# Bell numbers

David W. Wilson wilson at aprisma.com
Fri Jun 29 16:38:14 CEST 2001

```Olivier Gerard wrote:
>
> Jeffrey and seqfans,
>
> it was on Number-Theory:
>
> Delivery-date: Thu, 28 Jun 2001 14:54:18 +0200
> X-From_: owner-nmbrthry at LISTSERV.NODAK.EDU Thu Jun 28 14:51:57 2001
> Reply-To: Dan Fux <danfux at my-deja.com>
> From: Dan Fux <danfux at my-deja.com>
> Subject:      Bell numbers
> To: NMBRTHRY at LISTSERV.NODAK.EDU
>
> Ahmed Fares asked me to post this :
>
> I posted recently a new sequence to the EIS of Neil Sloane , here is
> the description:
>
> ID Number: A061462
> 1,1,2,1,1,4,1,1,4,1,1,2,1,1,2,1,1,4,...
> The exact power of 2 that divides the n-th Bell number (sequence A000110) .
> It looks like a(n) = 1 if n = 0 or 1 mod 3 and that a pattern exists also
> for n = 2 mod 3 but I don't have a formula yet .
>
> Can someone help about the formula for this sequence ?
>
> Thanks,Ahmed

I computed { Bell(n) mod 8 } for quite a distance.  It is obvious that the
sequence is periodic with period 24, the period being

(1 1 2 5 7 4 3 5 4 3 7 2 5 5 2 1 3 4 7 1 4 7 3 2)

I assume this follows (with a little work) from the binomial recursion for
Bell(n).

Given that this is true, none of the Bells is divisible by 8, hence the
highest power of 2 dividing a Bell number is 4.  So the highest power of 2
dividing Bell(n) is the highest power of 2 dividing { Bell(n) mod 8 },
which is the following sequence with period 12:

(1 1 2 1 1 4 1 1 4 1 1 2).

Empirical tests suggest that if k != 0 (mod 8), then k divides some Bell(n).

```