A033714 Number of zeros in numbers 0 to 999..9 (n digits).

1, 10, 190, 2890, 38890, 488890, 5888890, 68888890, 788888890, 8888888890, 98888888890, 1088888888890, 11888888888890, 128888888888890, 1388888888888890, 14888888888888890, 158888888888888890, 1688888888888888890, 17888888888888888890, 188888888888888888890

Offset: 1

Author

Olivier Gorin (gorin(AT)roazhon.inra.fr)

Keywords

• nonn
• base
• nice
• easy

Comments

This sequence also gives the total count of digits of n below 10^n. In such counts it makes sense to omit 10^0 as we are interested in having ten digits under each power of 10. For each power of 10 the total number of digits 0-9 is always the total of zeros for the next power. For example, at 10^1 there is 1 of each numeral 0-9, total 10 digits. At 10^2, the number of zeros is 10, with 20 each for the other 9 numerals and so on. - Enoch Haga, May 13 2006

Also the position of 10^n in Champernowne's constant (A033307). See Sikora, p. 3. - Robert G. Wilson v, Jun 29 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..100
• John K. Sikora, On the High Water Mark Convergents of Champernowne's Constant in Base Ten, arXiv:1210.1263 [math.NT], 2012.
• Index entries for linear recurrences with constant coefficients, signature (21,-120,100).

Crossrefs

Cf. A033713, A053541.

Cf. A212704 (first differences).

Programs

Magma

[(9*n*10^n-10*10^n+100)/90: n in [1..20]]; // Vincenzo Librandi, Jul 01 2014

Mathematica

a[1] = 1; a[n_] := a[n] = 9*10^(n-2)*(n-1) + a[n-1]; Table[a[n], {n, 1, 17}] (* Jean-François Alcover, Jul 13 2012 *)
f[n_] := 1 + Sum[9 m*10^(m - 1), {m, n}]; Array[f, 18, 0] (* Robert G. Wilson v, Jun 29 2014 *)
LinearRecurrence[{21,-120,100},{1,10,190},20] (* Harvey P. Dale, Dec 03 2021 *)

PARI

Vec(-x*(100*x^2-11*x+1)/((x-1)*(10*x-1)^2)  + O(x^100)) \\ Colin Barker, Jan 27 2015

Formula

a(n) = 10^(n-1)*n - (1/9)*10^n + 10/9. - Robert Israel, Jun 30 2014

G.f.: -x*(100*x^2-11*x+1) / ((x-1)*(10*x-1)^2). - Colin Barker, Jan 27 2015

From Bernard Schott, Nov 20 2022: (Start)

a(n) = A033713(n) + 1.

a(n+1) = a(n) + 9 * A053541(n). (End)

Extensions

More terms from Erich Friedman