A020665 a(n) is the (conjectured) maximal exponent k such that n^k does not contain a digit zero in its decimal expansion.

86, 68, 43, 58, 44, 35, 27, 34, 0, 41, 26, 14, 34, 27, 19, 27, 17, 44, 0, 13, 22, 10, 13, 29, 15, 9, 16, 14, 0, 16, 7, 23, 5, 17, 22, 16, 10, 19, 0, 9, 13, 10, 6, 39, 7, 8, 19, 5, 0, 19, 18, 7, 13, 11, 23, 7, 23, 14, 0, 16, 5, 14, 12, 3, 14, 14, 14, 12, 0, 8, 22, 6, 4, 19, 11, 12, 10, 9, 0

Offset: 2

Author

David W. Wilson

Keywords

• nonn
• base

Comments

Most of these values are not proved rigorously, but the search has been pushed very large (~ 10^9 or beyond for many n). See the OEIS wiki page for further reading. - M. F. Hasler, Mar 08 2014

From Bill McEachen, Apr 01 2015: (Start)

It appears that the values at square pointers will be no more than that of the base pointer. Specifically when the value at the base pointer is even, the value at the square will be 50%. For example, the sequence n=2,4,16 yields a(n)=86,43,19. The sequence n=3,9,81 yields a(n)=68,34,17.

Values at other than squares are less obvious. However, at some point, the run of the squares ends, implying remaining nonzero values should indicate either nonsquares or prime entries. (End)

Since (n^b)^j = n^(b*j), a(n) >= b*a(n^b); if a(n) is divisible by b then a(n^b) = a(n)/b. - Robert Israel, Apr 01 2015

Examples

			a(13) = 14 because 13^14 does not have a digit 0, but (it is conjectured that) for all k > 14, 13^k will have a digit 0. It is not excluded that there may be some k < a(n) for which n^k does have a digit 0, as is the case for 13^6. - _M. F. Hasler_, Mar 29 2015
		

Links

• Antoine Beaulieu, Table of n, a(n) for n = 2..200 (checked with PARI up to k=10^5)
• M. F. Hasler, Zeroless powers, OEIS Wiki, Mar 07 2014
• Eric Weisstein's World of Mathematics, Zero

Crossrefs

For the zeroless numbers (powers x^n), see A238938, A238939, A238940, A195948, A238936, A195908, A195946, A195945, A195942, A195943, A103662.

For the corresponding exponents, see A007377, A008839, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944.

For other related sequences, see A011540, A052382, A027870, A102483, A103663.

Programs

Maple

f:= proc(n)
  local p;
  if n mod 10 = 0 then return 0 fi;
  for p from 100 by -1 do
    if not has(convert(n^p,base,10),0) then return(p) fi
  od
0
end proc:
seq(f(n),n=2..80); # Robert Israel, Apr 01 2015

Mathematica

a = {}; Do[ If[ Mod[n, 10] == 0, b = 0; Continue]; Do[ If[ Count[ IntegerDigits[n^k], 0 ] == 0, b = k], {k, 1, 200} ]; a = Append[a, b], {n, 2, 81} ];

PARI

Nmax(x,L=99,m=0)=for(n=1,L,vecmin(digits(x^n))&&m=n);m \\ L=99 is enough to reproduce the known results, since no value > 86 is known; M. F. Hasler, Mar 08 2014

Formula

a(10n) = 0 for any n>0. - M. F. Hasler, Dec 17 2014

a(100n+1) = 0 for any n>0. - Robert Israel, Apr 01 2015

a(80*n+65) <= 3, because for k >= 4, (80*n+65)^k == 625 (mod 10000). - Robert Israel, Apr 02 2015

From Chai Wah Wu, Jan 08 2020: (Start)

The following values and bounds are for the actual maximal exponents (not conjectured).

a(A052382(n)) > 0 for n > 1.

a(225) = 1

a(225^k) = 0 for k > 1.

a(625) = 1.

a(625^k) = 0 for k > 1.

a(3126) = 2.

a(3126^2) = 1.

a(3126^k) = 0 for k > 2.

a(9376) = 1.

a(9376^k) = 0 for k > 1.

a(21876) = 2.

a(21876^2) = 1.

a(21876^k) = 0 for k > 2.

a(34376) = 1.

a(34376^k) = 0 for k > 1.

a(400*n + 225) <= 1, since for k >= 2, (400*n + 225)^k == 625 (mod 10000), i.e., if 400*n + 225 is in A052382, then a(400*n+225) = 1, otherwise it is 0.

a(25000*n + 34376) <= 1, since for k >= 2, (25000*n + 34376)^k == 9376 (mod 100000), i.e., if 25000*n + 34376 is in A052382, then a(25000*n + 34376) = 1, otherwise it is 0.

a(25000*n + 21876) <= 2, since for k >= 3, (25000*n + 21876)^k == 9376 (mod 100000).

a(12500*n + 3126) <= 4, since for k >= 5, (12500*n + 3126)^k == 9376 (mod 100000).

(End)