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%I #32 Sep 08 2022 08:45:15
%S 1,5,17,43,89,161,265,407,593,829,1121,1475,1897,2393,2969,3631,4385,
%T 5237,6193,7259,8441,9745,11177,12743,14449,16301,18305,20467,22793,
%U 25289,27961,30815,33857,37093,40529,44171,48025,52097,56393,60919,65681,70685,75937
%N a(n) = n^3 + (n+1)^2.
%C The finite simple continued fraction [1;n-1,n,n+1] has a numerator in the resulting rational number (n^3+2*n+n^2+1)/(n*(n^2+1)) that is the same as a(n). - _J. M. Bergot_, Sep 29 2011
%H Vincenzo Librandi, <a href="/A100705/b100705.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: ( 1+x+3*x^2+x^3 ) / (x-1)^4 . - _R. J. Mathar_, Sep 29 2011
%F a(n) = A053698(n)+n. - _Bruno Berselli_, Sep 30 2011
%t Table[n^3 + (n + 1)^2, {n, 0, 45}]
%t LinearRecurrence[{4,-6,4,-1},{1,5,17,43},50] (* _Harvey P. Dale_, Dec 03 2014 *)
%o (Magma) [n^3 + (n+1)^2: n in [0..50]]; // _Vincenzo Librandi_, Sep 30 2011
%o (PARI) a(n)=n^3+(n+1)^2 \\ _Charles R Greathouse IV_, Oct 07 2015
%K nonn,easy
%O 0,2
%A _Giovanni Teofilatto_, Jan 03 2005
%E More terms from Mark Hudson and _Farideh Firoozbakht_, Jan 04 2005
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