A052041 -id:A052041 - 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

A075415 Squares of A002280 or numbers (666...6)^2.

Original entry on oeis.org

0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556, 444444444444443555555555555556

Offset: 0

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 66^2 = 4356.
From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..500
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417.
Cf. A059988, A178630, A178631, A178632, A178633, A178634, A178635. [From Reinhard Zumkeller, May 31 2010]
Cf. A002275, A002278, A002279,  A002280, A002283, A002477, A052041, A102807, A309827.

Table[FromDigits[PadRight[{},n,6]]^2,{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{0,36,4356},20] (* Harvey P. Dale, May 20 2021 *)

a(n) = A002280(n)^2 = (6*A002275(n))^2 = 36*A002275(n)^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10 + 3)*10^(n-1) + A002279(n-1))*10 + 6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
From Elmo R. Oliveira, Jul 27 2025: (Start)
E.g.f.: 4*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 36*A002477(n). (End)

Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)

A052040 Numbers whose square is zeroless.

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, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 46, 54, 56, 57, 58, 59, 61, 62, 63, 65, 66, 67, 68, 69, 72, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 96

Offset: 1

Patrick De Geest, Dec 15 1999

This sequence is infinite, since 33...334^2 = 11...11155...556, for example. This answers an open problem stated in HAKMEM. - Karl W. Heuer, Aug 19 2015

			From _Jon E. Schoenfield_, Aug 16 2021: (Start)
31 is a term: 31^2 = 961 has no 0's among its digits.
32 is not a term, because 32^2 = 1024. (End)

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Michael Beeler, R. William Gosper, and Rich Schroeppel, "HAKMEM" Item 36, Memo 239, Artificial Intelligence Laboratory, Massachusetts Institute of Technology, Cambridge, Mass., 1972.

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

Cf. A052382.

[n: n in [1..100] | not 0 in Intseq(n^2)]; // Vincenzo Librandi, Feb 22 2015

Select[Range[0, 100], DigitCount[#^2, 10, 0]==0 &] (* Vincenzo Librandi, Feb 22 2015 *)

is(n)=vecmin(digits(n^2))>0 \\ Charles R Greathouse IV, Oct 11 2013

A052043 Squares of primes lacking the digit zero in their decimal expansion.

Original entry on oeis.org

4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 3481, 3721, 4489, 5329, 6241, 6889, 7921, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 24649, 26569, 27889, 29929, 32761, 36481, 37249, 44521, 49729, 51529, 52441, 54289

Offset: 1

Patrick De Geest, Dec 15 1999

Intersection of A052382 and A001248; A168046(a(n))*A064911(a(n))*A010052(a(n)) = 1. - Reinhard Zumkeller, Dec 01 2009

Zak Seidov, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Zerofree - _Reinhard Zumkeller_, Dec 01 2009

Squares: A052042, A052040, A052041.

Cubes: A051750, A051751, A051832, A051833.

Select[Prime[Range[100]]^2,DigitCount[#,10,0]==0&] (* Harvey P. Dale, Mar 18 2012 *)

is(n)=my(d=digits(n));vecsort(d)[1]&&issquare(n,&n)&&isprime(n) \\ Charles R Greathouse IV, Jun 05 2013

a(n) = A052042(n)^2. - R. J. Mathar, Jul 23 2025

A102807 a(n) is the square of one plus the number consisting of n 3's.

Original entry on oeis.org

1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556

Offset: 0

Ron Knott, Feb 27 2005

Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A052041, A093137, A075415, A109344, A104265.

a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2018

Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)

From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024

New name from Alois P. Heinz, Sep 03 2018

A052042 Primes that lack the digit zero in the decimal expansion of their squares.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 59, 61, 67, 73, 79, 83, 89, 107, 109, 113, 127, 131, 137, 139, 157, 163, 167, 173, 181, 191, 193, 211, 223, 227, 229, 233, 239, 263, 269, 271, 277, 281, 293, 307, 311, 313, 337, 359, 367, 373, 379, 383, 389, 409, 419, 421, 431

Offset: 1

Patrick De Geest, Dec 15 1999

			The primes 47, 53 and 71 are not in the sequence because 47^2=2209, 53^2=2809 and 71^2=5041 contain zeros in their decimal representation.

Zak Seidov, Table of n, a(n) for n = 1..10000

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

Cf. A245576.

fQ[n_] := DigitCount[n^2][[-1]] == 0; Select[Prime@ Range@ 80, fQ] (* Robert G. Wilson v, Aug 22 2012 *)

{p=2;for(k=1,10^2,if(vecmin(digits(p^2))>0,
print1(p", "));p=nextprime(1+p))}\\ Zak Seidov, Dec 24 2014

a(n) = sqrt(A052043(n)). - Zak Seidov, Dec 27 2014

A051833 Primes of form (210^(5n) - 10^(4n) + 210^(3n) + 10^(2n) + 10^n + 1)/3.

Original entry on oeis.org

2, 64037, 66666663333334000000033333336666667

Offset: 1

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

The Baxter-Hickerson function provides a number whose cube lacks zeros.

Next term has 665 digits and is in b-file.

Amiram Eldar, Table of n, a(n) for n = 1..4
Ed Pegg, Jr., Fun with numbers.
Eric Weisstein's World of Mathematics, Baxter-Hickerson Function.

Cubes: A051750, A051751, A051832, A052044, A052045, A052427.

Squares: A052040, A052041, A052042, A052043.

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 *)

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

A104264 Number of n-digit squares with no zero digits.

Original entry on oeis.org

3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359, 549697, 1563545, 4446173, 12650545, 35999714, 102439796, 291532841, 829634988, 2360947327, 6719171580, 19122499510, 54423038535, 154888366195

Offset: 1

Reinhard Zumkeller and Ron Knott, Feb 26 2005

Comments from David W. Wilson, Feb 26 2005: (Start)
"There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit squares. A random d-digit number has the probability p(d) = (9/10)^(d-1) of being zeroless (exponent d-1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless d-digit squares.
"For d = 1 through 12, we get (truncating): 1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117, ... The elements grow approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.
"The same naive estimate can easily be generalize to k-th powers, giving the estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit k-th powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless k-th powers when this ratio is >= 1, which it is for k <= 21. For k >= 22, the ratio is < 1 and we should expect a finite number of zeroless k-th powers." (End)

			a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.

Cf. A052041, A104265, A104266, A075415, A102807.

def aupton(terms):
  c, k, kk = [0 for i in range(terms)], 1, 1
  while kk < 10**terms:
    s = str(kk)
    c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
  return c
print(aupton(14)) # Michael S. Branicky, Mar 06 2021

a(14)-a(18) from Donovan Johnson, Nov 05 2009
a(19)-a(21) from Donovan Johnson, Mar 23 2011
a(22)-a(25) from Donovan Johnson, Jan 29 2013

cp's OEIS Frontend

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

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A075415 Squares of A002280 or numbers (666...6)^2.

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A052040 Numbers whose square is zeroless.

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A052043 Squares of primes lacking the digit zero in their decimal expansion.

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A102807 a(n) is the square of one plus the number consisting of n 3's.

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A052042 Primes that lack the digit zero in the decimal expansion of their squares.

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A051833 Primes of form (210^(5n) - 10^(4n) + 210^(3n) + 10^(2n) + 10^n + 1)/3.

A051832 Numbers k such that (210^(5k) - 10^(4k) + 210^(3k) + 10^(2k) + 10^k + 1)/3 is prime.