A244606 Least number k > 1 such that k^n contains the digit k k times, or 0 if no such digit exists.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 5, 3, 2, 7, 0, 0, 5, 3, 6, 4, 0, 3, 5, 3, 0, 2, 6, 5, 3, 0, 0, 0, 6, 2, 3, 4, 2, 4, 2, 2, 2, 3, 4, 3, 5, 2, 2, 4, 5, 2, 2, 0, 4, 0, 3, 7, 2, 5, 3, 4, 0, 4, 2, 4, 2, 7, 7, 7, 2, 0, 3, 2, 8, 6, 6, 2, 0, 3, 7, 2, 4, 0, 6, 0, 0, 0, 8, 5, 4, 3, 0, 0, 6, 5, 2, 5, 0, 8, 3

Offset

1,18

Comments

1 < a(n) < 10 if a(n) is not 0. Thus a(n) = 0 is definite for the n-values above.

Links

Table of n, a(n) for n=1..118.

Example

2^18, 2^19 and 2^21 all contain the digit 2 twice. So a(18) = a(19) = a(21) = 2.

Prog

(Python)
def tes(n):
..for k in range(2, 10):
....if str(k**n).count(str(k)) == k:
......return k
n = 1
while n < 200:
..if tes(n):
....print(tes(n), end=', ')
..else:
....print(0, end=', ')
..n += 1

Crossrefs

Cf. A244604, A244603.

Sequence in context: A141720 A353449 A248017 * A273127 A103272 A065710

Adjacent sequences: A244603 A244604 A244605 * A244607 A244608 A244609

Keyword

nonn,base,easy

Author

Derek Orr, Jul 01 2014

Status

approved