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%I #33 Aug 22 2018 07:50:21
%S 1,27,102,227,402,627,902,1227,1602,2027,2502,3027,3602,4227,4902,
%T 5627,6402,7227,8102,9027,10002,11027,12102,13227,14402,15627,16902,
%U 18227,19602,21027,22502,24027,25602,27227,28902,30627,32402,34227,36102,38027,40002
%N a(0) = 1, a(n) = 25*n^2 + 2 for n > 0.
%C Subsequence of A160842. - _Bruno Berselli_, Feb 06 2012
%C The identity (25*n^2 + 1)^2 - (25*n^2 + 2)*(5*n)^2 = 1 can be written as (A016850(n+1) + 1)^2 - a(n+1)*A008587(n+1)^2 = 1. - _Vincenzo Librandi_, Feb 08 2012
%H Bruno Berselli, <a href="/A010015/b010015.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: (1+x)*(1 + 23*x + x^2)/(1-x)^3. - _Bruno Berselli_, Feb 06 2012
%F E.g.f.: (x*(x+1)*25 + 2)*e^x - 1. - _Gopinath A. R._, Feb 14 2012
%t Join[{1}, 25 Range[40]^2 + 2] (* _Bruno Berselli_, Feb 06 2012 *)
%t Join[{1}, LinearRecurrence[{3, -3, 1}, {27, 102, 227}, 50]] (* _Vincenzo Librandi_, Feb 08 2012 *)
%o (PARI) A010015(n)=25*n^2+2-!n \\ _M. F. Hasler_, Feb 14 2012
%Y Cf. A206399.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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