A107846 Number of duplicate digits of n.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0

Offset: 0

Author

Rick L. Shepherd, May 24 2005

Keywords

• base
• easy
• nonn

Comments

a(A010784(n)) = 0; a(A109303(n)) > 0. - Reinhard Zumkeller, Jul 09 2013

Examples

			a(11) = 1 because 11 has two total decimal digits but only one distinct digit (1) and 2-1=1.
Similarly, a(3653135) = 7 (total digits) - 4 (distinct digits: 1,3,5,6) = 3 (There are three duplicate digits here, namely, 3, 3 and 5).
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A055642 (Total decimal digits of n), A043537 (Distinct decimal digits of n).

Programs

Haskell

import Data.List (sort, group)
a107846 = length . concatMap tail . group . sort . show :: Integer -> Int
-- Reinhard Zumkeller, Jul 09 2013

Mathematica

Table[Total[Select[DigitCount[n]-1,#>0&]],{n,0,120}] (* Harvey P. Dale, Jul 31 2013 *)

Python

def a(n): return len(s:=str(n)) - len(set(s))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jan 09 2023

Formula

a(n) = A055642(n) - A043537(n).