A055098 Number of distinct anagrams of digits of n without leading zeros.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4

Offset: 1

Author

Henry Bottomley, Apr 19 2000

Keywords

• base
• easy
• nice
• nonn

Examples

			a(101)=2 since the digits of 101 can be ordered 101 or 110 (but not 011).
		

Links

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

Crossrefs

Cf. A046810, A047726.

Programs

Haskell

import Data.List (permutations, nub)
a055098 n = length $ nub $ filter ((> '0') . head) $ permutations $ show n
-- Reinhard Zumkeller, Aug 14 2011

Mathematica

a[n_] := Length[ DeleteCases[ Permutations[ IntegerDigits[n]], {0 .., }]]; Table[a[n], {n, 1, 102}] (* Jean-François Alcover, Nov 30 2011 *)

PARI

a(n)={my(v=digits(n), f=vector(10), n=#v); for(i=1, #v, f[1+v[i]]++); (1 - f[1]/n) * n! / prod(i=1, #f, f[i]!)} \\ Andrew Howroyd, Jan 27 2020

Python

from math import factorial, prod
def a(n):
    s = str(n); d, c = len(s), [s.count(str(i)) for i in range(10)]
    return (d-c[0])*factorial(d-1)//prod(map(factorial, c))
print([a(n) for n in range(1, 50)]) # Michael S. Branicky, Aug 24 2022

Formula

a(n) = O(n/(log n)^(9/2)). - Charles R Greathouse IV, Aug 24 2022