[go: up one dir, main page]

login
A069716
Smallest number such that the LCM of the digits equals n, or 0 if no such number exists.
1
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
1,2
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
LINKS
Sean A. Irvine, Table of n, a(n) for n = 1..2520 (includes all nonzero terms)
EXAMPLE
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.
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
CROSSREFS
Sequence in context: A349279 A347189 A068189 * A095289 A174141 A343403
KEYWORD
easy,nonn,base
AUTHOR
Labos Elemer, Apr 02 2002
STATUS
approved