Program is in A307371.
From Bernard Schott, May 01 2019: (Start)
There are two nontrivial families in this sequence:
1st: 21, 215443, 2154434689, 2154434690, 21544346900318, ...
2nd: 463, 464, 4641588, 46415888335, 46415888336, ... (End)
From Jon E. Schoenfield, May 04 2019: (Start)
For each number k such that the digits of k^(1/m) begin with k, we have, for each m >= 2, floor(k^(1/m) * 10^d) = k for some integer d, so k^(1/m) * 10^d ~= k; solving for k gives k ~= 10^(d*m/(m1)).
In the m=4 case (this sequence), this gives k ~= 10^(d*4/3) so, as d is incremented by 1, 10^(d*4/3) increases by a factor of 10^(4/3) = 10000^1/3 = 21.5443469...:
.
d  10^(d*4/3)
+
0  1
1  21.544...
2  464.158...
3  10000
4  215443.469...
5  4641588.833...
6  100000000
7  2154434690.031...
8  46415888336.127...
9  1000000000000
.
Each nonnegative integer d corresponds to one or two terms in the sequence. Letting j = floor(10000^(d/3)), j is necessarily a term; j1 is also a term iff (j1)^(1/4)*10^d < j. This inequality is satisified
for d == 1 (mod 3) at d = 7, 13, 16, 34, 37, ...;
for d == 2 (mod 3) at d = 2, 8, 20, 29, 32, 35, ...;
and at every d == 0 (mod 3).
(The sequence contains no other terms than numbers k of the form j or j1 where j = floor(10000^(d/3)) for some nonnegative integer d.)
(End)
