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Numbers k such that the digits of k^(1/3) begin with k.
9

%I #31 Jan 18 2020 15:33:06

%S 0,1,31,999,1000,31622,999999,1000000,31622776,999999999,1000000000,

%T 31622776601,999999999999,1000000000000,31622776601683,

%U 999999999999999,1000000000000000,31622776601683792,31622776601683793,999999999999999999,1000000000000000000

%N Numbers k such that the digits of k^(1/3) begin with k.

%C Program is in A307371.

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

%H Chai Wah Wu, <a href="/A307588/b307588.txt">Table of n, a(n) for n = 1..1172</a>

%e 31622^(1/3) = 31.62251..., which begins with "31622", so 31622 is in the sequence.

%e The seeming pattern a(3k) = floor(10^(3k-3/2)), a(3k+1) = 10^(3k) - 1, a(3k+2) = 10^(3k), is broken at a(18) = a(19) - 1 = floor(10^(33/2)) - 1. - _Jon E. Schoenfield_, May 01 2019

%Y Cf. A307371, A307600.

%Y Cf. A052210 (analog for 3rd power instead of 1/3).

%K nonn,base

%O 1,3

%A _Dmitry Kamenetsky_, Apr 17 2019

%E a(10)-a(21) from _Jon E. Schoenfield_, May 01 2019