%I #10 Jun 13 2015 00:52:36
%S 101,11100111,1111100011111,111111100001111111,
%T 11111111100000111111111,1111111111100000011111111111,
%U 111111111111100000001111111111111,11111111111111100000000111111111111111
%N Concatenation of 2n-1 digits 1, n digits 0 and 2n-1 digits 1.
%C a(n) has 5n-2 digits.
%C a(n) is also A147540(n) written in base 2. [From _Omar E. Pol_, Nov 08 2008]
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (101101,-110201100,10110100000,-10000000000).
%F a(n) = (10^(2n-1)-1+10^(5n-2)-10^(3n-1))/9. [_R. J. Mathar_, Nov 07 2008, corrected Nov 09 2008]
%F G.f.: x*(1100000000*x^3-2000000*x^2+888910*x+101) / ((x-1)*(100*x-1)*(1000*x-1)*(100000*x-1)). - _Colin Barker_, Sep 16 2013
%e n ........... a(n)
%e 1 ........... 101
%e 2 ......... 11100111
%e 3 ....... 1111100011111
%e 4 ..... 111111100001111111
%e 5 ... 11111111100000111111111
%o (PARI) Vec(x*(1100000000*x^3-2000000*x^2+888910*x+101)/((x-1)*(100*x-1)*(1000*x-1)*(100000*x-1)) + O(x^100)) \\ _Colin Barker_, Sep 16 2013
%Y Cf. A138120, A138144, A138145, A138146, A138148, A138720, A138721, A138722, A147540.
%K base,easy,nonn
%O 1,1
%A _Omar E. Pol_, Apr 06 2008
%E Corrected typo in exponent of my formula. - _R. J. Mathar_, Nov 09 2008
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