A234967 Smallest zeroless number such that a(n)^n has at least one zero.

32, 16, 7, 4, 4, 8, 5, 6, 2, 2, 2, 4, 5, 3, 3, 2, 3, 4, 2, 2, 2, 2, 4, 3, 2, 4, 4, 2, 2, 4, 3, 3, 4, 3, 3, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2

Offset

2,1

Comments

It is very probable that a(n) = 2 for n > 87.

Links

Table of n, a(n) for n=2..74.

Example

7 is the smallest number with nonzero digits such that 7^4 has at least one zero, so a(4) = 7.

Mathematica

m[n_] := Min@ IntegerDigits@n; a[1]=0; a[n_] := Block[{k=2}, While[m[k] == 0 || m[k^n] > 0, k++]; k]; Array[a, 70] (* Giovanni Resta, Jan 11 2014 *)

Prog

(Python)
def f(x):
..for n in range(10**7):
....if str(n).find("0") == -1:
......if str(n**x).find("0") > -1:
........return n
x = 1
while x < 75:
..if f(x) == None:
....print(0)
..else:
....print(f(x))
..x += 1

Crossrefs

Cf. A052382.

Sequence in context: A097614 A069892 A070621 * A033352 A364904 A140387

Adjacent sequences: A234964 A234965 A234966 * A234968 A234969 A234970

Keyword

nonn,base

Author

Derek Orr, Jan 02 2014

Status

approved