A217111 Number of pandigital numbers <= 10^n.

0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 182891520, 5751285120, 134183589120, 2592611400960, 43947813288960, 676736110229760, 9685234777397760, 130890592784891520, 1689704521363998720, 21016063609130056320, 253507542701850904320, 2981020379966298432000

Offset: 1

Author

Hieronymus Fischer, Feb 13 2013

Keywords

• nonn
• base

Comments

The number of numbers with <= n digits which contain all decimal digits 0..9.

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".

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.

Partial sums of A217110.

Examples

			a(k) = 0, for k < 10 since there are no pandigital numbers <= 10^9, 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.
		

Links

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

Crossrefs

Cf. A171102, A050278, A011540, A002542, A053283, A217094, A217110.

Programs

Mathematica

3265920 Accumulate[StirlingS2[Range[25],10]] (* Harvey P. Dale, Oct 16 2022 *)

Formula

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).

Asymptotic behavior:

Limit_{n->oo} a(n)/10^n = 1.

G.f.: g(x) = 9*9!*x^10/((1-x)*Product_{j=1..10} (1-jx)).