%I #31 Jan 18 2020 15:31:20
%S 0,1,21,463,464,9999,10000,215443,4641588,99999999,100000000,
%T 2154434689,2154434690,46415888335,46415888336,999999999999,
%U 1000000000000,21544346900318,464158883361277,9999999999999999,10000000000000000,215443469003188371,215443469003188372
%N Numbers k such that the digits of k^(1/4) begin with k.
%C Program is in A307371.
%C From _Bernard Schott_, May 01 2019: (Start)
%C There are two nontrivial families in this sequence:
%C 1st: 21, 215443, 2154434689, 2154434690, 21544346900318, ...
%C 2nd: 463, 464, 4641588, 46415888335, 46415888336, ... (End)
%C From _Jon E. Schoenfield_, May 04 2019: (Start)
%C 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)).
%C 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...:
%C .
%C d | 10^(d*4/3)
%C ---+---------------------
%C 0 | 1
%C 1 | 21.544...
%C 2 | 464.158...
%C 3 | 10000
%C 4 | 215443.469...
%C 5 | 4641588.833...
%C 6 | 100000000
%C 7 | 2154434690.031...
%C 8 | 46415888336.127...
%C 9 | 1000000000000
%C .
%C 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
%C for d == 1 (mod 3) at d = 7, 13, 16, 34, 37, ...;
%C for d == 2 (mod 3) at d = 2, 8, 20, 29, 32, 35, ...;
%C and at every d == 0 (mod 3).
%C (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.)
%C (End)
%H Chai Wah Wu, <a href="/A307600/b307600.txt">Table of n, a(n) for n = 1..1172</a>
%e 215443^(1/4) = 21.544335..., which begins with "215443", so 215443 is in the sequence.
%Y Cf. A307371, A307588.
%Y Cf. A052211 (analog for 4th power instead of 1/4).
%K nonn,base
%O 1,3
%A _Dmitry Kamenetsky_, Apr 17 2019
%E a(12)-a(23) from _Jon E. Schoenfield_, May 01 2019