141,235]+[n*(n^2+3) for n in range(7,51)] # _G. C. Greubel_, Feb 04 2024

%Y Cf. A014056, A197985.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Oct 21 2011