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
KEYWORD
nonn,base
AUTHOR
Chai Wah Wu, Jan 18 2020
STATUS
approved