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A055098
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Number of distinct anagrams of digits of n without leading zeros.
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5
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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
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,12
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
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FORMULA
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a(n) = O(n/(log n)^(9/2)). - Charles R Greathouse IV, Aug 24 2022
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EXAMPLE
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a(101)=2 since the digits of 101 can be ordered 101 or 110 (but not 011).
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MATHEMATICA
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a[n_] := Length[ DeleteCases[ Permutations[ IntegerDigits[n]], {0 .., __}]]; Table[a[n], {n, 1, 102}] (* Jean-François Alcover, Nov 30 2011 *)
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PROG
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(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
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CROSSREFS
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Cf. A046810, A047726.
Sequence in context: A138471 A102669 A248327 * A297035 A055178 A174847
Adjacent sequences: A055095 A055096 A055097 * A055099 A055100 A055101
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KEYWORD
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base,easy,nice,nonn
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AUTHOR
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Henry Bottomley, Apr 19 2000
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STATUS
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approved
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