A112994 Numbers whose cubes are exclusionary: numbers k such that k has no repeated digits and k and k^3 have no digits in common.
2, 3, 7, 8, 27, 43, 47, 48, 52, 53, 63, 68, 92, 157, 172, 187, 192, 263, 378, 408, 423, 458, 468, 478, 487, 527, 587, 608, 648, 692, 823, 843, 918, 1457, 1587, 1592, 4657, 4732, 5692, 6058, 6378, 7658
Offset: 1
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
Programs
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Mathematica
Select[Range[8000],Max[DigitCount[#]]==1&&Intersection[IntegerDigits[ #],IntegerDigits[#^3]]=={}&] (* Harvey P. Dale, Sep 06 2021 *)
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PARI
isok(n) = my(digs = digits(n)); (#digs == #Set(digs)) && (#setintersect(Set(digs), Set(digits(n^3))) == 0); \\ Michel Marcus, Oct 26 2013
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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
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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
Extensions
Missing term 468 added by N. J. A. Sloane, May 22 2008
Definition clarified by Harvey P. Dale, Sep 06 2021
Comments