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

1,12

Links

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

Formula

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

Example

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

Mathematica

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

Prog

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

Crossrefs

Cf. A046810, A047726.

Sequence in context: A138471 A102669 A248327 * A297035 A055178 A174847

Adjacent sequences: A055095 A055096 A055097 * A055099 A055100 A055101

Keyword

base,easy,nice,nonn

Author

Henry Bottomley, Apr 19 2000

Status

approved