A382056 Remove every copy of the largest digit of n; if any digits remain, return the number formed by arranging the remaining digits in nondecreasing order. If no digits remain, return 0.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 2, 2, 2, 2, 2, 2, 2, 0, 1, 2, 0, 3, 3, 3, 3, 3, 3, 0, 1, 2, 3, 0, 4, 4, 4, 4, 4, 0, 1, 2, 3, 4, 0, 5, 5, 5, 5, 0, 1, 2, 3, 4, 5, 0, 6, 6, 6, 0, 1, 2, 3, 4, 5, 6, 0, 7, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0

Offset

1,23

Comments

This sequence is inspired by a joke from Finnish comedian ISMO: "The hardest part in English for me is silent letters. You write them down but never say. But since you have silent letters in the language, why don't you also have silent numbers? They would be far more useful. Like, I owe you 75 dollars, but the 7 is silent."

Links

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Formula

a(n) = 0 for n in A125289. - M. F. Hasler, Mar 14 2025

Example

For n = 132, we remove the largest digit and we get a(132) = 12. For n = 66, we remove the largest digit and we get a(66) = 0.

Maple

a:= n-> (l-> (m-> (h-> add(h[j]*10^(j-1), j=1..nops(h))
     )(subs(m=NULL, l)))(max(l)))(convert(n, base, 10)):
seq(a(n), n=1..102);  # Alois P. Heinz, Mar 14 2025

Mathematica

A382056[n_] := FromDigits[DeleteCases[#, Max[#]]] & [IntegerDigits[n]];
Array[A382056, 100] (* Paolo Xausa, Mar 23 2025 *)

Prog

(Python)
def a(n):
    s = str(n)
    m = max(s)
    r = s.replace(m, "")
    return int(r) if r != "" else 0
print([a(n) for n in range(1, 103)]) # Michael S. Branicky, Mar 13 2025
(PARI) apply( {A382056(n, m=vecmax(digits(n)))=fromdigits([d | d<-digits(n), d<m])}, [1..99]) \\ M. F. Hasler, Mar 14 2025

Crossrefs

Cf. A054055 (largest digit of n), A125289 (numbers with only one (distinct) digit > 0), A382401.

Sequence in context: A376680 A323356 A382102 * A319244 A309286 A102671

Adjacent sequences: A382053 A382054 A382055 * A382057 A382058 A382059

Keyword

nonn,look,base,easy

Author

Ali Sada, Mar 13 2025

Extensions

Definition clarified by N. J. A. Sloane, Mar 23 2025

Status

approved