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a(n) = Fibonacci(n) + n^3.
4

%I #14 Sep 08 2022 08:46:02

%S 0,2,9,29,67,130,224,356,533,763,1055,1420,1872,2430,3121,3985,5083,

%T 6510,8416,11040,14765,20207,28359,40824,60192,90650,138969,216101,

%U 339763,538618,859040,1376060,2211077,3560515,5742191,9270340,14977008,24208470,39143041

%N a(n) = Fibonacci(n) + n^3.

%H Bruno Berselli, <a href="/A212272/b212272.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,6,1,-3,1).

%F G.f.: x*(2-x+2*x^2-9*x^3)/((1-x-x^2)*(1-x)^4).

%t Table[Fibonacci[n] + n^3, {n, 0, 38}]

%o (PARI) for(n=0, 38, print1(fibonacci(n)+n^3", "));

%o (Magma) [Fibonacci(n)+n^3: n in [0..38]];

%Y Cf. A000045, A000578; A001611, A002062, A160536.

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, May 09 2012