OFFSET
1,1
COMMENTS
A number k with no repeated digits has an exclusionary cube k^3 if the latter is made up of digits not appearing in k. (This is a subsequence of A029785.) For the corresponding exclusionary cubes see A112993. Conjectured to be complete.
Data are complete: there are 42 terms. - Michael S. Branicky, Aug 27 2021
REFERENCES
H. Ibstedt, Solution to Problem 2623, "Exclusionary Powers", pp. 346-9, Journal of Recreational Mathematics, vol. 32 No.4 2003-4, Baywood NY.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
LINKS
Clifford A. Pickover, Extreme Challenges in Mathematics and Morals
MATHEMATICA
Select[Range[8000], Max[DigitCount[#]]==1&&Intersection[IntegerDigits[ #], IntegerDigits[#^3]]=={}&] (* Harvey P. Dale, Sep 06 2021 *)
PROG
(PARI) isok(n) = my(digs = digits(n)); (#digs == #Set(digs)) && (#setintersect(Set(digs), Set(digits(n^3))) == 0); \\ Michel Marcus, Oct 26 2013
(Python)
def ok(n):
s = str(n)
return len(s) == len(set(s)) and set(s) & set(str(n**3)) == set()
print([k for k in range(7659) if ok(k)]) # Michael S. Branicky, Aug 27 2021
(Python) # version for verifying full sequence
from itertools import permutations
def no_repeated_digits():
for d in range(1, 11):
for p in permutations("0123456789", d):
if p[0] == '0': continue
yield int("".join(p))
def afull():
alst = []
for k in no_repeated_digits():
if set(str(k)) & set(str(k**3)) == set():
alst.append(k)
return alst
print(afull()) # Michael S. Branicky, Aug 27 2021
CROSSREFS
KEYWORD
nonn,base,fini,full
AUTHOR
Lekraj Beedassy, Oct 13 2005
EXTENSIONS
Missing term 468 added by N. J. A. Sloane, May 22 2008
Definition clarified by Harvey P. Dale, Sep 06 2021
STATUS
approved