A382371 Remove all occurrences of a digit from n such that the resulting number, formed by the remaining digits in their original order, is as large as possible. If no digits remain, a(n)=0.
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 3, 4, 5, 6, 7, 8, 9, 2, 2, 0, 3, 4, 5, 6, 7, 8, 9, 3, 3, 3, 0, 4, 5, 6, 7, 8, 9, 4, 4, 4, 4, 0, 5, 6, 7, 8, 9, 5, 5, 5, 5, 5, 0, 6, 7, 8, 9, 6, 6, 6, 6, 6, 6, 0, 7, 8, 9, 7, 7, 7, 7, 7, 7, 7, 0, 8, 9, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1
Examples
For n = 95, we remove (all copies of) the digit 5 and we get a(n) = 9. For n = 88, we remove all copies of the digit 8 and we get a(88) = 0. For n = 1917, we remove all copies of the digit 7 and we get a(n) = 191.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
A382371[n_] := Max[Table[FromDigits[DeleteCases[#, d]], {d, DeleteDuplicates[#]}] & [IntegerDigits[n]]]; Array[A382371, 100] (* Paolo Xausa, Mar 23 2025 *)
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PARI
a(n, base = 10) = { my (d = digits(n, base), s = Set(d)); vecmax(apply(r -> fromdigits(select(t -> t!=r, d), base), s)) } -
Python
def f(s, r): return int(s) if (s:=s.replace(r, "")) != "" else 0 def a(n): s = str(n) return max(f(s, d) for d in set(s)) print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Mar 23 2025
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