A331492 Numbers k such that the digits of k^(1/5) begin with k.

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

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/(b-1)). Let s <= r be the largest integer >= 0 such that (r-s)*10^(bm) < (r-s+1)^b. Then r, r-1, ... r-s 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