%I #17 Aug 27 2021 02:00:20
%S 8,27,343,512,19683,79507,103823,110592,140608,148877,250047,314432,
%T 778688,3869893,5088448,6539203,7077888,18191447,54010152,67917312,
%U 75686967,96071912,102503232,109215352,115501303,146363183,202262003,224755712
%N Exclusionary cubes: cubes of the terms in A112994.
%C b-file is complete: there are 42 terms. - _Michael S. Branicky_, Aug 27 2021
%D H. Ibstedt, Solution to Problem 2623, "Exclusionary Powers", pp. 346-9, Journal of Recreational Mathematics, Vol. 32 No.4 2003-4, Baywood NY.
%D Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
%H N. J. A. Sloane, <a href="/A112993/b112993.txt">Table of n, a(n) for n = 1..42</a> [From the Clifford Pickover link. Conjectured to be the full list of terms.]
%H Clifford A. Pickover, <a href="http://sprott.physics.wisc.edu/Pickover/extremec.html">Extreme Challenges in Mathematics and Morals</a>
%o (Python)
%o def ok(n):
%o s = str(n)
%o return len(s) == len(set(s)) and set(s) & set(str(n**3)) == set()
%o print([k**3 for k in range(7659) if ok(k)]) # _Michael S. Branicky_, Aug 27 2021
%o (Python) # version for verifying full sequence
%o from itertools import permutations
%o def no_repeated_digits():
%o for d in range(1, 11):
%o for p in permutations("0123456789", d):
%o if p[0] == '0': continue
%o yield int("".join(p))
%o def afull():
%o alst = []
%o for k in no_repeated_digits():
%o if set(str(k)) & set(str(k**3)) == set():
%o alst.append(k**3)
%o return alst
%o print(afull()) # _Michael S. Branicky_, Aug 27 2021
%Y Cf. A112994.
%K nonn,base,fini,full
%O 1,1
%A _Lekraj Beedassy_, Oct 13 2005
%E Corrected by _N. J. A. Sloane_, May 22 2008