cp's OEIS Frontend

A229127 Number of n-digit numbers containing the digit '0'.

%I A229127 #30 Mar 11 2024 23:12:58
%S A229127 1,9,171,2439,30951,368559,4217031,46953279,512579511,5513215599,
%T A229127 58618940391,617570463519,6458134171671,67123207545039,
%U A229127 694108867905351,7146979811148159,73322818300333431,749905364703000879,7649148282327007911,77842334540943071199
%N A229127 Number of n-digit numbers containing the digit '0'.
%C A229127 Other than the number 0 itself, numbers with leading zeros are not allowed, so the general formula is a(n)=9*10^(n-1)-9^n, which is simply the number of n-digit numbers that begin with a nonzero digit (9*10^(n-1)) minus the number of n-digit numbers consisting only of nonzero digits (9^n). (Because of the 1-digit number 0 itself, the general formula does not apply at n=1.)
%C A229127 Other than the number 1, and 9 which is a semiprime, the minimum number of possible prime factors with multiplicity of a(n) = 3, which holds for 171 = 3^2 * 19; 2439 = 3^2 * 271; 46953279 = 3^2 * 5217031; 617570463519 = 3^2 * 68618940391; 77842334540943071199 = 3^2 * 8649148282327007911. - _Jonathan Vos Post_, Sep 16 2013
%H A229127 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (19,-90).
%F A229127 For n > 1, a(n) = 9*10^(n-1) - 9^n.
%F A229127 For n > 2, a(n) = 9*(a(n-1) + 10^(n-2)).
%F A229127 G.f.: x*(1-10*x+90*x^2)/((1-9*x)*(1-10*x)). - _R. J. Mathar_, Sep 14 2013
%F A229127 a(n) = A217094(n) - A217094(n-1), for n > 1. - _Hieronymus Fischer_, Dec 27 2013
%F A229127 E.g.f.: (9*exp(10*x) - 10*exp(9*x) + 10*x + 1)/10. - _Stefano Spezia_, Nov 15 2023
%e A229127 a(2) = 9, since there are 9 2-digit numbers that contain a '0'.
%t A229127 LinearRecurrence[{19,-90},{1,9,171},20] (* _Stefano Spezia_, Nov 15 2023 *)
%Y A229127 Cf. A011540, A050720, A217094.
%K A229127 nonn,base,easy
%O A229127 1,2
%A A229127 _Jon E. Schoenfield_, Sep 14 2013
%E A229127 Example added and g.f. corrected by _Hieronymus Fischer_, Dec 27 2013