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

Author

Chai Wah Wu, Jan 18 2020

Keywords

• nonn
• base

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^(b*m/(b-1))). Let s <= r be the largest integer >= 0 such that (r-s)*10^(b*m) < (r-s+1)^b. Then r, r-1, ... r-s are such numbers k and there are no other such numbers.

Examples

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

Links

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

Crossrefs

Cf. A307371, A307588, A307600.