A217094 -id:A217094 - cp's OEIS frontend

A011540 Numbers that contain a digit 0.

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

A002452 a(n) = (9^n - 1)/8.

Original entry on oeis.org

0, 1, 10, 91, 820, 7381, 66430, 597871, 5380840, 48427561, 435848050, 3922632451, 35303692060, 317733228541, 2859599056870, 25736391511831, 231627523606480, 2084647712458321, 18761829412124890, 168856464709124011, 1519708182382116100, 13677373641439044901, 123096362772951404110

Offset: 0

N. J. A. Sloane

From David W. Wilson: Numbers triangular, differences square.
To be precise, the differences are the squares of the powers of three with positive indices. Hence a(n+1) - a(n) = (A000244(n+1))^2 = A001019(n+1). [Added by Ant King, Jan 05 2011]
Partial sums of A001019. This is m-th triangular number, where m is partial sums of A000244. a(n) = A000217(A003462(n)). - Lekraj Beedassy, May 25 2004
With offset 0, binomial transform of A003951. - Philippe Deléham, Jul 22 2005
Numbers in base 9: 1, 11, 111, 1111, 11111, 111111, 1111111, etc. - Zerinvary Lajos, Apr 26 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 2, a(n-1) = (-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
From Hieronymus Fischer, Jan 30 2013: (Start)
Least index k such that A052382(k) >= 10^(n-1), for n > 0.
Also index k such that A052382(k) = (10^n-1)/9, n > 0.
A052382(a(n)) is the least zerofree number with n digits, for n > 0.
For n > 1: A052382(a(n)-1) is the greatest zerofree number with n-1 digits. (End)
For n > 0, 4*a(n) is the total number of holes in a certain triangle fractal (start with 9 triangles, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015
For n > 0, a(n) is the sum of the numerators and denominators of the reduced fractions 0 < (b/3^(n-1)) < 1 plus 1. Example for n=3 gives fractions 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, and 8/9 plus 1 has sum of numerators and denominators +1 = a(3) = 91. - J. M. Bergot, Jul 11 2015
Except for 0 and 1, all terms are Brazilian repunits numbers in base 9, so belong to A125134. All these terms are composite because a(n) is the ((3^n - 1)/2)-th triangular number. - Bernard Schott, Apr 23 2017
These are also the second steps after the junctions of the Collatz trajectories of 2^(2k-1)-1 and 2^2k-1. - David Rabahy, Nov 01 2017

			a(4) = (9^4 - 1)/8 = 820 = 1111_9 = (1/2) * 40 * 41 is the ((3^4 - 1)/2)-th = 40th triangular number. - _Bernard Schott_, Apr 23 2017

A. Fletcher, J. C. P. Miller, L. Rosenhead, and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
T. N. Thiele, Interpolationsrechnung. Teubner, Leipzig, 1909, p. 36.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Kival Ngaokrajang, Illustration of initial terms
Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
D. C. Santos, E. A. Costa, and P. M. M. C. Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
A. G. Shannon, Letter to N. J. A. Sloane, Dec 06 1974.
M. Ward, Note on divisibility sequences, Bull. Amer. Math. Soc., 42 (1936), 843-845.
Eric Weisstein's World of Mathematics, Repunit.
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Right-hand column 1 in triangle A008958.
Cf. A217094, A011540, A052382, A125857.
Cf. A125134, A000217.

[(9^n - 1)/8 : n in [0..25]]; // Vincenzo Librandi, Jun 01 2011

A002452 := 1/(9*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation

(9^# & /@ Range[0, 18] // Accumulate) (* Ant King, Jan 06 2011 *)
LinearRecurrence[{10,-9},{0,1},30] (* Harvey P. Dale, Sep 23 2018 *)

A002452(n):=floor((9^n-1)/8)$
makelist(A002452(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=9^n>>3 \\ Charles R Greathouse IV, Jul 25 2011

From Philippe Deléham, Mar 13 2004: (Start)
a(n) = 9*a(n-1) + 1; a(1) = 1.
G.f.: x / ((1-x)*(1-9*x)). (End)
a(n) = 10*a(n-1) - 9*a(n-2). - Ant King, Jan 05 2011
a(n) = floor(A000217(3^n)/4) - A033113(n-1). - Arkadiusz Wesolowski, Feb 14 2012
Sum_{n>0} a(n)*(-1)^(n+1)*x^(2*n+1)/(2*n+1)! = (1/6)*sin(x)^3. - Vladimir Kruchinin, Sep 30 2012
a(n) = A011540(A217094(n-1)) - A217094(n-1) + 2, n > 0. - Hieronymus Fischer, Jan 30 2013
a(n) = 10^(n-1) + 2 - A217094(n-1). - Hieronymus Fischer, Jan 30 2013
a(n) = det(|v(i+2,j+1)|, 1 <= i,j <= n-1), where v(n,k) are central factorial numbers of the first kind with odd indices (A008956) and n > 0. - Mircea Merca, Apr 06 2013
a(n) = Sum_{k=0..n-1} 9^k. - Doug Bell, May 26 2017
E.g.f.: exp(5*x)*sinh(4*x)/4. - Stefano Spezia, Mar 11 2023

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 08 2004

Offset changed from 1 to 0 and added 0 by Vincenzo Librandi, Jun 01 2011

A324160 Number of zero-containing nonnegative integers <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11

Offset: 0

Hieronymus Fischer, Feb 15 2019

This sequence represents the counting function of A011540.
n = n_max = 701 is the greatest number such that a(n) <= pi(n) [the number of primes <= n]. Thus, for all indices n > n_max, we have a(n) > pi(n). For n = n_max the number of primes is pi(n) = 126.
n = n_min = 510 is the least number (> 2) such that a(n) >= pi(n) [the number of primes <= n]. Thus, for all indices 2 < n < n_min, we have a(n) < pi(n). For n = n_min the number of primes is pi(n) = 97.

			a(10) = 2, since there are two numbers <= 10 which contain a '0'-digit (0 and 10).
a(100) = 11.
a(10^3) = 182.
a(10^4) = 2621.
a(10^5) = 33572.
a(10^6) = 402131.
a(10^7) = 4619162
a(10^8) = 51572441.
a(10^9) = 564151952.
a(10^10) = 6077367551.
a(10^20) = 86322626358560955101.
a(10^50) = 99420200289176487252583981229013676068210129037751 = 9.9420200289176... *10^49
a(10^100) = 9.9997011842625...13575501*10^99.
a(10^1000) = 9.999999999999...564125755001*10^999
(here, the first 45 digits are 9's).

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011540, A052382, A324161, A217094.

a(n) = 1 + sum(k=1, n, vecmin(digits(k)) == 0); \\ Michel Marcus, Mar 20 2019

With m := floor(log_10(n)); k := Max_{j | j = 1..m and (floor(n/10^j) mod 10)*j = 0} = digit position of the leftmost '0' in n counted from the right (starting with 0), k = 0 if there is no '0' digit; b(n,k):= floor(n/10^k)*10^k:
a(n) = 2 + Sum_{j = 1..m} floor((b(n,k+1)-1)/10^j)*9^(j-1), if k = 0 (valid for n > 9),
a(n) = 2 + n mod 10^k + Sum_{j = 1..m} floor((b(n,k)-1)/10^j)*9^(j-1), if k > 0 (valid for n > 0),
a(n) = 2 + n mod 10^k - ceiling(fract(n/10))*(1-ceiling(k/(m+1))) + Sum_{j = 1..m} floor((b(n,k)-1)/10^j)*9^(j-1) (all k, valid for n > 0).
a(n) + A324161(n) = n + 1
a(A011540(n)) = n.
A011540(a(n)) <= n, for n >= 0.
A011540(a(n)) = n, iff n is a zero-containing number.
a(10*n + k) <= 9*a(n) + n - 8, k = 0..9, equality holds for k = 9, and also, if n is a zerofree number (i.e., contains no '0'-digit).
a(10*A052382(n) + k) = 10*A052382(n) + 1 - 9*n, k = 0..9.
Values for special indices:
a(k*(10^n-1)/9 - j) = k*(8*10^n - 9*9^n + 1)/72 + 1, n > 0, k = 1..9, j = 0..k.
a(k*10^n - j) = k*(10^n - 9^n) + 1 - (9^n - 1)/8, n >= 0, k = 1..10, j = 1..10.
a(10^n) = 10^n + 2 - (9^(n+1) - 1)/8, n > 0.
a(k*10^n + j) = k*(10^n - 9^n) + j + 2 - (9^n - 1)/8, n > 0, k = 1..9, 0 <= j < (10^(n+1)-1)/9 - 10^n.
With: d := log_10(9) = 0.95424250943932...
Upper bound:
a(n) <= n + 2 - ((9*n + 10)^d - 1)/8,
equality holds for n = (10^k - 1)/9 - 1, k > 0.
Lower bound:
a(n) >= n + 2 - (9*(n + 1)^d - 1)/8,
equality holds for n = 10^k - 1, k >= 0.
Asymptotic behavior:
a(n) <= n + 2 + (1/8) - (9^d/8)*n^d*(1 + O(1/n)).
a(n) >= n + 2 + (1/8) - (9/8)*n^d*(1 + O(1/n))).
a(n) =  n*(1 + O(n^(d-1)) = n*(1 + O(1/n^0.045757490...)).
Lower and upper limits:
lim inf (a(n) - n)/n^d = -9/8, for n -> infinity.
lim sup (a(n) - n)/n^d = -9^d/8 = -1.0173931195971..., for n -> infinity.
From Hieronymus Fischer, Apr 04 2019: (Start)
Formulas for general bases b > 2:
With m := floor(log_b(n)); k := Max_{j | j=1..m and (floor(n/b^j) mod b)*j = 0} = digit position of the leftmost '0' in n counted from the right (starting with 0), k = 0 if there is no '0' digit; b(n,k):= floor(n/b^k)*b^k:
a(n) = 2 + Sum_{j=1..m} floor((b(n,k+1)-1)/b^j)*(b-1)^(j-1), if k = 0, valid for  n > b-1;
a(n) = 2 + n mod b^k + Sum_{j=1..m} floor((b(n,k)-1)/b^j)*(b-1)^(j-1), if k > 0, valid for n > 0;
a(n) = 2 + n mod b^k - ceiling(fract(n/b))*(1-ceiling(k/(m+1))) + Sum_{j=1..m} floor((b(n,k)-1)/b^j)* (b-1)^(j-1), all k, valid for n > 0.
Formula for base b = 2: a(n) = (n + 1 - floor(log_2(n + 1))).
With d := d(b) := log(b - 1)/log(b):
Upper bound (b = 10 for this sequence):
a(n) <= n + 2 - (((b - 1)*n + b)^d - 1)/(b - 2),
equality holds for n = (b^k - 1)/(b - 1) - 1, k > 0.
Lower bound (b = 10 for this sequence):
a(n) >= n + 2 - ((b - 1)*(n + 1)^d - 1)/(b - 2),
equality holds for n = b^k - 1, k >= 0.
Asymptotic behavior (b = 10 for this sequence):
a(n) =  n*(1 + O(n^(d(b)-1)), for b > 2,
a(n) =  n*(1 + O(log(n)/n)), for b = 2.
Lower and upper limits:
lim inf (a(n) - n)/n^d(b) = -(b - 1)/(b - 2), for n -> infinity, for b > 2.
lim sup (a(n) - n)/n^d(b) = -(b - 1)^d/(b - 2) for n -> infinity, for b > 2.
In case of b = 2:
lim (a(n) - n)/log(n) = -1/log(2), for n -> infinity.
(End)

The number of numbers between 10^(n-1) and 10^n which contain all decimal digits 0..9.
The ratio a(n)/(10^n-10^(n-1)) indicates the relative proportion of pandigital n-digit numbers compared to all n-digit numbers. Since that ratio converges to the limit 1 for n->oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/(10^n-10^(n-1)) = 0.99973439517775... for n = 100; in this case 99.9734...% of all 100-digit numbers contain all digits 0..9. Conversely, only the tiny proportion of 0.00026560482224... (< 0.03%) lacks one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers with n = 10 or 20 digits the relative proportion of numbers which do not contain all digits 0..9 is 99.96371...% or 78.52626...%, respectively. The least number of digits for which the pandigital numbers hold the majority is 27. Here, the proportion of numbers which do not contain all digits is 48.03664...%. So one could bet that a randomly chosen number with >= 27 digits contains all digits.

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

The number of numbers with <= n digits which contain all decimal digits 0..9.
The ratio a(n)/10^n indicates the relative proportion of pandigital numbers <= 10^n compared to all numbers <= 10^n. Since that ratio converges to the limit 1 for n -> oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/10^n = 0. 99973107526479... for n = 100; in this case 99.9731...% of all numbers <= 10^100 contain all digits 0..9. Conversely, only the tiny proportion of 0.000268924735210... (< 0.03%) lacks at least one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers <= 10^10 or <= 10^20 the relative proportion of numbers which do not contain all digits 0..9 is 99.96734...% or 78.98393...%, respectively. 10^27 is the least power of 10 such that the pandigital numbers hold the majority. Here, the proportion of pandigital numbers among all numbers <= 10^27 is 51.50961...%. So one could bet that a randomly chosen number <= 10^27 contains all digits.
Partial sums of A217110.

cp's OEIS Frontend

A011540 Numbers that contain a digit 0.

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A002452 a(n) = (9^n - 1)/8.

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A324160 Number of zero-containing nonnegative integers <= n.

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A217110 Number of pandigital numbers with n places.

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

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A217111 Number of pandigital numbers <= 10^n.

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