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

1, 9, 171, 2439, 30951, 368559, 4217031, 46953279, 512579511, 5513215599, 58618940391, 617570463519, 6458134171671, 67123207545039, 694108867905351, 7146979811148159, 73322818300333431, 749905364703000879, 7649148282327007911, 77842334540943071199

Offset: 1

Author

Jon E. Schoenfield, Sep 14 2013

Keywords

• nonn
• base
• easy

Comments

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.)

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

Examples

			a(2) = 9, since there are 9 2-digit numbers that contain a '0'.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (19,-90).

Crossrefs

Cf. A011540, A050720, A217094.

Programs

Mathematica

LinearRecurrence[{19,-90},{1,9,171},20] (* Stefano Spezia, Nov 15 2023 *)

Formula

For n > 1, a(n) = 9*10^(n-1) - 9^n.

For n > 2, a(n) = 9*(a(n-1) + 10^(n-2)).

G.f.: x*(1-10*x+90*x^2)/((1-9*x)*(1-10*x)). - R. J. Mathar, Sep 14 2013

a(n) = A217094(n) - A217094(n-1), for n > 1. - Hieronymus Fischer, Dec 27 2013

E.g.f.: (9*exp(10*x) - 10*exp(9*x) + 10*x + 1)/10. - Stefano Spezia, Nov 15 2023

Extensions

Example added and g.f. corrected by Hieronymus Fischer, Dec 27 2013