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

Author

Labos Elemer, Feb 19 2002

Keywords

• base
• nonn

Comments

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

Examples

			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
		

Links

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

Crossrefs

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

Cf. A085123, A280249.

Programs

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

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