A020665 -id:A020665 - cp's OEIS frontend

A175402 a(n) is the number of iterations of {r -> (((D_1^D_2)^D_3)^...)^D_k, where D_k is the k-th decimal digit of r} needed to reach a one-digit number, starting at r = n.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 4, 1, 1, 1, 3, 2, 3, 3, 3, 3, 2, 1, 1, 2, 3, 3, 2, 2, 2, 3, 3, 1, 1, 3, 2, 3, 3, 2, 3, 2, 2, 1, 1, 4, 4, 2, 3, 3, 3, 2, 2, 1, 1, 4, 4, 2, 2, 2, 3, 2, 2, 1, 1, 3, 4, 2, 3, 3, 2, 2, 2, 1, 1, 2, 3, 3, 2, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Jaroslav Krizek, May 01 2010

Conjecture: max(a(n)) = 4.

Assuming that A020665(2) = 86, A020665(3) = 68, A020665(5) = 58, and A020665(7) = 35, this conjecture is true, since in that case the largest power of a decimal digit that has no 0 is 7^35, and of those powers p none have a(p) > 3. The only n for which a(n) = 4 are those where one iteration goes to 6^2, 7^2, 6^3, 7^3, or 2^9.

			For n = 29: a(29) = 4 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}.

Cf. A175398, A175399, A175400, A175401, A175403, A175405.

iter(n)=my(v=eval(Vec(Str(n))));v[1]^prod(i=2,#v,v[i])
a(n)=my(k=0);while(n>9,k++;n=iter(n));k

Corrected, extended, comment, and program from Charles R Greathouse IV, Aug 03 2010

A175403 a(n) is the smallest number m requiring n iterations {((((D_1^D_2)^D_3)^D_4)^...)^D_k to reach a one-digit number starting at r = n, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)}.

Original entry on oeis.org

0, 10, 24, 26, 29

Offset: 0

Jaroslav Krizek, May 01 2010

Conjecture: sequence is finite.

Assuming that A020665(2) = 86, A020665(3) = 68, A020665(5) = 58 and A020665(7) = 35, a(4) is the last term; see A175402.

			For n = 4: a(4) = 29 because 29 is the smallest number with 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}.

Cf. A020665, A175398, A175399, A175400, A175401, A175402, A175405.

Comment and edits from Charles R Greathouse IV, Aug 03 2010

Further edits from N. J. A. Sloane, Aug 08 2010. I am still worried that n is mentioned too many times in the definition.

A062518 Conjectural largest exponent k such that n^k does not contain all of the digits 0 through 9 (in decimal notation) or 0 if no such k exists (for example if n is a power of 10).

Original entry on oeis.org

0, 168, 106, 84, 65, 64, 61, 56, 53, 0, 41, 51, 37, 34, 34, 42, 27, 25, 44, 168, 29, 24, 50, 23, 29, 31, 28, 28, 45, 106, 28, 18, 24, 34, 18, 32, 25, 17, 41, 84, 23, 19, 20, 29, 39, 32, 15, 29, 16, 65, 29, 29, 30, 18, 17, 33, 19, 31, 27, 64, 26, 19, 24, 28, 17, 15, 21, 25, 13

Offset: 1

Robert G. Wilson v, Jun 24 2001

I do not know how many of these terms have been proved to be correct. - N. J. A. Sloane
In particular, are the powers of 10 the only n with a(n) = 0?
Note that a(10n) = a(n) unless n^a(n) contains no 0 (i.e., a(n) = A020665(n)), in which case a(10n) < a(n). - Christopher J. Smyth, Aug 20 2014
From Robert G. Wilson v, Aug 22 2021: (Start)
Conjectured first occurrence of k for k >= 0: 1, 156224, 22148, 7342, 3376, 861, 609, 477, 295, 152, 153, 149, 138, 69, 139, 47, 49, 38, 32, 42, 43, 67, 92, 24, 22, 18, 61, 17, 27, 21, 53, 26, 36, 56, 14, 190, 271, 13, 110, 45, ?40?, 11, 16, ?43?, 19, 29, ..., .
Other integers which satisfy a(n) = 0 are 1023458769, 1023458967, 1023467895, 1023469875, 1023475986, 1023478695, .... These are all members of A171102.
(End)

			a(11) = 41 as 11^41 = 4978518112499354698647829163838661251242411 is the conjectural highest power of 11 not containing all ten digits.
a(110) = 38 as 110^38 does not contain the digit 2, while, conjecturally, all higher powers of 110 contain all ten digits. - _Christopher J. Smyth_, Aug 20 2014

Cf. A090493, A020665, A171102.

a(n^e) <= a(n)/e. - Robert G. Wilson v, Oct 02 2021

Definition corrected by Christopher J. Smyth, Aug 20 2014.

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.

A305946 Number of powers of 6 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

14, 10, 17, 16, 11, 14, 10, 8, 12, 19, 9, 16, 13, 11, 10, 10, 11, 10, 10, 17, 7, 15, 14, 16, 13, 22, 12, 17, 15, 17, 7, 6, 14, 22, 13, 19, 14, 12, 15, 7, 11, 14, 6, 12, 9, 12, 9, 14, 13, 15, 21

Offset: 0

M. F. Hasler, Jun 22 2018

a(0) = 14 is the number of terms in A030702 and in A195948, which includes the power 6^0 = 1.

These are the row lengths of A305926. 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 vanishing, 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. A030702 = row 0 of A305926: k such that 6^k has no 0's; A238936: these powers 6^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063596 = column 1 of A305926: least k such that 6^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

A305946(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(6^k))==n)

A305946_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(6^k)),nMax)]++);a[^-1]}

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])

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]}

A305944 Number of powers of 4 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

16, 22, 17, 14, 11, 19, 15, 15, 21, 20, 17, 22, 12, 13, 17, 24, 16, 19, 8, 17, 11, 15, 17, 15, 20, 17, 18, 20, 17, 21, 16, 19, 16, 14, 15, 19, 20, 24, 7, 16, 13, 14, 13, 14, 22, 22, 15, 18, 16, 16, 25

Offset: 0

M. F. Hasler, Jun 22 2018

a(0) = 16 is the number of terms in A030701 and in A238940, which includes the power 4^0 = 1.

These are the row lengths of A305924. It remains an open problem to provide a proof that these rows are complete (as are 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 vanishing, 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. A030701 = row 0 of A305924: k such that 4^k has no 0's; A238940: these powers 4^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063575 = column 1 of A305924: least k such that 4^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

A305944(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(4^k))==n)

A305944_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(4^k)),nMax)]++);a[^-1]}

A305945 Number of powers of 5 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

16, 16, 12, 11, 21, 12, 17, 14, 16, 17, 14, 13, 16, 18, 13, 14, 10, 10, 21, 7, 19, 13, 15, 13, 10, 15, 12, 15, 11, 11, 15, 10, 9, 15, 17, 16, 13, 12, 12, 11, 14, 9, 14, 15, 16, 14, 13, 14, 15, 24, 14

Offset: 0

M. F. Hasler, Jun 22 2018

a(0) = 16 is the number of terms in A008839 and in A195948, which includes the power 5^0 = 1.

These are the row lengths of A305925. It remains an open problem to provide a proof that these rows are complete (as are 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 vanishing, 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. A030701 (= row 0 of A305925): k such that 5^k has no 0's; A195948: these powers 4^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063585 (= column 1 of A305925): least k such that 5^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

A305945(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(5^k))==n)

A305945_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(5^k)),nMax)]++);a[^-1]}

cp's OEIS Frontend

A175402 a(n) is the number of iterations of {r -> (((D_1^D_2)^D_3)^...)^D_k, where D_k is the k-th decimal digit of r} needed to reach a one-digit number, starting at r = n.

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A175403 a(n) is the smallest number m requiring n iterations {((((D_1^D_2)^D_3)^D_4)^...)^D_k to reach a one-digit number starting at r = n, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)}.

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A062518 Conjectural largest exponent k such that n^k does not contain all of the digits 0 through 9 (in decimal notation) or 0 if no such k exists (for example if n is a power of 10).

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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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A305946 Number of powers of 6 having exactly n digits '0' (in base 10), conjectured.

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

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

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