A052423 Highest common factor of nonzero digits of n.

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

Offset: 1

Author

Henry Bottomley, Mar 17 2000

Keywords

• base
• easy
• nonn

Examples

			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.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007954.

Programs

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
-- Reinhard Zumkeller, Apr 14 2014

Maple

a:= n-> igcd(subs(0=[][], convert(n, base, 10))[]):
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 04 2020

Mathematica

Table[Apply[GCD, IntegerDigits[n]], {n, 100}] (* Alonso del Arte, Apr 02 2020 *)

PARI

a(n) = my(d=digits(n)); gcd(select(x->(x!=0), d)); \\ Michel Marcus, Apr 04 2020

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
(1 to 100).map(digitGCD()) // _Alonso del Arte, Apr 02 2020

Formula

a(A069715(n)) = 1. - Reinhard Zumkeller, Apr 14 2014