A323416 a(n) = (n-1)! * (10^n - 1) / 9.

1, 11, 222, 6666, 266664, 13333320, 799999920, 55999999440, 4479999995520, 403199999959680, 40319999999596800, 4435199999995564800, 532223999999946777600, 69189119999999308108800, 9686476799999990313523200, 1452971519999999854702848000, 232475443199999997675245568000, 39520825343999999960479174656000

Offset: 1

Author

David Cobac, Jan 13 2019

Keywords

• nonn
• easy
• base

Comments

Take an n-digit number with distinct digits, add all permutations of the digits, divide by the sum of the digits: the result is a(n).

Proof from David A. Corneth, Jan 14 2019: (Start)

Let m be an n-digit number (without leading 0, where n > 0). Then n! permutations of digits can be formed.

So each digit occurs n!/n = (n-1)! times in each position. Therefore the total sum is (10^n - 1) * (n - 1)! * s where s is the sum of digits of n. Dividing this product by s gives a(n) = (10^n - 1) * (n - 1)!. QED (End)

Examples

			Example for n = 3:
Take the number 569.
Sum the permutations of its digits: 569 + 596 + 659 + 695 + 956 + 965 = 4440.
Add all its digits: 5 + 6 + 9 = 20.
Divide: 4440 / 20 = 222.
General proof for n = 3:
Number: abc where a,b,c are distinct.
The sum of the permutations is 200*(a+b+c) + 20*(a+b+c) + 2*(a+b+c) = 222*(a+b+c), so a(3) = 222.
		

Links

• David Cobac, Table of n, a(n) for n = 1..325
• G. Villemin, Nombres permutés

Crossrefs

Cf. A000142, A002275, A071267. Sum of digits A110728.

Programs

Mathematica

Table[(n-1)! (10^n-1)/9,{n,20}] (* Harvey P. Dale, Mar 15 2024 *)

PARI

a(n) = (10^n - 1) / 9 * (n-1)! \\ David A. Corneth, Jan 13 2019

Python

f = lambda n:+(n==0) or n*f(n-1)
def seq(n):
   if n==0: return
   l = []
   for i in range(1, n + 1):
       # following line with a string repeat
       # s = int('1'*i)
       s = 0
       for j in range(i):
           s += 10 ** j
       l += [s*f(i-1)]
   return l

Formula

Recurrence relation: a(n+1) = n! * 10^n + n * a(n).

Proof: Assume R_n is a string of n 1's (repunit),

a(n) = (n-1)! * R_n so a(n+1) = n! * R_{n+1} = n! * (10^n + R_n);

Thus a(n+1) = n! * 10^n + n! * R_n = n! * 10^n + n * (n-1)! * R_n;

Hence a(n+1) = n! * 10^n + n * a(n).

Extensions

Edited by N. J. A. Sloane, Jan 19 2019