cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Lekraj Beedassy, Oct 13 2005

Keywords

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.

Crossrefs

Subsequence of A029785.
The corresponding cubes are in A112993.

Programs

  • Mathematica
    Select[Range[8000],Max[DigitCount[#]]==1&&Intersection[IntegerDigits[ #],IntegerDigits[#^3]]=={}&] (* Harvey P. Dale, Sep 06 2021 *)
  • 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

Extensions

Missing term 468 added by N. J. A. Sloane, May 22 2008
Definition clarified by Harvey P. Dale, Sep 06 2021