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.

A328095 Revenant numbers: numbers k such that k multiplied by the product of all its digits contains k as a substring.

Original entry on oeis.org

0, 1, 5, 6, 11, 25, 52, 77, 87, 111, 125, 152, 215, 251, 375, 376, 455, 512, 521, 545, 554, 736, 792, 1111, 1125, 1152, 1215, 1251, 1455, 1512, 1521, 1545, 1554, 2115, 2151, 2174, 2255, 2511, 2525, 2552, 4155, 4515, 4551, 5112, 5121, 5145, 5154, 5211, 5225, 5252, 5415, 5451, 5514, 5522, 5541, 5558, 5585, 5855, 8555, 8772, 9375
Offset: 1

Views

Author

N. J. A. Sloane, Oct 19 2019

Keywords

Comments

Sequence is infinite since 11...1 is always a member.
Numbers whose product of digits is a power of ten (and thus necessarily must only have 1,2,4,5,8 as digits) is a subsequence. - Chai Wah Wu, Oct 19 2019

Examples

			87 * 8 * 7 = 4872. As the string 87 is visible in the result, 87 is a revenant.
So is 792 because 792 * 7 * 9 * 2 = 99792.
And so is 9375 as 9375 * 9 * 3 * 7 * 5 = 8859375.
		

References

  • Eric Angelini, Posting to Sequence Fans Mailing List, Oct 19 2019

Crossrefs

Subsequences are: A328544, A328560, A328561.

Programs

  • Maple
    a:= proc(n) option remember; local k; if n=1 then 0 else
          for k from 1+a(n-1) while searchtext(cat(k), cat(k*
          mul(i, i=convert(k, base, 10))))=0 do od: k fi
        end:
    seq(a(n), n=1..75);  # Alois P. Heinz, Oct 19 2019
  • Mathematica
    Select[Range[0,10000],SequenceCount[IntegerDigits[#*(Times@@IntegerDigits[ #])],IntegerDigits[#]]>0&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 19 2019 *)
  • PARI
    is_A328095(n)={my(d,m); if(d=vecprod(digits(n))*n, m=10^logint(n, 10)*10; until(n>d\=10,d%m==n && return(1)),!n)} \\ M. F. Hasler, Oct 20 2019
  • Python
    from functools import reduce
    from operator import mul
    n, A328095_list = 0, []
    while len(A328095_list) < 10000:
        sn = str(n)
        if  sn in str(n*reduce(mul,(int(d) for d in sn))):
            A328095_list.append(n)
        n += 1 # Chai Wah Wu, Oct 19 2019
    

Formula