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a(n) = 1 + 111110*n.
1

%I #7 Oct 13 2016 03:39:36

%S 1,111111,222221,333331,444441,555551,666661,777771,888881,999991,

%T 1111101,1222211,1333321,1444431,1555541,1666651,1777761,1888871,

%U 1999981,2111091,2222201,2333311,2444421,2555531,2666641,2777751,2888861,2999971,3111081,3222191

%N a(n) = 1 + 111110*n.

%H G. C. Greubel, <a href="/A135403/b135403.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

%F From _G. C. Greubel_, Oct 12 2016: (Start)

%F G.f.: (1 + 111109*x)/(1 - x)^2.

%F E.g.f.: (1 + 111110*x)*exp(x).

%F a(n) = 2*a(n-1) - a(n-2). (End)

%t 1+111110*Range[0,30] (* _Harvey P. Dale_, Sep 21 2012 *)

%t LinearRecurrence[{2,-1},{1,111111}, 25] (* _G. C. Greubel_, Oct 12 2016 *)

%K nonn,easy

%O 0,2

%A Dohong Kim 20710847 (upossss(AT)naver.com), Dec 11 2007

%E a(12) and a(13) corrected by _T. D. Noe_, Mar 14 2008

%E More terms from _Harvey P. Dale_, Sep 21 2012