A068189 Smallest positive number whose product of digits equals n, or a(n)=0 if no such number exists, i.e. when n has a prime divisor greater than 7.

1, 2, 3, 4, 5, 6, 7, 8, 9, 25, 0, 26, 0, 27, 35, 28, 0, 29, 0, 45, 37, 0, 0, 38, 55, 0, 39, 47, 0, 56, 0, 48, 0, 0, 57, 49, 0, 0, 0, 58, 0, 67, 0, 0, 59, 0, 0, 68, 77, 255, 0, 0, 0, 69, 0, 78, 0, 0, 0, 256, 0, 0, 79, 88, 0, 0, 0, 0, 0, 257, 0, 89, 0, 0, 355, 0, 0, 0, 0, 258, 99, 0, 0, 267, 0

Offset

1,2

Comments

a(n) > 0 if and only if n is in A002473.

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Example

n=2,10,50,250 gives a(n)=2,25,255,2555; n=11,39,78, etc..a(n)=0.
10000 = 2 * 5 * 5 * 5 * 5 * 8. No product of two of these factors is less than 10 so a(10000) = 255558 (the concatenation of these factors in nondecreasing order). - David A. Corneth, Jul 31 2017

Mathematica

f[x_] := Apply[Times, IntegerDigits[x]] a = Table[0, {256} ]; Do[ b = f[n]; If[b < 257 && a[[b]] == 0, a[[b]] =n], {n, 1, 10000} ]; a

Prog

(PARI) a(n) = {if(n==1, return(1)); my(res = []); forstep(i=9, 2, -1, v = valuation(n, i); if(v > 0, res = concat(vector(v, j, i), res); n/=i^v)); if(n==1, fromdigits(res), 0)} \\ David A. Corneth, Jul 31 2017
(Python)
def convert(n):
    if n == 1:
        return 1
    result = 0
    cur = 1
    while n > 1:
        found = False
        for i in range(9, 1, -1):
            if n % i == 0:
                result += cur * i
                cur *= 10
                n //= i
                found = True
                break
        if not found:
            return 0
    return result
N = 256
for n in range(1, N):
    print(n, convert(n))
# Dmitry Kamenetsky, Oct 20 2008

Crossrefs

Cf. A001222, A002473, A007954, A067734, A068183-A068187, A068189-A068191.

Cf. A085123, A280249.

Sequence in context: A007532 A349279 A347189 * A069716 A095289 A174141

Adjacent sequences: A068186 A068187 A068188 * A068190 A068191 A068192

Keyword

base,nonn

Author

Labos Elemer, Feb 19 2002

Status

approved