OFFSET
1,2
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (20,-109,90).
FORMULA
a(n) = 10^n + 9/8 - 9^(1+n)/8.
G.f.: (1-19*x+99*x^2)/((1-x)*(1-10*x)*(1-9*x)). - Vincenzo Librandi, Aug 14 2013
a(n) = 20*a(n-1) - 109*a(n-2) + 90*a(n-3); a(0)=1, a(1)=10, a(2)=181. - Harvey P. Dale, Jun 20 2015
Limit_{n->oo} a(n+1)/a(n) = 10. - Bernard Schott, Feb 28 2023
EXAMPLE
a(3)=181 because among the integers from 0 to 999 there are 181 numbers which contain at least 1 zero.
MATHEMATICA
LinearRecurrence[{20, -109, 90}, {1, 10, 181}, 20] (* or *) Rest[ CoefficientList[ Series[(1-19x+99x^2)/((1-x)(1-10x)(1-9x)), {x, 0, 20}], x]] (* Harvey P. Dale, Jun 20 2015 *)
PROG
(PARI) a(n) = 10^n + 9/8 - 9^(1+n)/8; \\ Michel Marcus, Aug 13 2013
(Magma) [10^n + 9/8 - 9^(1+n)/8: n in [1..20]]; // Vincenzo Librandi, Aug 14 2013
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Michael Taktikos, Aug 25 2004
STATUS
approved