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).

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

1,2

Comments

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)

Links

Table of n, a(n) for n=1..69.

Formula

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

Example

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

Crossrefs

Cf. A090493, A020665, A171102.

Sequence in context: A289743 A308280 A259086 * A038823 A296890 A225535

Adjacent sequences: A062515 A062516 A062517 * A062519 A062520 A062521

Keyword

base,nonn

Author

Robert G. Wilson v, Jun 24 2001

Extensions

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

Status

approved