## Can n Divide !n ?

```For any integer n the "left factorial function" is defined as
!n = 0!+1!+2!+...+(n-1)!.   The conjecture that n does not divide !n
for any n>2 was listed as B44 in R. Guy's "Unsolved Problems in Number
Theory" back in 1981. As of 2015, no proof has been found. I don't
have a proof, but here's a geometrical approach to the problem,
followed by some generalized conjectures.

Consider a p-dimensional Euclidean space with orthogonal coordinates,
and take the two vectors

X = [ 0!, 1!, 2!,..., (p-1)! ]

Y = [ (p-1)!,..., 2!, 1!, 0! ]

The dot product (i.e., the scalar product) of X and Y is

X.Y = (0!)(p-1)! + (1!)(p-2)! + ... + (p-1)!(0!)

= -1 (mod p)

= |X| |Y| cos(alpha)

where alpha is the angle between X and Y.  Since X and Y have the same
components in reverse order, we know that |X| = |Y|, so we have

[ (0!)^2 + (1!)^2 + ... + ((p-1)!)^2 ] cos(alpha) =  -1  (mod p)

This relates the sum of the squares of the factorials 0! to (p-1)! with
the cosine of the angle between X and Y.  Of course, this doesn't prove
that the sum of squares is not divisible by p, it just rephrases the
question in terms of the cosine of the angle between two vectors.

For any positive integer n let ![n,k] denote the sum of the kth powers
of the factorials from 0! to (n-1)!.  For example, the preceding formula
can be written as ![p,2] cos(alpha) = -1 (mod p).

By Fermat's Little Theorem, the sequence of integers ![p,k] (mod p)
for k=0,1,2,... is periodic with a period dividing p-1.  Therefore,
we have
![p,(p-1)m] = 0   (mod p)

for any natural number m.  In general, we have ![p,k]=0 (mod p) if
and only k is of the form  (p-1)m + c  where c is one of a finite
set of values for any given prime p.  The question of whether !p
is ever divisibly by p is equivalent to asking if c can ever be 1.

The values of c for the first few primes are tabulated below:

p    c values        p     c values        p        c values
----  ----------     ----  -----------      ----  ---------------------
2   0               53   0                127   0  57
3   0               59   0  53            131   0 105
5   0  3            61   0  25            137   0  55
7   0               67   0                139   0
11   0               71   0                149   0   4  23  26 122 144
13   0 11            73   0  10  62        151   0
17   0  7            79   0                157   0
19   0               83   0  61            163   0
23   0  6 16         89   0  33            167   0
29   0  9 23         97   0                173   0
31   0              101   0                179   0
37   0 35           103   0  44  58        181   0   6 163 174
41   0              107   0                191   0
43   0              109   0  18  57  90    193   0 139
47   0  9 15        113   0  50  62        197   0

Some additional conjectures arise from the following table:

k       values of p^n such that ![p^n,k] = 0 (mod p^n)
-----    -----------------------------------------------
1       2
2       2, 3
3       2, 5, 25
4       2, 3, 5, 9, 149
5       2
6       2, 3, 7, 23, 181, 443
7       2, 5, 17
8       2, 3, 5, 367
9       2, 29, 47
10       2, 3, 11, 73
11       2, 5, 13
12       2, 3, 5, 7, 13

Note that if a1, a2,... are the highest powers of the primes p1,
p2,... such that (pj)^(aj) divides ![(pj)^(aj),k], then the integers
N that divide ![N,k] are all those of the form

N = (p1)^(b1) (p2)^(b2)...

with bj <= aj.

Hare-brained Conjectures:
1.  ![n,1] is not divisible by n for any n>2. (This is the original)
2.  ![n,5] is not divisible by n for any n>2. (verified for p<1000)
(See update below.)
3.  For any k the number of primes p that divide !(p,k) is finite.
4.  There is at least one odd prime p that divides !(p,k) for each
k except k=1 and k=5. (See update below.)

Update: In May of 2015, Vladica Andrejic and Milos Tatarevic found
that ![n,5] is divisible by n for the prime n = 9632267.
```