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%I #23 Jun 29 2023 18:23:03
%S 21,291,2991,29991,299991,2999991,29999991,299999991,2999999991,
%T 29999999991,299999999991,2999999999991,29999999999991,
%U 299999999999991,2999999999999991,29999999999999991,299999999999999991
%N Numbers k such that Reverse(k+9) = 3.
%C Reverse(a(n)) = A086948(n).
%C If k is in this sequence then Reverse(k) = (2/3)*k - 2. Also A101703 is the sequence of all numbers k such that Reverse(k) = (2/3)*k - 2. So this sequence is a subsequence of A101703. - _Farideh Firoozbakht_, Dec 30 2004
%H Vincenzo Librandi, <a href="/A086947/b086947.txt">Table of n, a(n) for n = 1..300</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, -10).
%F a(n) = 3*(10^n - 3).
%F From _Chai Wah Wu_, Aug 01 2020: (Start)
%F a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.
%F G.f.: x*(60*x + 21)/((x - 1)*(10*x - 1)). (End)
%t Table[3*(10^n-3), {n, 17}]
%t Table[FromDigits[PadRight[{3},n,0]],{n,2,20}]-9 (* _Harvey P. Dale_, Nov 27 2012 *)
%o (Magma) [3*(10^n-3): n in [1..25] ]; // _Vincenzo Librandi_, Aug 22 2011
%Y Cf. A085331, A086948, A101700, A101703.
%K nonn,base
%O 1,1
%A _Ray Chandler_, Jul 24 2003
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