A112993 Exclusionary cubes: cubes of the terms in A112994.

8, 27, 343, 512, 19683, 79507, 103823, 110592, 140608, 148877, 250047, 314432, 778688, 3869893, 5088448, 6539203, 7077888, 18191447, 54010152, 67917312, 75686967, 96071912, 102503232, 109215352, 115501303, 146363183, 202262003, 224755712

Offset: 1

Author

Lekraj Beedassy, Oct 13 2005

Keywords

• nonn
• base
• fini
• full

Comments

b-file is 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

• N. J. A. Sloane, Table of n, a(n) for n = 1..42 [From the Clifford Pickover link. Conjectured to be the full list of terms.]
• Clifford A. Pickover, Extreme Challenges in Mathematics and Morals

Crossrefs

Cf. A112994.

Programs

Python

def ok(n):
    s = str(n)
    return len(s) == len(set(s)) and set(s) & set(str(n**3)) == set()
print([k**3 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**3)
    return alst
print(afull()) # Michael S. Branicky, Aug 27 2021

Extensions

Corrected by N. J. A. Sloane, May 22 2008