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

Author

Rémy Sigrist, Mar 23 2025

Keywords

• nonn
• base
• look

Comments

A variant of A382102.

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

Crossrefs

Cf. A010785, A382102.

Programs

Mathematica

A382371[n_] := Max[Table[FromDigits[DeleteCases[#, d]], {d, DeleteDuplicates[#]}] & [IntegerDigits[n]]];
Array[A382371, 100] (* Paolo Xausa, Mar 23 2025 *)

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

Formula

a(n) >= A382102(n).

a(n) = 0 iff n belongs to A010785.