A043489 Numbers having one 0 in base 10.

0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 220, 230, 240, 250, 260, 270, 280, 290, 301, 302, 303

Offset

1,2

Comments

From Hieronymus Fischer, May 28 2014: (Start)

Inversion:

Given a term m, the index n such that a(n) = m can be calculated by the following procedure [see Prog section with an implementation in Smalltalk]. With k := floor(log_10(m)), z = digit position of the '0' in m counted from the right (starting with 0).

Case 1: A043489_inverse(m) = 1 + Sum_{j=1..k} A052382_inverse(floor(m/10^j))*9^(j-1), if z = 0.

Case 2: A043489_inverse(m) = 1 + A043489_inverse(m - c - m mod 10^z) + A052382_inverse(m mod 10^z)) - (9^z - 1)/8, if z > 0, where c := 1, if the digit at position z+1 of m is ‘1’ and k > z + 1, otherwise c := 10.

Example 1: m = 990, k = 2, z = 0 (Case 1), A043489_inverse(990) = 1 + A052382_inverse(99))*1 + A052382_inverse(9))*9 = 1 + 90 + 81 = 172.

Example 2: m = 1099, k = 3, z = 2 (Case 2), A043489_inverse(1099) = 1 + A043489_inverse(990) + A052382_inverse(99)) - 10 = 1 + A043489_inverse(990) + 80 = 1 + 172 + 80 = 253.

(End)

Links

Enrique Pérez Herrero and Hieronymus Fischer [terms 1..2000 from Enrique Pérez Herrero], Table of n, a(n) for n = 1..10000

Index entries for 10-automatic sequences.

Formula

From Hieronymus Fischer, May 28 2014: (Start)

a(1 + Sum_{j=1..n} j*9^j) = 10*(10^n - 1).

a(2 + Sum_{j=1..n} j*9^j) = 10^(n+1) + (10^n - 1)/9 = (91*10^n - 1)/9.

a((9^(n+1) - 1)/8 + 1 + Sum_{j=1..n} j*9^j) = 10*(10^(n+1) - 1)/9, where Sum_{j=1..n} j*9^j = (1-(n+1)*9^n+n*9^(n+1))*9/64.

Iterative calculation:

With i := digit position of the '0' in a(n) counted from the right (starting with 0), j = number of contiguous '9' digits in a(n) counted from position 1, if i = 0, and counted from position 0, if i > 0 (0 if none)

a(n+1) = a(n) + 10 + (10^j - 1)/9, if i = 0.

a(n+1) = a(n) + 1 + (10^(j-1) - 1)/9, if i = j > 0.

a(n+1) = a(n) + 1 + (10^j - 1)/9, if i > j.

[see Prog section for an implementation in Smalltalk].

Direct calculation:

Set j := max( m | (Sum_{i=1..m} i*9^i) < n) and c(1) := n - 2 - Sum_{i=1..j} i*9^i. Define successively,

c(i+1) = c(i) mod ((j-i+2)*9^(j-i+1)) - 9^(j-i+1) while this value is >= 0, and set k := i for the last such index for which c(i) >= 0.

Then a(n) = A052382(c(k) mod ((j-k+2)*9^(j-k+1)) + (9^(j-k+1)-1)/8) + Sum_{i=1..k} ((floor(c(i)/((j-i+2)*9^(j-i+1))) + 1) * 10^(j-i+2)). [see Prog section for an implementation in Smalltalk].

Behavior for large n:

a(n) = O(n^(log(10)/log(9))/log(n)).

a(n) = O(n^1.047951651.../log(n)).

Inequalities:

a(n) < 2*(8n)^log_9(10)/(log_9(8n)*log_9(10)).

a(n) < (8n)^log_9(10)/(log_9(8n)*log_9(10)), for large n (n > 10^50).

a(n) > 0.9*(8n)^log_9(10)/(log_9(8n)*log_9(10)), for 2 < n < 10^50.

a(n) >= A011540(n), equality holds for n <= 10.

(End)

Example

a(10^1)= 90.
a(10^2)= 590.
a(10^3)= 4190.
a(10^4)= 35209.
a(10^5)= 308949.
a(10^6)= 2901559.
a(10^7)= 27250269.
a(10^8)= 263280979.
a(10^9)= 2591064889.
a(10^10)= 25822705899.
a(10^20)= 366116331598324670219.
a(10^50)= 3.7349122484477433715662812...*10^51
a(10^100)= 4.4588697999177752943575344...*10^103.
a(10^1000)= 5.5729817962143143812258616...*10^1045.
[Examples by Hieronymus Fischer, May 28 2014]

Mathematica

Select[Range[0, 9000], DigitCount[#, 10, 0]==1&] (* Enrique Pérez Herrero, Nov 29 2013 *)

Prog

(Smalltalk)
A043489_nextTerm
  "Answers the minimal number > m which contains exactly 1 zero digit (in base 10), where m is the receiver.
  Usage: a(n) A043489_nextTerm
  Answer: a(n+1)"
  | d d0 s n p |
  n := self.
  p := 1.
  s := n.
  (d0 := n // p \\ 10) = 0
     ifTrue:
          [p := 10 * p.
          s := s + 1].
  [(d := n // p \\ 10) = 9] whileTrue:
          [s := s - (8 * p).
          p := 10 * p].
  (d = 0 or: [d0 = 0]) ifTrue: [s := s - (p // 10)].
  ^s + p
[by Hieronymus Fischer, May 28 2014]
------------------
(Smalltalk)
A043489
"Answers the n-th number such that number of 0's in base 10 is 1, where n is the receiver. Uses the method zerofree: base from A052382.
  Usage: n A043489
  Answer: a(n)"
  | n a b dj cj gj ej j r |
  n := self.
  n <= 1 ifTrue: [^r := 0].
  n <= 10 ifTrue: [^r := (n - 1) * 10].
  j := n invGeometricSum2: 9.
  b := j geometricSum2: 9.
  cj := 9 ** j.
  dj := (j + 1) * cj.
  gj := (cj - 1) / 8.
  ej := 10 ** j.
  a := n - b - 2.
  b := a \\ dj.
  r := (a // dj + 1) * ej * 10.
  [b >= cj] whileTrue:
          [a := b - cj.
          cj := cj // 9.
          dj := j * cj.
          b := a \\ dj.
          r := (a // dj + 1) * ej + r.
          gj := gj - cj.
          ej := ej // 10.
          j := j - 1].
  r := (b + gj zerofree: 10) + r.
  ^r
[by Hieronymus Fischer, May 28 2014]
------------------
(Smalltalk)
A043489_inverse
  "Answers the index n such that A043489(n) = m, where m is the receiver. Uses A052382_inverse from A052382.
  Usage: n zerofree_inverse: b [b = 10 for this sequence]
  Answer: a(n)"
  | m p q s r m1 mr |
  m := self.
  m < 100 ifTrue: [^m // 10 + 1].
  p := q := 1.
  s := 0.
  [m // p \\ 10 = 0] whileFalse:
     [p := 10 * p.
     s := s + q.
     q := 9 * q].
  p > 1
     ifTrue:
     [r := m \\ p.
     p := 10 * p.
     m1 := m // p.
     (m1 \\ 10 = 1 and: [m1 > 10])
          ifTrue: [mr := m - r - 1]
          ifFalse: [mr := m - r - 10].
     ^mr A043489_inverse + r A052382_inverse - s + 1]
     ifFalse:
     [s := 1.
     p := 10.
     q := 1.
     [p < m] whileTrue:
          [s := (m // p) A052382_inverse * q + s.
          p := 10 * p.
          q := 9 * q].
     ^s]
[by Hieronymus Fischer, May 28 2014]
(PARI) is(n)=#select(d->d==0, digits(n))==1 \\ Charles R Greathouse IV, Oct 06 2016

Crossrefs

Cf. A043493, A043497, A043501, A043505, A043509, A043513, A043517, A043521, A043525, A011540, A052382.

Sequence in context: A118959 A273239 A343452 * A352438 A028440 A332046

Adjacent sequences: A043486 A043487 A043488 * A043490 A043491 A043492

Keyword

nonn,base,easy

Author

Clark Kimberling

Status

approved