A043489 -id:A043489 - cp's OEIS frontend

A052382 Numbers without 0 in the decimal expansion, colloquial 'zeroless numbers'.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 111, 112, 113

Offset: 1

Henry Bottomley, Mar 13 2000

The entries 1 to 79 match the corresponding subsequence of A043095, but then 81, 91-98, 100, 102, etc. are only in one of the two sequences. - R. J. Mathar, Oct 13 2008
Complement of A011540; A168046(a(n)) = 1; A054054(a(n)) > 0; A007602, A038186, A038618, A052041, A052043, and A052045 are subsequences. - Reinhard Zumkeller, Apr 25 2012, Apr 07 2011, Dec 01 2009
a(n) = n written in base 9 where zeros are not allowed but nines are. The nine distinct digits used are 1, 2, 3, ..., 9 instead of 0, 1, 2, ..., 8. To obtain this sequence from the "canonical" base 9 sequence with zeros allowed, just replace any 0 with a 9 and then subtract one from the group of digits situated on the left. For example, 9^3 = 729 (10) (in base 10) = 1000 (9) (in base 9) = 889 (9-{0}) (in base 9 without zeros) because 100 (9) = [9-1]9 = 89 (9-{0}) and thus 1000 (9) = [89-1]9 = 889 (9-{0}). - Robin Garcia, Jan 15 2014
From Hieronymus Fischer, May 28 2014: (Start)
Inversion: Given a term m, the index n such that a(n) = m can be calculated by A052382_inverse(m) = m - sum_{1<=j<=k} floor(m/10^j)*9^(j-1), where k := floor(log_10(m)) [see Prog section for an implementation in Smalltalk].
Example 1: A052382_inverse(137) = 137 - (floor(137/10) + floor(137/100)*9) = 137 - (13*1 + 1*9) = 137 - 22 = 115.
Example 2: A052382_inverse(4321) = 4321 - (floor(4321/10) + floor(4321/100)*9 + floor(4321/1000)*81) = 4321 - (432*1 + 43*9 + 4*81) = 4321 - (432 + 387 + 324) = 3178. (End)
The sum of the reciprocals of these numbers from a(1)=1 to infinity, called the Kempner series, is convergent towards a limit: 23.103447... whose decimal expansion is in A082839. - Bernard Schott, Feb 23 2019
Integer n > 0 is encoded using bijective base-9 numeration, see Wikipedia link below. - Alois P. Heinz, Feb 16 2020

			For k >= 0, a(10^k) = (1, 11, 121, 1331, 14641, 162151, 1783661, 19731371, ...) = A325203(k). - _Hieronymus Fischer_, May 30 2012 and Jun 06 2012; edited by _M. F. Hasler_, Jan 13 2020

Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, p. 258.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
K. Mahler, On the generating function of the integers with a missing digit, J. Indian Math. Soc. 15A (1951), 34-40.
Eric Weisstein's World of Mathematics, Kempner Series.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration
Index entries for 10-automatic sequences.

Cf. A004719, A052040, different from A067251.
Cf. A055640, A046034, A007931, A007932, A084544, A084545.
Column k=9 of A214676.
Cf. A011540 (complement), A043489, A054054, A168046.
Cf. A052383 (without 1), A052404 (without 2), A052405 (without 3), A052406 (without 4), A052413 (without 5), A052414 (without 6), A052419 (without 7), A052421 (without 8), A007095 (without 9).
Zeroless numbers in some other bases <= 10: A000042 (base 2), A032924 (base 3), A023705 (base 4), A248910 (base 6), A255805 (base 8), A255808 (base 9).
Cf. A082839 (sum of reciprocals).
Cf. A038618 (subset of primes)
Subsequences: A007602, A038186, A052041, A052043, A052045, A059405, A066484, A099542.

a052382 n = a052382_list !! (n-1)
a052382_list = iterate f 1 where
f x = 1 + if r < 9 then x else 10 * f x' where (x', r) = divMod x 10
-- Reinhard Zumkeller, Mar 08 2015, Apr 07 2011

[ n: n in [1..114] | not 0 in Intseq(n) ]; // Bruno Berselli, May 28 2011

a:= proc(n) local d, l, m; m:= n; l:= NULL;
      while m>0 do d:= irem(m, 9, 'm');
        if d=0 then d:=9; m:= m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 11 2015
is_zeroless := n -> not is(0 in convert(n, base, 10)):
select(is_zeroless, [seq(1..113)]);  # Peter Luschny, Jun 20 2025

A052382 = Select[Range[100], DigitCount[#, 10, 0] == 0 &] (* Alonso del Arte, Mar 10 2011 *)

select( {is_A052382(n)=n&&vecmin(digits(n))}, [0..111]) \\ actually: is_A052382 = (bool) A054054. - M. F. Hasler, Jan 23 2013, edited Jan 13 2020

a(n) = for (w=0, oo, if (n >= 9^w, n -= 9^w, return ((10^w-1)/9 + fromdigits(digits(n, 9))))) \\ Rémy Sigrist, Jul 26 2017

apply( {A052382(n,L=logint(n,9))=fromdigits(digits(n-9^L>>3,9))+10^L\9}, [1..100])
next_A052382(n, d=digits(n+=1))={for(i=1, #d, d[i]|| return(n-n%(d=10^(#d-i+1))+d\9)); n} \\ least a(k) > n. Used in A038618.
( {A052382_vec(n,M=1)=M--;vector(n, i, M=next_A052382(M))} )(99) \\ n terms >= M
\\ See OEIS Wiki page (cf. LINKS) for more programs. - M. F. Hasler, Jan 11 2020

A052382 = [n for n in range(1,10**5) if not str(n).count('0')]
# Chai Wah Wu, Aug 26 2014

from sympy import integer_log
def A052382(n):
    m = integer_log(k:=(n<<3)+1,9)[0]
    return sum((1+(k-9**m)//(9**j<<3)%9)*10**j for j in range(m)) # Chai Wah Wu, Jun 27 2025

A052382
"Answers the n-th term of A052382, where n is the receiver."
^self zerofree: 10
A052382_inverse
"Answers that index n which satisfy A052382(n) = m, where m is the receiver.”
^self zerofree_inverse: 10
zerofree: base
"Answers the n-th zerofree number in base base, where n is the receiver. Valid for base > 2.
Usage: n zerofree: b [b = 10 for this sequence]
Answer: a(n)"
| n m s c bi ci d |
n := self.
c := base - 1.
m := (base - 2) * n + 1 integerFloorLog: c.
d := n - (((c raisedToInteger: m) - 1)//(base - 2)).
bi := 1.
ci := 1.
s := 0.
1 to: m
do:
[:i |
s := (d // ci \\ c + 1) * bi + s.
bi := base * bi.
ci := c * ci].
^s
zerofree_inverse: base
"Answers the index n such that the n-th zerofree number in base base is = m, where m is the receiver. Valid for base > 2.
Usage: m zerofree_inverse: b [b = 10 for this sequence]
Answer: n"
| m p q s |
m := self.
s := 0.
p := base.
q := 1.
[p < m] whileTrue:
[s := m // p * q + s.
p := base * p.
q := (base - 1) * q].
^m - s
"by Hieronymus Fischer, May 28 2014"

seq 0 1000 | grep -v 0; # Joerg Arndt, May 29 2011

a(n+1) = f(a(n)) with f(x) = 1 + if x mod 10 < 9 then x else 10*f([x/10]). - Reinhard Zumkeller, Nov 15 2009
From Hieronymus Fischer, Apr 30, May 30, Jun 08 2012, Feb 17 2019: (Start)
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 9)*10^j, where m = floor(log_9(8*n + 1)), b(j) = floor((8*n + 1 - 9^m)/(8*9^j)).
Also: a(n) = Sum_{j=0..m-1} (1 + A010878(b(j)))*10^j.
a(9*n + k) = 10*a(n) + k, k=1..9.
Special values:
a(k*(9^n - 1)/8) = k*(10^n - 1)/9, k=1..9.
a((17*9^n - 9)/8) = 2*10^n - 1.
a((9^n - 1)/8 - 1) = 10^(n-1) - 1, n > 1.
Inequalities:
a(n) <= (1/9)*((8*n+1)^(1/log_10(9)) - 1), equality holds for n=(9^k-1)/8, k>0.
a(n) > (1/10)*((8*n+1)^(1/log_10(9)) - 1), n > 0.
Lower and upper limits:
lim inf a(n)/10^log_9(8*n) = 1/10, for n -> infinity.
lim inf a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/10, for n -> infinity.
lim sup a(n)/10^log_9(8*n) = 1/9, for n -> infinity.
lim sup a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/9, for n -> infinity.
G.f.: g(x) = (x^(1/8)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(9/8)*(1 - 10z(j)^9 + 9z(j)^10)/((1-z(j))(1-z(j)^9)), where z(j) = x^9^j.
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1 - 10(x^9^j)^9 + 9(x^9^j)^10)*x^9^j*f_j(x)/(1-x^9^j), where f_j(x) = 10^j*x^((9^j-1)/8)/(1-(x^9^j)^9). Here, the f_j obey the recurrence f_0(x) = 1/(1-x^9), f_(j+1)(x) = 10x*f_j(x^9).
Also: g(x) = (1/(1-x))*((Sum{k=0..8} h_(9,k)(x)) - 9*h_(9,9)(x)), where h_(9,k)(x) = Sum_{j>=0} 10^j*x^((9^(j+1)-1)/8)*x^(k*9^j)/(1-x^9^(j+1)).
Generic formulas for analogous sequences with numbers expressed in base p and only using the digits 1, 2, 3, ... d, where 1 < d < p:
a(n) = Sum_{j=0..m-1} (1 + b(j) mod d)*p^j, where m = floor(log_d((d-1)*n+1)), b(j) = floor(((d-1)*n+1-d^m)/((d-1)*d^j)).
Special values:
a(k*(d^n-1)/(d-1)) = k*(10^n-1)/9, k=1..d.
a(d*((2d-1)*d^(n-1)-1)/(d-1)) = ((d+9)*10^n-d)/9 = 10^n + d*(10^n-1)/9.
a((d^n-1)/(d-1)-1) = d*(10^(n-1)-1)/9, n > 1.
Inequalities:
a(n) <= (10^log_d((d-1)*n+1)-1)/9, equality holds for n = (d^k-1)/(d-1), k > 0.
a(n) > (d/10)*(10^log_d((d-1)*n+1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_d((d-1)*n) = d/90, for n -> infinity.
lim sup a(n)/10^log_d((d-1)*n) = 1/9, for n -> infinity.
G.f.: g(x) = (1/(1-x)) Sum_{j>=0} (1 - (d+1)(x^d^j)^d + d(x^d^j)^(d+1))*x^d^j*f_j(x)/(1-x^d^j), where f_j(x) = p^j*x^((d^j-1)/(d-1))/(1-(x^d^j)^d). Here, the f_j obey the recursion f_0(x) = 1/(1-x^d), f_(j+1)(x) = px*f_j(x^d).
(End)
A052382 = { n | A054054(n) > 0 }. - M. F. Hasler, Jan 23 2013
From Hieronymus Fischer, Feb 20 2019: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 0.696899720...
Sum_{n>=1} 1/a(n)^2 = 1.6269683705819...
Sum_{n>=1} 1/a(n) = 23.1034479... = A082839. This so-called Kempner series converges very slowly. For the calculation of the sum, it is helpful to use the following fraction of partial sums, which converges rapidly:
lim_{n->infinity} (Sum_{k=p(n)..p(n+1)-1} 1/a(k)) / (Sum_{k=p(n-1)..p(n)-1} 1/a(k)) = 9/10, where p(n) = (9^n-1)/8, n > 1.
(End)

Typos in formula section corrected by Hieronymus Fischer, May 30 2012

Name clarified by Peter Luschny, Jun 20 2025

A011540 Numbers that contain a digit 0.

Original entry on oeis.org

0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 301, 302

Offset: 1

N. J. A. Sloane

Complement of A052382.
A168046(a(n)) = 0; A054054(a(n)) = 0; A055640(a(n)) = 0 for n = 1 and A055640(a(n)) > 0 for n > 1; A055641(a(n)) > 0; subsequence of A188643. - Reinhard Zumkeller, Apr 25 2012, Apr 07 2011; corrected by Hieronymus Fischer, Jan 13 2013
A067898(a(n)) > 0. - Reinhard Zumkeller, May 04 2012; corrected by Hieronymus Fischer, Jan 13 2013
From Hieronymus Fischer, Jan 13 2013, May 28 2014; edited by M. F. Hasler; edited by Hieronymus Fischer, Dec 27 2018: (Start)
Zerofree floor: The greatest zerofree number < a(n) is A052382(a(n) + 1 - n).
The greatest zero-containing number (i.e., non-zerofree number, or term of this sequence) less than a given zerofree number A052382(n) is a(A052382(n) + 1 - n).
The ratio n/(a(n) + 1) indicates the relative proportion of zero-containing numbers less than or equal to a(n) compared to all numbers less than or equal to a(n). Since Lim_{n -> infinity} a(n)/n = 1, this can be expressed as "Almost all numbers contain a 0" (in a slightly informal manner).
As an example, for n = 10^100, n/(a(n) + 1) = 0.9999701184..., i.e., 99.997...% of all numbers between 0 and 10^100 contain a zero digit. Only the tiny proportion of 0.0000298816... (less than 0.003%) contain no zero digit. This is in contrast to the behavior for small indices, where the relative portion of numbers that contain no zero digit is significant: for n = 10^3 and even n = 10^7, the proportion of numbers less than or equal to n that contain no zero digit exceeds 81% and 53%, respectively.
Inversion: Given a number z that contains a zero digit, the index n for which a(n) = z is n = (z+1)*probability that a randomly chosen number k from the range 0..z contains a zero digit.
Example 1: z = 10; the probability that a randomly chosen number less than or equal to 10 contains no zero digit is 9/11. The probability that it contains a zero digit is p = 2/11. Thus, n = (z+1)*p = 2 and a(2) = 10.
Example 2: z = 10^6; the probability that a randomly chosen number with m > 1 digits contains no zero digit is (9/10)^(m-1). For m = 1 the probability is 9/10. The probability that a randomly chosen number with 1..m digits contains no zero digit is q = (9/10)*10/(10^m+1) + Sum_{i = 2..m} (9/10)^(i-1)*(10^i - 10^(i-1))/(10^m+1) = (72 + 81*(9^(m-1) - 1))/(8*(10^m+1)). Hence, the probability that the chosen number with 1..m digits contains a zero digit is p = 1 - q = (8*10^m - 9*9^m + 17)/(8*(10^m + 1)). Thus, p = 402131/1000001 (for z = 10^6) and so n = (z+1)*p = 402131, which implies a(402131) = 10^6.
The number of terms z such that k*10^m <= z < (k+1)*10^m is 10^m - 9^m, where 1 <= k < 10 and m >= 0.
The number of terms z such that 10^m <= z < 10^(m+1) is 9*(10^m - 9^m), where m >= 0.
The number of terms z <= 10^m is (8*10^m - 9*9^m + 17)/8 where m>=1 (cf. A217094).
Infinitely many terms are primes, and most primes are zero-containing numbers. Sketch of a proof: The number of zero-containing numbers less than or equal to a(n) is n. Hence there are a(n) + 1 - n zerofree numbers less than or equal to a(n). From the asymptotic behavior of a(n) (see formula section) it follows a(n) + 1 - n < (5/4)*n^log_10(9) for sufficiently large n. By the prime number theorem we have for each fixed d > 0 the relation pi(n) [number of primes less than or equal to n] > (1 - d/4)*(n/log(n)) for sufficiently large n. Thus, for the number of primes less than or equal to a(n) which contain a zero digit [hereafter denoted as P_0(a(n))] we have P_0(a(n)) > pi(a(n)) - (a(n) + 1 - n) > (1 - d/4)*a(n)/log(a(n)) - (5/4)*n^log_10(9) > (1-d/4)*n/log(n) - (5/4)*n^log_10(9) = (1-d/4)*n/log(n) * (1 - (5/4)*(1/(1-d/4))*(1/n) * n^(log_10(9))*log(n)) > (1-d/2)*n/log(n) for sufficiently large n. Because of a(n) = n + o(n) this also implies P_0(a(n)) > (1 - d)*a(n)/log(a(n)) for sufficiently large n. Thus, the proportion of primes less than or equal to a(n) which contain a zero digit compared to the total number of primes less than or equal to a(n) is arbitrarily near to 1 for sufficiently large n.
Sequence inversion:
Given a term m > 0, the index n such that a(n) = m can be calculated with the following procedure: Define k := floor(log_10(m)) and i := digit position of the leftmost '0' in m counted from the right (starting with 0), then:
A011540_inverse(m) = 2 + m mod 10^i + Sum_{j = 1..k} floor((m - 1 - m mod 10^i)/10^j)*9^(j-1) [see PROG section for an implementation in Smalltalk].
Example: m = 905, k = 2, i = 1, A011540_inverse(905) = 2 + 905 mod 10 + floor((905 - 1 - 905 mod 10)/10)*1 + floor((905 - 1 - 905 mod 10)/100)*9 = 2 + 5 + floor(899/10)*1 + floor(899/100)*9 = 2 + 5 + 89*1 + 8*9 = 168.
(End)
For the number of k-digit numbers containing the digit '0', see A229127. - Jon E. Schoenfield, Sep 14 2013
The above "sketch of proof" only compares the relative densities, and since the density of this sequence is 1, the result is "obvious". But the nontrivial part is that there is no correlation between the absence of a digit '0' and primality of the number (cf. A038618). Indeed, consider the set S defined to be the set of primes with all digits '0' replaced by the smallest possible nonzero digit while avoiding duplicates. Having exactly the same density as the set of primes, the argument of the proof applies in the same way and leads to the same conclusion for the number of zero-containing terms; however, there is none in the set S. - M. F. Hasler, Oct 11 2015, example added Feb 11 2019

			a(10)      = 90.
a(100)     = 540.
a(10^3)    = 4005.
a(10^4)    = 30501.
a(10^5)    = 253503.
a(10^6)    = 2165031.
a(10^7)    = 20163807
a(10^8)    = 182915091.
a(10^9)    = 1688534028.
a(10^10)   = 15749319096.
a(10^20)   = 114131439770460123393.
a(10^50)   = 10057979971082351274741...89870962249 = 1.0057979971082...*10^50
a(10^100)  = 10000298815737485...786424499 = 1.0000298815737...*10^100.
a(10^1000) = 1...(45 zeros)...196635515818798306...4244999 (1001 digits), using recursive calculation. - _Hieronymus Fischer_, Jan 13 2013

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.

A192825 is a subsequence.

Cf. A007954, A043489, A052382, A055641, A168046, A217094, A217096, A229127.

a011540 n = a011540_list !! (n-1)
a011540_list = filter ((== 0) . a168046) [0..]
-- Reinhard Zumkeller, Apr 07 2011

[0] cat [ n: n in [0..350] | 0 in Intseq(n) ]; // Vincenzo Librandi, Oct 12 2015

Select[Range[0, 299], DigitCount[#, 10, 0] > 0 &] (* Alonso del Arte, Mar 10 2011 *)
Select[Range[0, 299], Times@@IntegerDigits[#] == 0 &] (* Alonso del Arte, Aug 29 2014 *)

is(n)=!n||!vecmin(digits(n)) \\ M. F. Hasler, Feb 28 2018, replacing an earlier version from Charles R Greathouse IV, Aug 09 2011

A011540(n)=my(m=log(n+.5)\log(10)+1, f(m)=n-10^m+(9*9^m-17)/8, j=(sign(f(m)+1)+1)\2+m-1, c=[f(j)], k=1); while(c[k]>0,c=concat(c,c[k] % (10^(j-k+1) - 9^(j-k+1)) - 10^(j-k));k++); k>1&&k--||n>1||return(0); c[k]%(10^(j-k+1) - 9^(j-k+1)) + sum(i=1,k, (c[i]\(10^(j-i+1) - 9^(j-i+1)) + 1)*10^(j-i+1)) \\ Uses the "Direct calculation" formula given by H. Fischer. - M. F. Hasler, Oct 11 2015

A011540_list = [n for n in range(10**3) if '0' in str(n)] # Chai Wah Wu, Mar 26 2021

A011540
"Calculates the n-th number with zero digits recursively - not optimized"
| n j m b d p r |
n := self.
n < 2 ifTrue: [^r := 0].
m := (n integerFloorLog: 10) + 1.
j := (n + 1 - ((10 raisedToInteger: m) - (((9 raisedToInteger: (m + 1)) - 17) // 8))) sign + 1 // 2 + m - 1.
d := (10 raisedToInteger: j) - (9 raisedToInteger: j).
b := ((10 raisedToInteger: j) - (((9 raisedToInteger: (j + 1)) - 17) // 8)).
(((n - b) \\ d > (10 raisedToInteger: (j - 1))) and: [n >= 19])
ifTrue:
[p := (((n - b) \\ d + b - d) A011540)].
(n - b) \\ d > (10 raisedToInteger: (j - 1))
ifFalse: [p := (n - b) \\ d].
r := (((n - b) // d + 1) * (10 raisedToInteger: j)) + p.
^r "Hieronymus Fischer, Jan 13 2013"

A011540_inverse
"Version 1: Answers the index n such that A011540(n) = m, where m is the receiver.
Usage: m A011540_inverse
Answer: n"
| m p q s r d |
m := self.
m < 10 ifTrue: [^1].
p := q := 1.
[p < m] whileTrue:
[d := m // p \\ 10.
d = 0 ifTrue: [q := p].
p := 10 * p].
r := m \\ q.
s := r + 2.
p := 10.
q := 1.
m := m - r - 1.
[p < m] whileTrue:
[s := m // p * q + s.
p := 10 * p.
q := 9 * q].
^s
"Hieronymus Fischer, May 28 2014"

A011540_inverse
"Version 2: Answers the index n such that A011540(n) = m, where m is the receiver.
Uses A052382_inverse from A052382.
Usage: m A011540_inverse
Answer: n"
| m p q d |
m := self.
m < 10 ifTrue: [^1].
p := q := 1.
[p < m] whileTrue:
[d := m // p \\ 10.
d = 0 ifTrue: [q := p].
p := 10 * p].
^m + 1 - (m - 1 - (m \\ q)) A052382_inverse
"Hieronymus Fischer, May 28 2014"

From Hieronymus Fischer, Jan 13 2013: (Start)
Inequalities:
a(n) <= 10*(n - 1), equality holds for 1 <= n <= 11.
a(n) <= 9*n, for n <> 11.
a(n) < n + 10 * n^log_10(9).
a(n) < n + 2 * n^log_10(9), for n > 6*10^8.
a(n) > n + 9^log_10(9)/8 * n^log_10(9).
a(n) < A043489(n), for n > 10.
Iterative calculation:
a(n+1) = a(n) + 1 + 9*sign(A007954(a(n)+1)).
Recursive calculation (for n > 1):
Set m := floor(log_10(n)) + 1), j := floor(sign(n+1 - (8*10^m - 9*9^m + 17)/8) + 1)/2) + m - 1, d := 10^j - 9^j, b := (8*10^j - 9*9^j + 17)/8, and determine r(n) as follows:
Case 1: r(n) = a(b - d + (n - b) mod d), if (n - b) mod d > 10^(j-1) and n >= 19
Case 2: r(n) = (n - b) mod d, if (n - b) mod d <= 10^(j-1).
Then a(n) = (floor((n - b)/d) + 1)*10^j + r(n).
Direct calculation (for n>1):
Set m := floor(log_10(n)) + 1), j := floor((sign(n+1 - (8*10^m - 9*9^m + 17)/8) + 1)/2) + m - 1, and determine k and c(i) as follows:
c(1) = n - (8*10^j - 9*9^j + 17)/8, then define successively for i = 1, 2, ...,
c(i+1) = (c(i) mod (10^(j-i+1) - 9^(j-i+1))) - 10^(j-i) while this value is > 0, and set k := i for the last such index for which c(i) > 0 (in any case k is k<=j).
Then a(n) = c(k) mod (10^(j-k+1) - 9^(j-k+1)) + sum_{i=1..k}(floor(c(i)/(10^(j-i+1) - 9^(j-i+1))) + 1)*10^(j-i+1).
Asymptotic behavior:
a(n) = n + O(n^log_10(9)) = n*(1+ O(1/n^0.04575749056...)).
lim a(n)/n = 1 for n -> infinity.
lim inf (a(n) - n)/n^log_10(9) = 9^log_10(9)/8 = 1.017393081085670008926619124438...
lim sup (a(n) - n)/n^log_10(9) = 9/8 = 1.125.
Sums:
Sum_{n >= 2} (-1)^n/a(n) = 0.0693489578....
Sum_{n >= 2} 1/a(n)^2 = 0.0179656962...
Sum_{n >= 2} 1/a(n) diverges, because of a(n) < 10*n.
Sum_{n >= 1} a(n)/n^2 diverges too.
Sum_{n >= 2} 1/a(n)^2 + Sum_{n >= 1} 1/A052382(n)^2 = Pi^2/6.
Generating function:
g(x) = Sum_{k >= 1} g_k(x), where the terms g_k(x) obey the following recurrence relation:
g_k(x) = 10^k*x^b(k) * (1 - 10x^(9d(k)) + 9x^(10d(k)))/((1-x^d(k))(1-x)) + (x*x^b(k) * (1 - 10^(k-1)*x^(10^(k-1)-1) + (10^(k-1)-1)*x^10^(k-1))/((1-x)^2) + g_(k-1)(x)*x^d(k)) * (1-x^(9d(k)))/(1-x^d(k)),
where b(k) := 2 + 10^k - 9^k - (9^k-1)/8,
d(k) := 10^k - 9^k, and g_0(x) = 0.
Explicit representation of g_k(x):
g_k(x) = (10^k*x^b(k)*(1 - 10x^(9d(k)) + 9x^(10d(k)))/(1-x^d(k)) + sum_{j=1..k-1} ((10^j*x^b(j) * (1 - 10x^(9d(j)) + 9x^(10d(j)))/(1-x^d(j)) + x^(b(j)-10^j+1) * (1 - 10^j*x^(10^j-1) + (10^j-1)*x^10^j)/(1-x)) * Product_{i=j+1..k} x^d(i)*(1-x^(9d(i)))/(1-x^d(i)))/(1-x).
A summation term g_k(x) of the g.f. represents all the sequence terms >= 10^k and < 10^(k+1).
Example 1: g_1(x) = 10*x^2*(1 - 10x^9 + 9x^10)/(1-x)^2 represents the g.f. fragment 10x^2 + 20x^3 + ... + 90x^10 and so generates the terms a(2)=10 ... a(10)=90.
Example 2: g_2(x) = 10^2*x^11*(1 - 10x^(9*19) + 9x^(10*19))/((1-x)(1-x^19)) + 10*x^21 * (1 - 10x^9 + 9x^10)/((1-x)^2) * (1-x^(9*19))/(1-x^19)) + x^11*x * (1 - 10x^9 + 9x^10)/((1-x)^2) * (1-x^(9*19))/(1-x^19) represents the g.f. fragment 100x^11 + 101x^12 + ... + 109x^20 + 110x^21 + 120x^22 + ... + 190x^29 + 200x^30 + 201x^31 + ... + 210x^40 + ... + 990x^181 and so generates the terms a(11) = 100 ... a(181) = 990.
(End)
From Hieronymus Fischer, Feb 12 2019: (Start)
The number C(n) of zero-containing numbers <= n (counting function) is given by C(n) = A011540_inverse(n), if n is a zero-containing number, and C(n) = A011540_inverse(A052382(a(n) + 1 - n)), if n is a zerofree number.
Upper bound:
  C(n) <= n+1-((9*n+1)^d-1)/8.
Lower bound:
  C(n) > n+1-((10*n+1)^d-1)/8
where d = log_10(9) = 0.95424250943932...
(see A324160).
(End)

Edited by M. F. Hasler, Oct 11 2015

cp's OEIS Frontend

A155483 a(n)=A000040(A043489(n+1)).

Views

Author

Keywords

Extensions

A052382 Numbers without 0 in the decimal expansion, colloquial 'zeroless numbers'.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

Python

Python

Smalltalk

sh

Formula

Extensions

A011540 Numbers that contain a digit 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

PARI

Python

Smalltalk

Smalltalk

Smalltalk

Formula

Extensions

A043493 Numbers that contain a single 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A043509 Numbers having exactly one 5 in base 10.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A043513 Numbers having one 6 in base 10.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A043521 Numbers having one 8 in base 10.

Views

Author

Keywords

Links