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%I #8 Jan 18 2020 18:55:29
%S 0,1,17,315,316,5623,99999,100000,1778279,31622776,562341324,
%T 562341325,9999999999,10000000000,177827941003,3162277660168,
%U 56234132519034,999999999999999,1000000000000000,17782794100389227,17782794100389228,316227766016837932,316227766016837933
%N Numbers k such that the digits of k^(1/5) begin with k.
%C 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.
%H Chai Wah Wu, <a href="/A331492/b331492.txt">Table of n, a(n) for n = 1..1140</a>
%e 5623^(1/5) = 5.6233305990931... which starts with the digits 5623, so 5623 is in the sequence.
%Y Cf. A307371, A307588, A307600.
%K nonn,base
%O 1,3
%A _Chai Wah Wu_, Jan 18 2020
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