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%I #26 Sep 08 2022 08:45:19
%S 5,29,69,125,197,285,389,509,645,797,965,1149,1349,1565,1797,2045,
%T 2309,2589,2885,3197,3525,3869,4229,4605,4997,5405,5829,6269,6725,
%U 7197,7685,8189,8709,9245,9797,10365,10949,11549,12165,12797,13445,14109,14789
%N a(n) = 8*n^2 - 3.
%C Sequence found by reading the segment (5, 29) together with the line from 29, in the direction 29, 69,..., in the square spiral whose vertices are the triangular numbers A000217. - _Omar E. Pol_, Sep 04 2011
%H Vincenzo Librandi, <a href="/A108928/b108928.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 2*(2*n-1)*(2*n+1)-1.
%F a(1)=5, a(2)=29, a(3)=69, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). - _Harvey P. Dale_, Jul 21 2012
%F From _G. C. Greubel_, Jul 15 2017:(Start)
%F G.f.: x*(-5 - 14*x + 3*x^2)/(-1 + x)^3.
%F E.g.f.: (8*x^2 + 8*x - 3)*exp(x) + 3. (End)
%e (1*3 = 3)+2 = 5; (3*5 = 15)+14 = 29; (5*7 = 35)+34 = 69; (7*9 = 63)+62 = 125; ...
%p seq(8*n^2-3,n=1..50); # _Emeric Deutsch_, Aug 01 2005
%t 8*Range[50]^2-3 (* or *) LinearRecurrence[{3,-3,1},{5,29,69},50] (* _Harvey P. Dale_, Jul 21 2012 *)
%o (PARI) a(n)=8*n^2-3 \\ _Charles R Greathouse IV_, Sep 04 2011
%o (Magma) [8*n^2 - 3: n in [1..50]]; // _Vincenzo Librandi_, Sep 05 2011
%K easy,nonn
%O 1,1
%A Marcel Hetkowski Fabeny (marcelfabeny(AT)yahoo.com.br), Jul 19 2005
%E More terms from _Emeric Deutsch_, Aug 01 2005