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
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
Programs
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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
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