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

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, 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, 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

Offset: 1

Author

Labos Elemer, Apr 02 2002

Keywords

• easy
• nonn
• base

Comments

If n is a prime with more than one digit, a(n) = 0. - Alonso del Arte, Dec 20 2015

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

Examples

			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.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..2520 (includes all nonzero terms)

Crossrefs

Cf. A068189, A069716.

Programs

Mathematica

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