(2! + 3! + 5! + ... + 53!) = prime * 2^4

[Martin Fuller]

1,2,3,5,6,16,27 and no others.  I think it is not
worth submitting, because there are no big primes and
the sequence is fairly easy to calculate (less than a
week using basic PARI/GP).  But feel free to use the
result if you can find a use for it.

Martin

--- Jonathan Post <jvospost3 at gmail.com> wrote:

> Thank you, Dean Hickerson. That's why it was nagging
> at me.  Tony Noe and I
> had discussed the "n>=762, prime(1)! + ... +
> prime(n)! is divisible by
> prime(763) = 5813" fact years ago, when we were both
> doing things with sums
> of factorials. I agree that it is somewhat
> contrived, and I stopped
> calculating after the sum included 101!, as the
> increasing factorization
> times did not look like a good investment on my
> slow, ancient PC. But, now
> that you correctly advise that the sequence is
> finite, I wonder how many
> solutions there are, and what the largest is. It is
> clearly much smaller
> than any of the huge primes being discovered in the
> past decade. Artificial
> as it is, at least it is not base!
> 
> On 12/17/06, Dean Hickerson <dean at math.ucdavis.edu>
> wrote:
> >
> > Jonathan Post wrote:
> >
> > > Numbers n such that the sum of the first n
> factorials
> > > of primes is a power of 2, or a prime times a
> power of 2.
> > ...
> > > a(n) = 1, 2, 3, 5, 6, 16, ...
> > >
> > > I looked at sums up through 101!
> >
> > I hesitate to comment, since this seems like
> another artifical sequence
> > to me.  However:
> >
> > First note that from 2!+3!+5!+7! on, the power of
> 2 will always be 2^4.
> > So after the first 3 terms, an equivalent
> definition is that the sum
> > has the form 16p for some prime p.  Also, the
> sequence is finite:  For all
> > n>=762, prime(1)! + ... + prime(n)! is divisible
> by prime(763) = 5813,
> > and is therefore not of the form 16p.
> >
> > Dean Hickerson
> > dean at math.ucdavis.edu
> >
>