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A163663 a(0) = 0 and A059153(n-1) written in base 2 otherwise.

%I A163663 #16 Aug 02 2017 05:05:25
%S A163663 0,100,11000,1110000,111100000,11111000000,1111110000000,
%T A163663 111111100000000,11111111000000000,1111111110000000000,
%U A163663 111111111100000000000,11111111111000000000000,1111111111110000000000000,111111111111100000000000000,11111111111111000000000000000
%N A163663 a(0) = 0 and A059153(n-1) written in base 2 otherwise.
%C A163663 The digit pattern of a(n) is: n times 1 and (n+1) times 0.
%H A163663 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (110, -1000).
%F A163663 From _R. J. Mathar_, Aug 12 2009: (Start)
%F A163663 a(n) = 110*a(n-1) - 1000*a(n-2).
%F A163663 a(n) = 100*A109241(n-1).
%F A163663 G.f.: 100*x/((100*x-1) * (10*x-1)). (End)
%F A163663 From _Wesley Ivan Hurt_, Jun 22 2013: (Start)
%F A163663 a(n) = (10/9) * (10^n) * (10^n - 1).
%F A163663 a(n) = A002275(n) * 10^(n+1). (End)
%F A163663 E.g.f.: (10/9)*(exp(100*x) - exp(10*x)). - _G. C. Greubel_, Aug 01 2017
%t A163663 Table[FromDigits[Join[PadRight[{},n-1,1],PadRight[{},n,0]]],{n,20}] (* _Harvey P. Dale_, Aug 29 2015 *)
%o A163663 (PARI) x='x+O('x^50); Vec(100*x/((100*x-1)*(10*x-1))) \\ _G. C. Greubel_, Aug 01 2017
%K A163663 nonn,base
%O A163663 0,2
%A A163663 _Jaroslav Krizek_, Aug 02 2009
%E A163663 Edited by _R. J. Mathar_, Aug 12 2009