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

1,2

Links

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

Formula

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

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))[]):
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 *)

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
-- Reinhard Zumkeller, Apr 14 2014
(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
(PARI) a(n) = my(d=digits(n)); gcd(select(x->(x!=0), d)); \\ Michel Marcus, Apr 04 2020

Crossrefs

Cf. A007954.

Sequence in context: A075877 A133500 A256229 * A126616 A121042 A369529

Adjacent sequences: A052420 A052421 A052422 * A052424 A052425 A052426

Keyword

base,easy,nonn

Author

Henry Bottomley, Mar 17 2000

Status

approved