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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Henry Bottomley, Mar 13 2000

Keywords

Comments

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

Examples

			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

References

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

Links

Crossrefs

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.

Programs

  • Haskell
    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
  • Magma
    [ n: n in [1..114] | not 0 in Intseq(n) ]; // Bruno Berselli, May 28 2011
  • Maple
    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
  • Mathematica
    A052382 = Select[Range[100], DigitCount[#, 10, 0] == 0 &] (* Alonso del Arte, Mar 10 2011 *)
  • PARI
    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
  • PARI
    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
  • PARI
    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
  • Python
    A052382 = [n for n in range(1,10**5) if not str(n).count('0')]
# Chai Wah Wu, Aug 26 2014
  • Python
    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
  • Smalltalk
    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"
  • sh
    seq 0 1000 | grep -v 0; # Joerg Arndt, May 29 2011

Formula

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)

Extensions

Typos in formula section corrected by Hieronymus Fischer, May 30 2012
Name clarified by Peter Luschny, Jun 20 2025

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