

A114260


Numbers k such that the 4th power of k contains exactly 4 copies of each digit of k.


8



5702631489, 7264103985, 7602314895, 7824061395, 8105793624, 8174035962, 8304269175, 8904623175, 8923670541, 9451360827, 9785261403, 9804753612, 9846032571, 57026314890, 59730829461, 60947591328, 64017823995, 65190218436, 67024081935, 70645192839, 72641039850, 74991208356
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OFFSET

1,1


COMMENTS

First 13 terms of the sequence are also pandigital, i.e., they contain all the 10 digits at least once. This is probably accidental, but quite curious.
If a(n) is in the sequence, then 10*a(n) is also in the sequence. So 10^k*a(n) is also in the sequence for positive integers k. Thus this sequence differs from A365144.  Ray Chandler, Aug 23 2023
Not all terms are pandigital, for example 65190218436, 75932056341 and 83581076421 are not.
For any k ends in, say, 425742 (which has six digits) the last six digits of k^4 are fixed. So if k ends in 425742 then k^4 ends in 425742^2 mod 10^6 = 318096 and so it must have a 3, 1, 8, 0, 9 and 6 none of which are contained in 425742.
Therefore if k ends in 425742 then it must have at least 12 digits. In a search for terms <= 10^11 this eases the search, in a search for terms <= 10^12 this leaves only 6! * 5 / 6 = 600 cases to check instead of 10^6.
An additional idea one might use is that the number of digits of k^4 must be a multiple of 4. I.e. 10^(4*m + 3) <= k^4 <= 10^(4*m + 4) so 10^m * 10^0.75 < m < 10^(m + 1) (all strict inequalities as 10^0.75 (cf. A210522) is irrational) which tells us a bunch about the leading digits of k.
Checking pandigital numbers seperately might ease the search. That way if the union of some k and last q digits of k^4 is all decimal digits one could end the search there. This goes for example for 100426. If a term ends in 100426 then it has all decimal digits. (END)
All terms are divisible by 9. First decimal digit of a term is 5 or larger.  Chai Wah Wu, Feb 27 2024


LINKS



EXAMPLE

5702631489 is a term since its 4th power 1057550783692741389295697108242363408641 contains four 5's, four 7's, four 0's and so on.


PROG

(Python)
from itertools import count, islice
from sympy import integer_nthroot
def A114260_gen(): # generator of terms
for l in count(1):
a = integer_nthroot(10**(4*l1), 4)[0]
if (a9:=a%9):
a += 9a9
for b in range(a, 10**l, 9):
if sorted(str(b)*4)==sorted(str(b**4)):
yield b


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



