A217094 Least index k such that A011540(k) >= 10^n.
2, 2, 11, 182, 2621, 33572, 402131, 4619162, 51572441, 564151952, 6077367551, 64696307942, 682266771461, 7140400943132, 74263608488171, 768372476393522, 7915352287541681, 81238170587875112, 831143535290875991, 8480291817617883902, 86322626358560955101
Offset: 0
Examples
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.
Links
- Hieronymus Fischer, Table of n, a(n) for n = 0..200
- Index entries for linear recurrences with constant coefficients, signature (20,-109,90).
Programs
-
Magma
[2 +10^n -9^n -(9^n -1)/8: n in [0..50]]; // G. C. Greubel, Apr 18 2018
-
Mathematica
LinearRecurrence[{20,-109,90},{2,2,11},30] (* Harvey P. Dale, Aug 02 2015 *)
-
PARI
for(n=0,50, print1(2 +10^n -9^n -(9^n -1)/8, ", ")) \\ G. C. Greubel, Apr 18 2018
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) = 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
Comments