

A307588


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


9



0, 1, 31, 999, 1000, 31622, 999999, 1000000, 31622776, 999999999, 1000000000, 31622776601, 999999999999, 1000000000000, 31622776601683, 999999999999999, 1000000000000000, 31622776601683792, 31622776601683793, 999999999999999999, 1000000000000000000
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OFFSET

1,3


COMMENTS

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


LINKS



EXAMPLE

31622^(1/3) = 31.62251..., which begins with "31622", so 31622 is in the sequence.
The seeming pattern a(3k) = floor(10^(3k3/2)), 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


CROSSREFS

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


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



