|
|
A052423
|
|
Highest common factor of nonzero digits of n.
|
|
6
|
|
|
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 5, 1, 1, 1, 1, 5, 1, 1, 1, 1, 6, 1, 2, 3, 2, 1, 6, 1, 2, 3, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 8, 1, 2, 1, 4, 1, 2, 1, 8, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
a(46) = 2 because the highest common factor of 4 and 6 is 2.
a(47) = 1 because the highest common factor of 4 and 7 is 1.
|
|
MAPLE
|
a:= n-> igcd(subs(0=[][], convert(n, base, 10))[]):
|
|
MATHEMATICA
|
Table[Apply[GCD, IntegerDigits[n]], {n, 100}] (* Alonso del Arte, Apr 02 2020 *)
|
|
PROG
|
(Haskell)
a052423 n = f n n where
f x 1 = 1
f x y | x < 10 = gcd x y
| otherwise = if d == 1 then 1 else f x' (gcd d y)
where (x', d) = divMod x 10
(Scala) def euclGCD(a: Int, b: Int): Int = b match { case 0 => a; case n => Math.abs(euclGCD(b, a % b)) }
def digitGCD(n: Int) = n.toString.toCharArray.map(_ - 48).scanLeft(0)(euclGCD(_, _)).last
(PARI) a(n) = my(d=digits(n)); gcd(select(x->(x!=0), d)); \\ Michel Marcus, Apr 04 2020
|
|
CROSSREFS
|
|
|
KEYWORD
|
base,easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|