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 A068189 Smallest positive number whose product of digits equals n, or a(n)=0 if no such number exists, e.g. when n has prime-factor larger than 7. 12
 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 (list; graph; refs; listen; history; text; internal format)
 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: A134703 A061862 A007532 * A069716 A095289 A174141 Adjacent sequences:  A068186 A068187 A068188 * A068190 A068191 A068192 KEYWORD base,nonn AUTHOR Labos Elemer, Feb 19 2002 STATUS approved

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Last modified December 5 06:51 EST 2020. Contains 338944 sequences. (Running on oeis4.)