OFFSET
1,2
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)
LINKS
David Cobac, Table of n, a(n) for n = 1..325
G. Villemin, Nombres permutés
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).
EXAMPLE
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.
MATHEMATICA
Table[(n-1)! (10^n-1)/9, {n, 20}] (* Harvey P. Dale, Mar 15 2024 *)
PROG
(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
(PARI) a(n) = (10^n - 1) / 9 * (n-1)! \\ David A. Corneth, Jan 13 2019
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
David Cobac, Jan 13 2019
EXTENSIONS
Edited by N. J. A. Sloane, Jan 19 2019
STATUS
approved