A348834 Zeroless numbers that have the largest possible remainder when divided by their digit product.

1, 2, 3, 4, 5, 6, 7, 8, 9, 96, 969, 987, 9997, 99998, 999999, 9899997, 99999997, 999999998, 9999999999, 98999999998, 999999999998, 9999999999999, 99999999999998, 999999999999999, 9999999999999997, 99999999999999999, 699999999999999999, 9799999999999999997

Offset

1,2

Comments

For the largest remainder of an n-digit number when divided by its product of digits, see A348833.

Starting from 9997, each term appears to have 1 more decimal digit than the previous term. - Chai Wah Wu, Nov 08 2021

Links

Chai Wah Wu, Table of n, a(n) for n = 1..61

Example

1-digit number 3 == 0 mod (3).
2-digit number 96 == 42 mod (9*6).
3-digit numbers 969 == 483 mod (9*6*9) and 987 == 483 mod (9*8*7).
4-digit number 9997 == 4894 mod (9*9*9*7).
5-digit number 99998 == 47510 mod (9*9*9*9*8).

Prog

(Python)
from math import prod
from itertools import product
def auptod(maxdigits):
    alst = []
    for d in range(1, maxdigits+1):
        maxr, argmaxr = 0, []
        for p in product("123456789", repeat=d):
            t = int("".join(p))
            r = t%prod(map(int, p))
            if r == maxr:
                argmaxr.append(t)
            elif r > maxr:
                maxr = r
                argmaxr = [t]
        alst.extend(argmaxr)
    return alst
print(auptod(7)) # Michael S. Branicky, Nov 07 2021

Crossrefs

Cf. A052382, A007602, A348799, A348833.

Sequence in context: A061219 A071271 A066492 * A342978 A281091 A271569

Adjacent sequences: A348831 A348832 A348833 * A348835 A348836 A348837

Keyword

nonn,base

Author

Bernard Schott, Nov 07 2021

Extensions

a(18)-a(21) from Michael S. Branicky, Nov 07 2021

a(22)-a(28) from Chai Wah Wu, Nov 08 2021

Status

approved