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

[Jonathan Post]

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
>
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