A307600 Numbers k such that the digits of k^(1/4) begin with k.

0, 1, 21, 463, 464, 9999, 10000, 215443, 4641588, 99999999, 100000000, 2154434689, 2154434690, 46415888335, 46415888336, 999999999999, 1000000000000, 21544346900318, 464158883361277, 9999999999999999, 10000000000000000, 215443469003188371, 215443469003188372

Offset: 1

Author

Dmitry Kamenetsky, Apr 17 2019

Keywords

• nonn
• base

Comments

Program is in A307371.

From Bernard Schott, May 01 2019: (Start)

There are two nontrivial families in this sequence:

1st: 21, 215443, 2154434689, 2154434690, 21544346900318, ...

2nd: 463, 464, 4641588, 46415888335, 46415888336, ... (End)

From Jon E. Schoenfield, May 04 2019: (Start)

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)).

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...:

.

d | 10^(d*4/3)

---+---------------------

0 | 1

1 | 21.544...

2 | 464.158...

3 | 10000

4 | 215443.469...

5 | 4641588.833...

6 | 100000000

7 | 2154434690.031...

8 | 46415888336.127...

9 | 1000000000000

.

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

for d == 1 (mod 3) at d = 7, 13, 16, 34, 37, ...;

for d == 2 (mod 3) at d = 2, 8, 20, 29, 32, 35, ...;

and at every d == 0 (mod 3).

(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.)

(End)

Examples

			215443^(1/4) = 21.544335..., which begins with "215443", so 215443 is in the sequence.
		

Links

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

Crossrefs

Cf. A307371, A307588.

Cf. A052211 (analog for 4th power instead of 1/4).

Extensions

a(12)-a(23) from Jon E. Schoenfield, May 01 2019