%I #33 Mar 12 2024 14:19:19

%S 0,9,90,819,7380,66429,597870,5380839,48427560,435848049,3922632450,

%T 35303692059,317733228540,2859599056869,25736391511830,

%U 231627523606479,2084647712458320,18761829412124889,168856464709124010,1519708182382116099,13677373641439044900

%N Number of integers from 1 to 10^n-1 that lack 0 as a digit.

%H Vincenzo Librandi, <a href="/A052386/b052386.txt">Table of n, a(n) for n = 0..500</a>

%H Peter D. Loly and Ian D. Cameron, <a href="https://arxiv.org/abs/2008.11020">Frierson's 1907 Parameterization of Compound Magic Squares Extended to Orders 3^L, L = 1, 2, 3, ..., with Information Entropy</a>, arXiv:2008.11020 [math.HO], 2020.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-9).

%F a(n) = 9*a(n-1) + 9.

%F a(n) = 9*(9^n-1)/8 = sum_{j=1..n} 9^j = a(n-1)+9^n = 9*A002452(n) = A002452(n+1)-1; write A000918(n+1) in base 2 and read as if written in base 9. - _Henry Bottomley_, Aug 30 2001

%F a(n) = 10*a(n-1)-9*a(n-2). G.f.: 9*x / ((x-1)*(9*x-1)). - _Colin Barker_, Sep 26 2013

%e For n=2, the numbers from 1 to 99 which *have* 0 as a digit are the 9 numbers 10, 20, 30, ..., 90. So a(1) = 99 - 9 = 90.

%t Table[9(9^n - 1)/8, {n, 0, 20}]

%t LinearRecurrence[{10,-9},{0,9},30] (* _Harvey P. Dale_, Mar 22 2019 *)

%o (Magma) [9*(9^n-1)/8: n in [0..20]]; // _Vincenzo Librandi_, Jul 04 2011

%o (PARI) a(n)=9^(n+1)\8 \\ _Charles R Greathouse IV_, Aug 25 2014

%Y Cf. A024101, A052379.

%Y Row n=9 of A228275.

%K easy,nonn,base

%O 0,2

%A _Odimar Fabeny_, Mar 10 2000

%E More terms and revised description from _James A. Sellers_, Mar 13 2000

%E More terms and revised description from _Robert G. Wilson v_, Apr 14 2003

%E More terms from _Colin Barker_, Sep 26 2013