cp's OEIS Frontend

A069716 Smallest number such that the LCM of the digits equals n, or 0 if no such number exists.

%I A069716 #19 May 16 2024 08:41:51
%S A069716 1,2,3,4,5,6,7,8,9,25,0,34,0,27,35,0,0,29,0,45,37,0,0,38,0,0,0,47,0,
%T A069716 56,0,0,0,0,57,49,0,0,0,58,0,67,0,0,59,0,0,0,0,0,0,0,0,0,0,78,0,0,0,
%U A069716 345,0,0,79,0,0,0,0,0,0,257,0,89,0,0,0,0,0,0,0,0,0,0,0,347,0,0,0,0,0,259,0,0
%N A069716 Smallest number such that the LCM of the digits equals n, or 0 if no such number exists.
%C A069716 If n is a prime with more than one digit, a(n) = 0. - _Alonso del Arte_, Dec 20 2015
%C A069716 More generally, if prime p >= 11 divides n then a(n) = 0, if 7^2 | n or 5^2 | n or 3^3 | n or 2^4 | n, then a(n) = 0. Consequently, a(n) = 0 for all n > 2520. This arises naturally by noting lcm{1,2,...,9} = 2520. - _Sean A. Irvine_, May 15 2024
%H A069716 Sean A. Irvine, <a href="/A069716/b069716.txt">Table of n, a(n) for n = 1..2520</a> (includes all nonzero terms)
%e A069716 a(20) = 45 because lcm(4, 5) = 20. If one solution exists, then an infinite number of solutions exist. For n = 20, e.g., 455, 445555555, 545544 etc. are also solutions.
%t A069716 digLCMSeek[x_] := Apply[LCM, IntegerDigits[x]]; A069716 = Table[0, {256}]; Do[s = digLCMSeek[n]; If[s < 257 && A069716[[s]] == 0, A069716[[s]] = n], {n, 10000}]; A069716
%Y A069716 Cf. A068189, A069716.
%K A069716 easy,nonn,base
%O A069716 1,2
%A A069716 _Labos Elemer_, Apr 02 2002