%I #26 Sep 08 2022 08:46:04
%S 2,2,11,182,2621,33572,402131,4619162,51572441,564151952,6077367551,
%T 64696307942,682266771461,7140400943132,74263608488171,
%U 768372476393522,7915352287541681,81238170587875112,831143535290875991,8480291817617883902,86322626358560955101
%N Least index k such that A011540(k) >= 10^n.
%C For n>0 also index k such that A011540(k) = 10^n.
%C For n>1: A011540(a(n)) is the least number with n zero digits.
%C For n>0: a(n) - 1 is the number of numbers with <= n digits which contain the digit '0'. - _Hieronymus Fischer_, Dec 27 2013
%H Hieronymus Fischer, <a href="/A217094/b217094.txt">Table of n, a(n) for n = 0..200</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (20,-109,90).
%F a(n+1) = 10*a(n) - 9*a(n-1) + 9*10^(n-1), n>0.
%F a(n) = 2 + 10^n - 9^n - (9^n - 1)/8.
%F A011540(a(n)) = 10^n, for n>0.
%F a(n) = A053283(A002452(n+1) - 1) - A002452(n+1) + 3, n>0.
%F a(n) = 10^n + 2 - A002452(n+1).
%F G.f.: (189*x^2 - 38*x + 2)/((1-x)*(1-9*x)*(1-10*x)).
%F a(n) = 1 + sum_{1<=k<=n} A229127(k), for n>0. - _Hieronymus Fischer_, Dec 27 2013
%e a(0) = 2, since A011540(2) = 10 >= 10^0.
%e a(1) = 2, since A011540(2) = 10 >= 10^1.
%e a(2) = 11, since A011540(11) = 100 >= 10^2, but A011540(10) = 90 < 10^2.
%t LinearRecurrence[{20,-109,90},{2,2,11},30] (* _Harvey P. Dale_, Aug 02 2015 *)
%o (PARI) for(n=0,50, print1(2 +10^n -9^n -(9^n -1)/8, ", ")) \\ _G. C. Greubel_, Apr 18 2018
%o (Magma) [2 +10^n -9^n -(9^n -1)/8: n in [0..50]]; // _G. C. Greubel_, Apr 18 2018
%Y Cf. A011540, A053283, A002452, A229127.
%K nonn,easy
%O 0,1
%A _Hieronymus Fischer_, Jan 23 2013