A307588 Numbers k such that the digits of k^(1/3) begin with k.

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

Offset

1,3

Comments

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

Links

Chai Wah Wu, Table of n, a(n) for n = 1..1172

Example

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

Crossrefs

Cf. A307371, A307600.

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

Sequence in context: A138958 A158675 A154808 * A218285 A138861 A015257

Adjacent sequences: A307585 A307586 A307587 * A307589 A307590 A307591

Keyword

nonn,base

Author

Dmitry Kamenetsky, Apr 17 2019

Extensions

a(10)-a(21) from Jon E. Schoenfield, May 01 2019

Status

approved