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Program is in A307371.

The subsequence {31, 31622, 31622776, 31622776601, 31622776601683, ...} looks like this subsequence of A052210 {32, 31623, 316228, 3162278, 31622777, ..., 316227766016838, ...}. - Bernard Schott, May 04 2019

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/(m-1)).

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; j-1 is also a term iff (j-1)^(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 j-1 where j = floor(10000^(d/3)) for some nonnegative integer d.)

(End)

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.