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A217111 Number of pandigital numbers <= 10^n.

%I A217111 #19 Aug 07 2023 05:22:49
%S A217111 0,0,0,0,0,0,0,0,0,3265920,182891520,5751285120,134183589120,
%T A217111 2592611400960,43947813288960,676736110229760,9685234777397760,
%U A217111 130890592784891520,1689704521363998720,21016063609130056320,253507542701850904320,2981020379966298432000
%N A217111 Number of pandigital numbers <= 10^n.
%C A217111 The number of numbers with <= n digits which contain all decimal digits 0..9.
%C A217111 The ratio a(n)/10^n indicates the relative proportion of pandigital numbers <= 10^n compared to all numbers <= 10^n. Since that ratio converges to the limit 1 for n -> oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
%C A217111 Example: a(n)/10^n = 0. 99973107526479... for n = 100; in this case 99.9731...% of all numbers <= 10^100 contain all digits 0..9. Conversely, only the tiny proportion of 0.000268924735210... (< 0.03%) lacks at least 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 <= 10^10 or <= 10^20 the relative proportion of numbers which do not contain all digits 0..9 is 99.96734...% or 78.98393...%, respectively. 10^27 is the least power of 10 such that the pandigital numbers hold the majority. Here, the proportion of pandigital numbers among all numbers <= 10^27 is 51.50961...%. So one could bet that a randomly chosen number <= 10^27 contains all digits.
%C A217111 Partial sums of A217110.
%H A217111 Hieronymus Fischer, <a href="/A217111/b217111.txt">Table of n, a(n) for n = 1..200</a>
%F A217111 a(n) = 9*9!*Sum_{j=1..n} S2(j,10), where the S2(j,10) are the Stirling numbers of the second kind (cf. triangle A008277).
%F A217111 Asymptotic behavior:
%F A217111 Limit_{n->oo} a(n)/10^n = 1.
%F A217111 G.f.: g(x) = 9*9!*x^10/((1-x)*Product_{j=1..10} (1-jx)).
%e A217111 a(k) = 0, for k < 10 since there are no pandigital numbers <= 10^9, trivially.
%e A217111 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.
%t A217111 3265920 Accumulate[StirlingS2[Range[25],10]] (* _Harvey P. Dale_, Oct 16 2022 *)
%Y A217111 Cf. A171102, A050278, A011540, A002542, A053283, A217094, A217110.
%K A217111 nonn,base
%O A217111 1,10
%A A217111 _Hieronymus Fischer_, Feb 13 2013