A113119 Total number of digits in all n-digit nonnegative integers.

10, 180, 2700, 36000, 450000, 5400000, 63000000, 720000000, 8100000000, 90000000000, 990000000000, 10800000000000, 117000000000000, 1260000000000000, 13500000000000000, 144000000000000000, 1530000000000000000, 16200000000000000000, 171000000000000000000

Offset

1,1

Links

Colin Barker, Table of n, a(n) for n = 1..950

The Math Forum, Ask Dr. Math: Really Counting to One Billion, 2001.

Index entries for linear recurrences with constant coefficients, signature (20,-100).

Formula

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

From Colin Barker, Aug 05 2016: (Start)

a(n) = 20*a(n-1) - 100*a(n-2) for n > 3.

G.f.: 10*x*(1 - 2*x + 10*x^2) / (1-10*x)^2.

(End)

From Bernard Schott, Nov 14 2022: (Start)

a(n) = A212704(n) for n > 1.

a(n) = 9 * A053541(n) for n > 1.

a(n) = 9 * A081045(n-1) + A212704(n-1), for n > 1 (means a(n) = number of nonzero digits + number of zero digits). (End)

E.g.f.: x*(1 + 9*exp(10*x)). - Stefano Spezia, Dec 24 2022

Example

a(1)=10 because there are ten one-digit numbers (including the 0).
a(2)=180 because there are 100-10=90 two-digit numbers, for a total of 90*2=180 digits.

Mathematica

LinearRecurrence[{20, -100}, {10, 180, 2700}, 20] (* Harvey P. Dale, Dec 09 2021 *)

Prog

(PARI) Vec(10*x*(1-2*x+10*x^2)/(1-10*x)^2 + O(x^20)) \\ Colin Barker, Aug 05 2016
(Python)
def a(n): return 10 if n == 1 else 9*n*10**(n-1)
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Nov 14 2022

Crossrefs

Cf. A053541, A081045.

Essentially the same as A212704.

Sequence in context: A057122 A261177 A367986 * A067416 A113671 A001762

Adjacent sequences: A113116 A113117 A113118 * A113120 A113121 A113122

Keyword

base,easy,nonn

Author

Alexandre Wajnberg, Jan 03 2006

Extensions

More terms from Joshua Zucker, May 08 2006

a(17) corrected by Colin Barker, Aug 05 2016

Status

approved