

A331492


Numbers k such that the digits of k^(1/5) begin with k.


6



0, 1, 17, 315, 316, 5623, 99999, 100000, 1778279, 31622776, 562341324, 562341325, 9999999999, 10000000000, 177827941003, 3162277660168, 56234132519034, 999999999999999, 1000000000000000, 17782794100389227, 17782794100389228, 316227766016837932, 316227766016837933
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OFFSET

1,3


COMMENTS

The following algorithm will generate all numbers k such that the digits of k^(1/b) begins with k: For each integer m >= 0, compute r = floor(10^(bm/(b1)). Let s <= r be the largest integer >= 0 such that (rs)*10^(bm) < (rs+1)^b. Then r, r1, ... rs are such numbers k and there are no other such numbers.


LINKS

Chai Wah Wu, Table of n, a(n) for n = 1..1140


EXAMPLE

5623^(1/5) = 5.6233305990931... which starts with the digits 5623, so 5623 is in the sequence.


CROSSREFS

Cf. A307371, A307588, A307600.
Sequence in context: A197526 A282965 A142428 * A244764 A089571 A196455
Adjacent sequences: A331489 A331490 A331491 * A331493 A331494 A331495


KEYWORD

nonn,base


AUTHOR

Chai Wah Wu, Jan 18 2020


STATUS

approved



