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a(n) = 31*n^2 + 1.
1

%I #16 Nov 29 2024 19:50:45

%S 1,32,125,280,497,776,1117,1520,1985,2512,3101,3752,4465,5240,6077,

%T 6976,7937,8960,10045,11192,12401,13672,15005,16400,17857,19376,20957,

%U 22600,24305,26072,27901,29792,31745,33760,35837,37976,40177,42440,44765,47152,49601

%N a(n) = 31*n^2 + 1.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(0)=1, a(1)=32, a(2)=125, a(3)=280, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Harvey P. Dale_, Apr 28 2016

%F From _Elmo R. Oliveira_, Nov 29 2024: (Start)

%F G.f.: (32*x^2+29*x+1)/(1-x)^3.

%F E.g.f.: exp(x)*(1 + 31*x + 31*x^2).

%F a(n) = A010020(n) - 1 for n>=1. (End)

%t 31*Range[100]^2+1 (* or *) LinearRecurrence[{3,-3,1},{32,125,280},100] (* _Harvey P. Dale_, Apr 28 2016 *)

%o (PARI) a(n)=31*n^2+1 \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. A010020.

%K nonn,easy

%O 0,2

%A _Garrett Backer_, Nov 21 2014