cp's OEIS Frontend

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.

Showing 1-8 of 8 results.

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

Views

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

A051833 Primes of form (2*10^(5n) - 10^(4n) + 2*10^(3n) + 10^(2n) + 10^n + 1)/3.

Original entry on oeis.org

2, 64037, 66666663333334000000033333336666667
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 11 1999

Keywords

Comments

The Baxter-Hickerson function provides a number whose cube lacks zeros.
Next term has 665 digits and is in b-file.

Links

Crossrefs

Cubes: A051750, A051751, A051832, A052044, A052045, A052427.
Squares: A052040, A052041, A052042, A052043.

Programs

  • Mathematica
    Select[Table[(2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3, {n, 0, 150}], PrimeQ] (* Amiram Eldar, Jul 18 2025 *)

Formula

a(n) = A052427(A051832(n)). - Amiram Eldar, Jul 18 2025

A051750 Primes whose cubes lack zeros.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 53, 61, 71, 83, 97, 113, 137, 139, 151, 157, 167, 173, 179, 181, 191, 197, 211, 233, 239, 241, 251, 257, 263, 277, 283, 293, 307, 331, 337, 347, 353, 359, 373, 379, 383, 389, 409, 421, 433, 457, 461, 463, 499, 503
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 07 1999

Keywords

Links

Crossrefs

Cf. A000040, A000578.
Cubes: A052044, A052045, A051751, A051832, A051833. Squares: A052040, A052041, A052042, A052043.

Programs

  • Mathematica
    Select[Prime[Range[100]],FreeQ[IntegerDigits[#^3],0]&] (* Harvey P. Dale, Jan 06 2017 *)
  • PARI
    isok(n) = isprime(n) && vecmin(digits(n^3)); \\ Michel Marcus, Jan 06 2014

Extensions

More terms from Michel Marcus, Jan 06 2014

A051832 Numbers k such that (2*10^(5*k) - 10^(4*k) + 2*10^(3*k) + 10^(2*k) + 10^k + 1)/3 is prime.

Original entry on oeis.org

0, 1, 7, 133
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 11 1999

Keywords

Comments

The Baxter-Hickerson function provides a number whose cube lacks zeros.
The next term is > 4400. - Jason Earls, Sep 10 2005
The next term is > 20000 (found using pfgw64). - Patrick De Geest, Jul 22 2012

Links

Crossrefs

Cubes: A052044, A052045, A051750, A051751, A051833. Squares: A052040, A052041, A052042, A052043.

Programs

  • Maple
    f := n->(2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3;
  • PARI
    is(n)=isprime((2*10^(5*n)-10^(4*n)+2*10^(3*n)+10^(2*n)+10^n+1)/3) \\ Charles R Greathouse IV, Feb 17 2017

A052044 Numbers k such that k^3 lacks the digit zero in its decimal expansion.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 17, 18, 19, 21, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 44, 45, 46, 49, 51, 53, 54, 55, 56, 57, 58, 61, 62, 64, 65, 66, 68, 71, 72, 75, 76, 77, 78, 81, 82, 83, 85, 88, 91, 92, 95, 96, 97, 98, 104, 105, 108, 111
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite since A052427 is a subsequence. - Amiram Eldar, Nov 23 2020

Links

Crossrefs

Cubes: A052045, A051750, A051751, A051832, A051833, A052427.
Squares: A052040, A052041, A052042, A052043.

Programs

Formula

a(n) = A052045(n)^(1/3). - Amiram Eldar, Nov 23 2020

A051751 Cubes arising in A051750.

Original entry on oeis.org

8, 27, 125, 343, 1331, 2197, 4913, 6859, 12167, 24389, 29791, 68921, 148877, 226981, 357911, 571787, 912673, 1442897, 2571353, 2685619, 3442951, 3869893, 4657463, 5177717, 5735339, 5929741, 6967871, 7645373, 9393931, 12649337
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 07 1999

Keywords

Crossrefs

Cubes: A052044, A052045, A051750, A051832, A051833. Squares: A052040, A052041, A052042, A052043.

Formula

A030078 INTERSECT A052382. - R. J. Mathar, Mar 23 2007

A052427 Baxter-Hickerson numbers.

Original entry on oeis.org

2, 64037, 6634003367, 666334000333667, 66663334000033336667, 6666633334000003333366667, 666666333334000000333333666667, 66666663333334000000033333336666667
Offset: 0

Views

Author

Eric W. Weisstein

Keywords

Comments

From Amiram Eldar, Nov 23 2020: (Start)
Named after Lew Baxter and Dean Hickerson.
Pegg (1999) conjectured that the sequence of zeroless cubes (A052045) is finite. On April 19, 1999, Hickerson gave the counterexample: if n == 2 (mod 3) and n >= 5, then the cube of (2*10^(5*n) - 10^(4*n) + 17*10^(3*n-1) + 10^(2*n) + 10^n - 2)/3 is zeroless. Three days later, Baxter gave a simpler variation which is valid for all n>=0 and is given in the Formula section. (End)

References

  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005. See p. 109.

Links

Crossrefs

Subsequence of A052044.
Cf. A016789, A051832, A051833, A052045.

Programs

  • Maple
    a(0) = 2, and 2^3 = 8 is zeroless.
a(1) = 64037, and 64037^3 = 262598918898653 is zeroless.
  • Mathematica
    a[n_] := (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3; Array[a, 10, 0] (* Amiram Eldar, Nov 23 2020 *)

Formula

a(n) = (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3 (Baxter, 1999). - Amiram Eldar, Nov 23 2020

Extensions

Offset changed to 0 by Amiram Eldar, Nov 23 2020

A067070 Cubes whose product of digits is a cube > 0.

Original entry on oeis.org

1, 8, 24389, 226981, 9393931, 11239424, 17373979, 36264691, 66923416, 94818816, 348913664, 435519512, 463684824, 549353259, 555412248, 743677416, 3929352552, 4982686912, 5526456832, 11329982936, 12374478297, 12681938368, 15142552424
Offset: 1

Views

Author

Amarnath Murthy, Jan 05 2002

Keywords

Examples

			24389 is in the sequence because (1) it is a cube and (2) the product of its digits is 2*4*3*8*9, = 1728 which is a cube > 0.

References

  • Felice Russo, A set of new Smarandache Functions, Sequences and conjectures in number theory, American Research Press, Lupton USA.

Links

Crossrefs

Intersection of A237767 and A000578.
Cf. A007954, A052045, A052382, A067071.

Programs

  • Mathematica
    pdcQ[n_]:=Module[{pd=Times@@IntegerDigits[n]},pd>0&&IntegerQ[ Surd[ pd,3]]]; Select[Range[3000]^3,pdcQ] (* Harvey P. Dale, Jun 01 2015 *)
  • PARI
    isA237767(k)={my(p=vecprod(digits(k))); p && ispower(p,3)}
{ for (m=1, 2500, if(isA237767(m^3), print1(m^3, ", "))) } \\ Harry J. Smith, May 04 2010
  • PARI
    first(n) = {
     my(res = List(), c, vp, i);
     for(i = 1, oo,
          c = i^3;
          vp = vecprod(digits(c));
          if(vp && ispower(vp,3),
               listput(res, c);
               if(#res >= n,
                    return(Vec(res))
               )
          )
     )
} \\ David A. Corneth, Dec 01 2023

Formula

a(n) = A067071(n)^3. - Andrew Howroyd, Dec 05 2024

Extensions

More terms from Sascha Kurz, Mar 23 2002
One further term from Luc Stevens (lms022(AT)yahoo.com), May 03 2006
Edited by R. J. Mathar, Aug 08 2008
Offset changed from 0 to 1 by Harry J. Smith, May 04 2010
Showing 1-8 of 8 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.