

A217110


Number of pandigital numbers with n places.


2



0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 179625600, 5568393600, 128432304000, 2458427811840, 41355201888000, 632788296940800, 9008498667168000, 121205358007493760, 1558813928579107200, 19326359087766057600, 232491479092720848000, 2727512837264447527680, 31331281164921975283200, 353549170783043484480000
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OFFSET

1,10


COMMENTS

The number of numbers between 10^(n1) and 10^n which contain all decimal digits 0..9.
The ratio a(n)/(10^n10^(n1)) indicates the relative proportion of pandigital numbers with n places compared to all numbers with n places. Since that ratio converges to the limit 1 for n>infinity this can be expressed for large numbers as follows (in a slightly popular manner): ‘Almost all numbers contain all decimal digits 0..9’.
Example: a(n)/(10^n10^(n1)) = 0.99973439517775... for n = 100; in this case 99.9734...% of all 100digit numbers contain all digits 0..9. Conversely, only the tiny proportion of 0.00026560482224... (<0.03%) lacks one digit. That’s astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n<10 the portion is 100%, and even for numbers with n = 10 or 20 places the relative proportion of numbers which contain not all digits 0..9 is 99.96371...% or 78.52626...%, respectively. The least number of places for which the pandigital numbers hold the majority is 27. Here, the proportion of numbers which contain not all digits is 48.03664...%. So one could bet that a randomly chosen number with >= 27 places contains all digits.


LINKS

Hieronymus Fischer, Table of n, a(n) for n = 1..200


FORMULA

a(n) = 9*9!*S2(n,10), where the S2(n,10) are the Stirling numbers of the second kind (cf. triangle A008277).
Asymptotic behavior:
lim a(n)/10^n = 9/10, for n > infinity.
G.f.: g(x) = 9*9!*x^10/(product_{j=1..10} (1jx)).
E.g.f. g(x) = (9/10) * (e^x  1)^10.


EXAMPLE

a(k) = 0, for k < 10 since there a no pandigital numbers with < 10 places, trivially.
a(10) = 9*9!, since the first digit can be in the range 1..9 and for the following 9 digits there are 9, 8, 7, ..., 1 possible values.


CROSSREFS

A171102, A050278, A011540, A002542, A053283, A217094.
Sequence in context: A206083 A199634 A203987 * A217111 A036471 A206316
Adjacent sequences: A217107 A217108 A217109 * A217111 A217112 A217113


KEYWORD

nonn,base


AUTHOR

Hieronymus Fischer, Feb 13 2013


STATUS

approved



