A305933 -id:A305933 - cp's OEIS frontend

A305929 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 9^k has n digits '0' (conjectured).

0, 1, 2, 3, 4, 6, 7, 12, 13, 14, 17, 34, 5, 8, 9, 10, 25, 26, 36, 11, 15, 16, 18, 19, 20, 21, 22, 23, 24, 28, 29, 30, 31, 32, 48, 54, 68, 41, 45, 56, 33, 35, 37, 44, 49, 53, 58, 64, 65, 38, 39, 40, 43, 46, 51, 52, 59, 61, 67, 82, 83, 106, 42, 47, 62, 66, 69, 72, 73, 76, 84, 89, 144, 27, 50

Offset: 0

M. F. Hasler, Jun 19 2018

The set of (nonempty) rows forms a partition of the nonnegative integers.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (9, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but it is no longer guaranteed to have a partition of the integers.
The author finds this sequence "nice", i.e., appealing (as well as, e.g., the variant A305933 for basis 3) in view of the idea of partitioning the integers in such an elementary yet highly nontrivial way, and the remarkable fact that the rows are just roughly one line long. Will this property remain for large n, or else, how will the row lengths evolve?

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 4, 6, 7, 12, 13, 14, 17, 34 (= A030705)
1 : 5, 8, 9, 10, 25, 26, 36
2 : 11, 15, 16, 18, 19, 20, 21, 22, 23, 24, 28, 29, 30, 31, 32, 48, 54, 68
3 : 41, 45, 56
4 : 33, 35, 37, 44, 49, 53, 58, 64, 65
5 : 38, 39, 40, 43, 46, 51, 52, 59, 61, 67, 82, 83, 106
...
Column 0 is A063626: least k such that 9^k has n digits '0' in base 10.
Row lengths are 12, 7, 18, 3, 9, 13, 11, 11, 6, 9, 17, 15, 12, 9, 11, 6, 9, 9, ... (A305939).
Last element of the rows (largest exponent such that 9^k has exactly n digits 0) are (34, 36, 68, 56, 65, 106, 144, 134, 119, 138, 154, ...), A306119.
Inverse permutation is (0, 1, 2, 3, 4, 12, 5, 6, 13, 14, 15, 19, 7, 8, 9, 20, 21, 10, 22, 23, 24, 25, 26, 27, 28, 16, 17, 73, 29, 30, 31, 32, ...), not in OEIS.

M. F. Hasler, Zeroless powers.. OEIS Wiki, March 2014.

Cf. A030705, A063626.

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305928 (analog for 8^k).

mx = 1000; g[n_] := g[n] = DigitCount[9^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)

apply( A305929_row(n,M=50*(n+1))=select(k->#select(d->!d,digits(9^k))==n,[0..M]), [0..10])
print(apply(t->#t,%)"\n"apply(vecmax,%)"\n"apply(t->t-1,Vec(vecsort(concat(%),,1)[1..99]))) \\ to show row lengths, last terms and the inverse permutation

Row n consists of the integers in (row n of A305933 divided by 2).

A305932 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 2^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86, 10, 11, 12, 17, 20, 21, 22, 23, 26, 29, 30, 38, 40, 41, 44, 45, 46, 47, 48, 50, 57, 58, 65, 66, 68, 71, 73, 74, 75, 84, 85, 95, 96, 122, 124, 129, 130, 149, 151, 184, 43, 53, 61, 69, 70

Offset: 0

M. F. Hasler, Jun 14 2018

A partition of the nonnegative integers (the rows being the subsets).
Although it remains an open problem to provide a proof that the rows are complete (as are all terms of A020665), we can assume it for the purpose of this sequence.
Read as a flattened sequence, a permutation of the nonnegative integers.

			The table reads:
n \ k's
0 : 0, 1, ..., 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, (...), 81, 86 (cf. A007377)
1 : 10, 11, 12, 17, 20, 21, 22, 23, 26, 29, 30, 38, 40, 41, 44, (...), 151, 184
2 : 42, 52, 54, 55, 56, 59, 60, 62, 63, 64, 78, 80, 82, 92, 107, (...), 171, 231
3 : 43, 53, 61, 69, 70, 83, 87, 89, 90, 93, 109, 112, 114, 115, (...), 221, 359
4 : 79, 91, 94, 97, 106, 118, 126, 127, 137, 139, 157, 159, 170, (...), 241, 283
5 : 88, 98, 99, 103, 104, 113, 120, 143, 144, 146, 152, 158, 160, (...), 343, 357
...
Column 0 is A031146: least k such that 2^k has n digits '0' in base 10.
Row lengths = number of powers of 2 with exactly n '0's = (36, 41, 31, 34, 25, 32, 37, 23, 43, 47, 33, 35, 29, 27, 27, 39, 34, 34, 28, 29, ...): not in the OEIS.
Largest number in row n = (86, 229, 231, 359, 283, 357, 475, 476, 649, 733, 648, 696, 824, 634, 732, 890, 895, 848, 823, 929, 1092, ...): not in the OEIS.
Row number of n = Number of '0's in 2^n = A027870: (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, ...).
Inverse permutation (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 36, 37, 38, 10, 11, 12, 13, 39, 14, 15, 40, 41, 42, 43, 16, 17, 44, 18, 19, 45, 46, 20, 21, ...) is not in the OEIS.

M. F. Hasler, Zeroless powers, OEIS Wiki, March 2014.

Cf. A007377, A031146.
Sequence A027870 yields the row number of a given integer.
Cf. A305933 (analog for 3^n), A305924 (for 4^n), ..., A305929 (for 9^n).

mx = 1000; g[n_] := g[n] = DigitCount[2^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)

apply( A305932_row(n,M=200*(n+1))=select(k->A027870(k)==n,[0..M]), [0..20]) \\ A027870(k)=#select(d->!d, digits(2^k))

Row n = { k >= 0 | A027870(k) = n }.

A071531 Smallest exponent r such that n^r contains at least one zero digit (in base 10).

Original entry on oeis.org

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

Offset: 2

Paul Stoeber (paul.stoeber(AT)stud.tu-ilmenau.de), Jun 02 2002

For all n, a(n) is at most 40000, as shown below. Is 10 an upper bound?
If n has d digits, the numbers n, n^2, ..., n^k have a total of about N = k*(k+1)*d/2, and if these were chosen randomly the probability of having no zeros would be (9/10)^N. The expected number of d-digit numbers n with f(n)>k would be 9*10^(d-1)*(9/10)^N. If k >= 7, (9/10)^(k*(k+1)/2)*10 < 1 so we would expect heuristically that there should be only finitely many n with f(n) > 7. - Robert Israel, Jan 15 2015
The similar definition using "...exactly one digit 0..." would be ill-defined for all multiples of 100 and others (1001, ...). - M. F. Hasler, Jun 25 2018
When r=40000, one of the last five digits of n^r is always 0. Working modulo 10^5, we have 2^r=9736 and 5^r=90625, and both of these are idempotent; also, if gcd(n,10)=1, then n^r=1, and if 10|n, then n^r=0. Therefore the last five digits of n^r are always either 00000, 00001, 09736, or 90625. In particular, a(n) <= 40000. - Mikhail Lavrov, Nov 18 2021

			a(4)=5 because 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024 (has zero digit).

Robert Israel, Table of n, a(n) for n = 2..10000
OEIS Wiki, Zeroless powers (2014).

Cf. A305941 for the actual powers n^k.
Cf. A007377, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944: decimal expansion of k^n contains no zeros, k = 2, 3, 4, ...
Cf. A305932, A305933, A305924, ..., A305929: row n = {k: x^k has n 0's}, x = 2, 3, ..., 9.
Cf. A305942, ..., A305947, A305938, A305939: #{k: x^k has n 0's}, x = 2, 3, ..., 9.
Cf. A306112, ..., A306119: largest k: x^k has n 0's; x = 2, 3, ..., 9.

f:= proc(n) local j;
for j from 1 do if has(convert(n^j,base,10),0) then return j fi od:
end proc:
seq(f(n),n=2..100); # Robert Israel, Jan 15 2015

zd[n_]:=Module[{r=1},While[DigitCount[n^r,10,0]==0,r++];r]; Array[zd,110,2] (* Harvey P. Dale, Apr 15 2012 *)

A071531(n)=for(k=1, oo, vecmin(digits(n^k))||return(k)) \\ M. F. Hasler, Jun 23 2018

def a(n):
    r, p = 1, n
    while 1:
        if "0" in str(p):
            return r
        r += 1
        p *= n
[a(n) for n in range(2, 100)] # Tim Peters, May 19 2005

a(n) >= 1 with equality iff n is in A011540 \ {0} = {10, 20, ..., 100, 101, ...}. - M. F. Hasler, Jun 23 2018

A305924 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 4^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 8, 9, 12, 14, 16, 17, 18, 36, 38, 43, 5, 6, 10, 11, 13, 15, 19, 20, 22, 23, 24, 25, 29, 33, 34, 37, 42, 48, 61, 62, 65, 92, 21, 26, 27, 28, 30, 31, 32, 39, 40, 41, 46, 54, 58, 68, 74, 75, 77, 35, 45, 56, 57, 64, 66, 67, 70, 71, 78, 82, 83, 87, 88, 47, 53, 59, 63, 85, 89, 91, 93, 98

Offset: 0

M. F. Hasler, Jun 14 2018

A partition of the nonnegative integers, the rows being the subsets.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (4, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but it is no longer guaranteed to have a partition of the integers.
The author finds "nice", i.e., appealing, the idea of partitioning the integers in such an elementary yet highly nontrivial way, and the remarkable fact that the rows are just roughly one line long. Will this property remain for large n, or else, how will the row lengths evolve?

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 4, 7, 8, 9, 12, 14, 16, 17, 18, 36, 38, 43 (= A030701)
1 : 5, 6, 10, 11, 13, 15, 19, 20, 22, 23, 24, 25, 29, 33, 34, 37, 42, 48, 61, 62, 65, 92
2 : 21, 26, 27, 28, 30, 31, 32, 39, 40, 41, 46, 54, 58, 68, 74, 75, 77
3 : 35, 45, 56, 57, 64, 66, 67, 70, 71, 78, 82, 83, 87, 88
4 : 47, 53, 59, 63, 85, 89, 91, 93, 98, 104, 115
5 : 44, 49, 52, 60, 72, 73, 76, 79, 80, 84, 90, 96, 109, 110, 114, 116, 120, 129, 171
...
Column 0 is A063575: least k such that 4^k has n digits '0' in base 10.
Row lengths are 16, 22, 17, 14, 11, ... = A305944.
Largest terms of the rows are 43, 92, 77, 88, 115, ... = A306114.
The inverse permutation is (0, 1, 2, 3, 4, 16, 17, 5, 6, 7, 18, 19, 8, 20, 9, 21, 10, 11, 12, 22, 23, 38, 24, 25, 26, 27, 39, 40, 41, 28, 42, 43, ...), not in OEIS.

M. F. Hasler, Zeroless powers.. OEIS Wiki, March 2014.

Cf. A030701, A063575.

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305925 (analog for 5^k), ..., A305929 (analog for 9^k).

mx = 1000; g[n_] := g[n] = DigitCount[4^n, 10, 0]; f[n_] := Select[Range@ mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018*)

apply( A305924_row(n,M=50*(n+1))=select(k->#select(d->!d,digits(4^k))==n,[0..M]), [0..19])
print(apply(t->#t,%)"\n"apply(vecmax,%)"\n"apply(t->t-1,Vec(vecsort( concat(%),,1)[1..99]))) \\ to show row lengths, last terms & inverse permutation

Row n is given by the even terms of row n of A305932, divided by 2.

A305925 Irregular table read by rows in which row n >= 0 lists all k >= 0 such that the decimal representation of 5^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58, 8, 12, 14, 15, 16, 19, 20, 21, 22, 25, 26, 31, 41, 42, 43, 85, 13, 23, 24, 27, 28, 29, 32, 36, 37, 56, 57, 107, 34, 35, 38, 39, 50, 54, 59, 74, 75, 84, 112, 40, 44, 46, 47, 49, 51, 60, 73, 78, 79, 82, 83, 86, 88, 89, 95, 96, 97, 106, 113, 127

Offset: 0

M. F. Hasler, Jun 19 2018

The set of (nonempty) rows is a partition of the nonnegative integers.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (5, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but we are no more guaranteed to get a partition of the integers.
The author finds this sequence "nice", i.e., appealing (as well as, e.g., the variant A305933 for basis 3) in view of the idea of partitioning the integers in such an elementary yet highly nontrivial way, and the remarkable fact that the rows are just roughly one line long. Will this property remain for large n, or else, how will the row lengths evolve?

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58 (cf. A008839)
1 : 8, 12, 14, 15, 16, 19, 20, 21, 22, 25, 26, 31, 41, 42, 43, 85
2 : 13, 23, 24, 27, 28, 29, 32, 36, 37, 56, 57, 107
3 : 34, 35, 38, 39, 50, 54, 59, 74, 75, 84, 112
4 : 40, 44, 46, 47, 49, 51, 60, 73, 78, 79, 82, 83, 86, 88, 89, 95, 96, 97, 106, 113, 127
5 : 48, 55, 61, 67, 77, 91, 102, 110, 111, 126, 148, 157
...
The first column is A063585: least k such that 5^k has n digits '0' in base 10.
Row lengths are 16, 16, 12, 11, 21, 12, 17, 14, 16, 17, 14, 13, 16, 18, 13, 14, 10, 10, 21, 7,... (A305945).
Last terms of the rows are (58, 85, 107, 112, 127, 157, 155, 194, 198, 238, 323, 237, 218, 301, 303, 324, 339, 476, 321, 284, ...), A306115.
The inverse permutation is (0, 1, 2, 3, 4, 5, 6, 7, 16, 8, 9, 10, 17, 32, 18, 19, 20, 11, 12, 21, 22, 23, 24, 33, 34, 25, 26, 35, 36, 37, 13, 27,...), not in OEIS.

Cf. A008839, A063585.

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305929 (analog for 9^k).

mx = 1000; g[n_] := g[n] = DigitCount[5^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)

apply( A305925_row(n,M=60*(n+1))=select(k->#select(d->!d,digits(5^k))==n,[0..M]), [0..19])

A305928 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 8^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 9, 11, 12, 13, 17, 24, 27, 4, 7, 10, 15, 16, 19, 22, 25, 28, 32, 43, 14, 18, 20, 21, 26, 36, 37, 39, 45, 47, 49, 50, 55, 57, 77, 23, 29, 30, 31, 38, 41, 44, 51, 52, 58, 61, 42, 53, 59, 62, 65, 69, 33, 40, 48, 56, 60, 64, 73, 76, 80, 86, 114, 119, 35, 46

Offset: 0

M. F. Hasler, Jun 19 2018

The set of (nonempty) rows forms a partition of the nonnegative integers.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (8, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but it is no longer guaranteed to have a partition of the integers.

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 5, 6, 8, 9, 11, 12, 13, 17, 24, 27 (= A030704)
1 : 4, 7, 10, 15, 16, 19, 22, 25, 28, 32, 43
2 : 14, 18, 20, 21, 26, 36, 37, 39, 45, 47, 49, 50, 55, 57, 77
3 : 23, 29, 30, 31, 38, 41, 44, 51, 52, 58, 61
4 : 42, 53, 59, 62, 65, 69
5 : 33, 40, 48, 56, 60, 64, 73, 76, 80, 86, 114, 119
...
Column 0 is A063596: least k such that 8^k has n digits '0' in base 10.
Row lengths are 14, 11, 15, 11, 6, 12, 10, 7, 14, 21, 9, 9, 15, 8, 6, 10, 8, 13, ... (not in the OEIS).
The inverse permutation is (0, 1, 2, 3, 14, 4, 5, 15, 6, 7, 16, 8, 9, 10, 25, 17, 18, 11, 26, 19, 27, 28, 20, 40, 12, 21, 29, 13, 22, ...), also not in the OEIS.

M. F. Hasler, Zeroless powers.. OEIS Wiki, March 2014

Cf. A030704, A063596.

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305929 (analog for 9^k).

mx = 1000; g[n_] := g[n] = DigitCount[8^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)

apply( A305928_row(n,M=50*(n+1))=select(k->#select(d->!d,digits(8^k))==n,[0..M]), [0..7])

Row n consists of integers in row n of A305932 divided by 3.

A305934 Powers of 3 that have exactly one digit '0' in base 10.

Original entry on oeis.org

59049, 14348907, 43046721, 129140163, 387420489, 3486784401, 847288609443, 68630377364883, 328256967394537077627, 26588814358957503287787, 717897987691852588770249, 6461081889226673298932241, 1144561273430837494885949696427, 22528399544939174411840147874772641, 67585198634817523235520443624317923

Offset: 1

M. F. Hasler (following a suggestion by Zak Seidov), Jun 14 2018

Motivated by A030700: decimal expansion of 3^n contains no zeros (probably finite).
It appears that this sequence is finite. Is a(15) = 3^73 the last term?
There are no more terms through at least 3^(10^7) (which is a 4771213-digit number). It seems nearly certain that no power of 3 containing this many or more decimal digits could have fewer than two '0' digits. (Among numbers of the form 3^k with 73 < k <= 10^7, the only one having fewer than two '0' digits among its final 200 digits is 3^5028978.) - Jon E. Schoenfield, Jun 24 2018
The first 6 terms coincide with A305931: powers of 3 having at least one digit 0, with complement A238939 (within A000244: powers of 3) conjectured to be finite, too. Then, a(7..8) = A305931(9..10), etc.

Cf. A030700: decimal expansion of 3^n contains no zeros (probably finite), A238939: powers of 3 with no digit '0' in their decimal expansion, A000244: powers of 3.
Subsequence of A305931: powers of 3 having at least one '0'.
Cf. A305933: row n = { k | 3^k has n digits '0' }.

Select[3^Range[120], DigitCount[#, 10, 0] == 1 &] (* Michael De Vlieger, Jul 01 2018 *)

for(n=1,99, #select(t->!t,digits(3^n))==1&& print1(3^n","))

a(n) = 3^A305933(1,n).

A305926 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 6^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 12, 17, 24, 29, 44, 10, 11, 14, 15, 18, 22, 28, 40, 42, 59, 9, 16, 20, 21, 26, 30, 31, 33, 37, 38, 39, 45, 46, 49, 51, 53, 63, 13, 23, 25, 27, 32, 34, 35, 36, 47, 48, 54, 61, 72, 73, 76, 82, 19, 52, 60, 64, 65, 70, 71, 83, 91, 93, 98, 43, 50

Offset: 0

M. F. Hasler, Jun 19 2018

The set of (nonempty) rows forms a partition of the nonnegative integers.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (6, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but it is no longer guaranteed to have a partition of the integers.
The author finds this sequence "nice", i.e., appealing (as well as, e.g., the variant A305933 for basis 3) in view of the idea of partitioning the integers in such an elementary yet highly nontrivial way, and the remarkable fact that the rows are just roughly one line long. Will this property remain for large n, or else, how will the row lengths evolve?

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 12, 17, 24, 29, 44 (= A030702)
1 : 10, 11, 14, 15, 18, 22, 28, 40, 42, 59
2 : 9, 16, 20, 21, 26, 30, 31, 33, 37, 38, 39, 45, 46, 49, 51, 53, 63
3 : 13, 23, 25, 27, 32, 34, 35, 36, 47, 48, 54, 61, 72, 73, 76, 82
4 : 19, 52, 60, 64, 65, 70, 71, 83, 91, 93, 98
5 : 43, 50, 55, 58, 62, 66, 67, 75, 77, 78, 101, 106, 129, 134
...
Column 0 is A063596: least k such that 6^k has n digits '0' in base 10.
Row lengths are 14, 10, 17, 16, 11, 14, 10, 8, 12, 19, 9, 16, 13, 11, 10, 10, 11, 10, 10, 17, ... (A305946).
Last terms of the rows yield (44, 59, 63, 82, 98, 134, 108, 123, 199, 189, 192, 200, 275, 282, 267, 307, 298, 296, 391, 338, ...), A306116.
The inverse permutation is (0, 1, 2, 3, 4, 5, 6, 7, 8, 24, 14, 15, 9, 41, 16, 17, 25, 10, 18, 57, 26, 27, 19, 42, 11, 43, 28, 44, 20, 12, 29, 30, ...), not in OEIS.

M. F. Hasler, Zeroless powers.. OEIS Wiki, March 2014.

Cf. A030702, A063596.

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305929 (analog for 9^k).

mx = 1000; g[n_] := g[n] = DigitCount[6^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)

apply( A305926_row(n,M=50*(n+1))=select(k->#select(d->!d,digits(6^k))==n,[0..M]), [0..19])

A305931 Powers of 3 having at least one digit '0' in their decimal representation.

Original entry on oeis.org

59049, 14348907, 43046721, 129140163, 387420489, 3486784401, 10460353203, 31381059609, 847288609443, 68630377364883, 205891132094649, 1853020188851841, 5559060566555523, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801

Offset: 1

M. F. Hasler, Jun 15 2018

The analog of A298607 for 3^k instead of 2^k.

The complement A238939 is conjectured to have only 23 elements, the largest being 3^68. Thus, all larger powers of 3 are (conjectured to be) in this sequence. Each of the subsequences "powers of 3 with exactly n digits 0" is conjectured to be finite. Provided there is at least one such element for each n >= 0, this leads to a partition of the integers, given in A305933.

Cf. A030700 = row 0 of A305933: decimal expansion of 3^n contains no zeros.
Complement (within A000244: powers of 3) of A238939: powers of 3 with no digit '0' in their decimal expansion.
Analog of A298607: powers of 2 with the digit '0' in their decimal expansion.
The first six terms coincide with the finite sequence A305934: powers of 3 having exactly one digit 0.

Select[3^Range[0,40],DigitCount[#,10,0]>0&] (* Harvey P. Dale, May 30 2020 *)

for(k=0,69, vecmin(digits(3^k))|| print1(3^k","))

select( t->!vecmin(digits(t)), apply( k->3^k, [0..40]))

A305943 Number of powers of 3 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

23, 15, 31, 13, 18, 23, 23, 25, 16, 17, 28, 25, 22, 20, 18, 21, 19, 19, 18, 24, 33, 17, 17, 18, 17, 14, 21, 26, 25, 23, 24, 29, 17, 22, 18, 21, 27, 26, 20, 21, 13, 27, 24, 12, 18, 24, 16, 17, 15, 30, 24, 32, 24, 12, 16, 16, 23, 23, 20, 23, 19, 23, 10, 21, 20, 21, 23, 20, 19, 23, 23, 22, 16, 18, 20, 20, 13, 15, 25, 24, 28, 24, 21, 16, 14, 23, 21, 19, 23, 19, 27, 26, 22, 18, 27, 16, 31, 21, 18, 25, 24

Offset: 0

M. F. Hasler, Jun 22 2018

a(0) = 23 is the number of terms in A030700 and in A238939, which include the power 3^0 = 1.

These are the row lengths of A305933. It remains an open problem to provide a proof that these rows are complete (as for all terms of A020665), but the search has been pushed to many orders of magnitude beyond the largest known term, and the probability of finding an additional term is vanishingly small, cf. Khovanova link.

M. F. Hasler, Zeroless powers, OEIS Wiki, March 2014, updated 2018.
T. Khovanova, The 86-conjecture, Tanya Khovanova's Math Blog, Feb. 2011.
W. Schneider, No Zeros, 2000, updated 2003. (On web.archive.org--see A007496 for a cached copy.)

Cf. A030700 = row 0 of A305933: k s.th. 3^k has no '0'; A238939: these powers 3^k.
Cf. A305931, A305934: powers of 3 with at least / exactly one '0'.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063555 = column 1 of A305933: least k such that 3^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

A305943(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(3^k))==n)

A305943_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(3^k)),nMax)]++);a[^-1]}

cp's OEIS Frontend

A305929 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 9^k has n digits '0' (conjectured).

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A305932 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 2^k has n digits '0' (conjectured).

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A071531 Smallest exponent r such that n^r contains at least one zero digit (in base 10).

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A305924 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 4^k has n digits '0' (conjectured).

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A305925 Irregular table read by rows in which row n >= 0 lists all k >= 0 such that the decimal representation of 5^k has n digits '0' (conjectured).

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A305928 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 8^k has n digits '0' (conjectured).

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A305934 Powers of 3 that have exactly one digit '0' in base 10.

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