[seqfan] Re: A100083

[David Wilson]

As I remarked in an earlier post, A100083 includes the numbers n that
satisfy

	!n == 2 (mod n)

where !n = A003422(n) = 0! + 1! + ... + n!.

Suppose for a moment that there were a number F which was the sum of all the
factorials:

	F = 0! + 1! + 2! + ...

This F clearly cannot be an integer, but let's overlook this for the moment.

By the definition of !n, we have

	F = !n + (n+1)! + (n+2)! + (n+3)! + ...

On the right side, the all the terms after the first are divisible by n, so
again closing our eyes, we conclude

	F == !n (mod n)

So we could describe A100083 as the numbers n satisfying

	F == 2 (mod n).

As I said, F is not an integer, but we can compute F in the p-adics by
summing the factorials.
For example, in the 2-adics, we get

	F = ...1000011010.

And we can likewise compute p-adic F for any prime p.

Returning to the 2-adics, we compute

	F - 2 = ...1000011000.

We might interpret the last three zeros to mean that 2^3 is the highest
power of 2 dividing F.

If we similarly compute F - 2 in the p-adics for other prime p < 1000, we
find that F - 2 ends in a single zero for p = 31 and p = 373 and does not
end in zero for any other p.

If looks as if we could interpret this to meain that F has the divisors 2^3,
31^1 and 373^1, and that A100083 is listing the divisors of F - 2.

To me, it looks as if there is a generalization of the integers to a broader
class of numbers (like F) that have p-adic representations for all primes p.
This latter class of numbers seems to have some interesting divisibility
properties.
Since I have only a very tenuous grasp of p-adic theory, I have no idea how
to develop this idea.

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of David
> Wilson
> Sent: Friday, June 14, 2013 10:39 PM
> To: 'Sequence Fanatics Discussion list'
> Subject: [seqfan] Re: A100083
> 
> Let
> 
> 	a_k = {n : !n == k (mod n)}
> 
> where !n = A003422(n).
> 
> Then
> 	a_1 = (1, 3, 9, 11, 33, 99, ...) = A057245
> 	a_2 = (1, 2, 4, 8, 31, 62, 124, 248, ...) = A100083
> 	a_3 = (1, 467, ...),
> 	a_4 = (1, 2, 3, 5, 6, 10, 15, 30, 41, 82, 123, ...)
> 	a_5 = (1, 37, ...),
> 	a_6 = (1, 2, 4, 7, 14, 28, ...)
> 	a_7 = (1, 3, 277, 831, 7717, 23151, ...)
> 	a_8 = (1, 2, ...)
> 	a_9 = (1, 5, 19, 95, 197, 985, 3743, 18715, ...)
> 	a_10 = (1, 2, 3, 4, 5, 6, 8, 9, 12, 13, 16, 18, 24, ...)
> 
> I'm guessing that all a_k share the same nice properties we observed for
a_2
> = A100083:
> 
> 	If n in a_k and  m | n, then m in a_k
> 	If n, m in a_k and (n, m) = 1, then nm in a_k.
> 
> 
> > -----Original Message-----
> > From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> > Giovanni Resta
> > Sent: Wednesday, June 12, 2013 5:10 AM
> > To: Sequence Fanatics Discussion list
> > Subject: [seqfan] Re: FW: A100083
> >
> > On 06/12/2013 03:22 AM, Robert G. Wilson v wrote:
> > > Et al,
> > >
> > > 	YES a(16) does indeed equal 92504. I am now looking for a(17).
> >
> >
> > Charles Greathouse and I already found a(16) yesterday, thanks for re-
> > checking.
> >
> > About a(17), currently I can only say that a(17) > 1.2*10^7.
> >
> > Giovanni
> >
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> 
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