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 A217094 Least index k such that A011540(k) >= 10^n. 7
 2, 2, 11, 182, 2621, 33572, 402131, 4619162, 51572441, 564151952, 6077367551, 64696307942, 682266771461, 7140400943132, 74263608488171, 768372476393522, 7915352287541681, 81238170587875112, 831143535290875991, 8480291817617883902, 86322626358560955101 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS For n>0 also index k such that A011540(k) = 10^n. For n>1: A011540(a(n)) is the least number with n zero digits. For n>0: a(n) - 1 is the number of numbers with <= n digits which contain the digit '0'. - Hieronymus Fischer, Dec 27 2013 LINKS Hieronymus Fischer, Table of n, a(n) for n = 0..200 Index entries for linear recurrences with constant coefficients, signature (20,-109,90). FORMULA a(n+1) = 10*a(n) - 9*a(n-1) + 9*10^(n-1), n>0. a(n) = 2 + 10^n - 9^n - (9^n - 1)/8. A011540(a(n)) = 10^n, for n>0. a(n) = A053283(A002452(n+1) - 1) - A002452(n+1) + 3, n>0. a(n) = 10^n + 2 - A002452(n+1). G.f.: (189*x^2 - 38*x + 2)/((1-x)*(1-9*x)*(1-10*x)). a(n) = 1 + sum_{1<=k<=n} A229127(k), for n>0. - Hieronymus Fischer, Dec 27 2013 EXAMPLE a(0) = 2, since A011540(2) = 10 >= 10^0. a(1) = 2, since A011540(2) = 10 >= 10^1. a(2) = 11, since A011540(11) = 100 >= 10^2, but A011540(10) = 90 < 10^2. MATHEMATICA LinearRecurrence[{20, -109, 90}, {2, 2, 11}, 30] (* Harvey P. Dale, Aug 02 2015 *) PROG (PARI) for(n=0, 50, print1(2 +10^n -9^n -(9^n -1)/8, ", ")) \\ G. C. Greubel, Apr 18 2018 (MAGMA) [2 +10^n -9^n -(9^n -1)/8: n in [0..50]]; // G. C. Greubel, Apr 18 2018 CROSSREFS Cf. A011540, A053283, A002452, A229127. Sequence in context: A175976 A095215 A264712 * A076976 A291757 A286464 Adjacent sequences:  A217091 A217092 A217093 * A217095 A217096 A217097 KEYWORD nonn,easy AUTHOR Hieronymus Fischer, Jan 23 2013 STATUS approved

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Last modified August 8 05:50 EDT 2020. Contains 336290 sequences. (Running on oeis4.)