# continuation of sequence

Sat May 24 00:54:01 CEST 2003

```Joshua Zucker <Joshua_Zucker at castilleja.org> schrieb am 23.05.2003,
07:25:46:
> OK, SOme quick hacking with mathematica extends the sequence at
> http://primes.utm.edu/curios/page.php?short=539633
>
> For the n, the base, I have
> 12, 88, 207
> and for the prime "image" of n, sum of p^n over all prime factors p of n.
> 539633 = 2^12 + 2^12 + 3^12
>
> 43909277783870034878569768760415886733743786946105343887995366054267119200102384004474562849
> = 2^88 + 2^88 + 2^88 + 11^88
>
> 7545048844883559926134754437031975993054726674236856637164268202580382602383411898370028958674896550375094045606979132820374507241375063026190221072559862927227552429910707339260761215523069351700952640157359769228765406964459882330366607234722984341
> 91295086941047447198115204429821 = 3^207 + 3^207 + 23^207
>
> I haven't done any verification that these latter numbers are prime except
> Mathematica's PrimeQ function.
>
> And my algorithm was a VERY unsophisticated brute force search.  I'm sure
> that anyone with a bit of intelligence can speed up the algorithm enough
> to search some larger n.  I've checked up to n = 400 and found only these.
>
> --Joshua Zucker
> Castilleja School
> joshua.zucker at stanfordalumni.org

Joshua, Neil, SeqFans,

all the terms given above are prime. (Checked with
Dario Alperns ECM Java Applet
attempts up to n=638, without finding more terms,
but I couldn't check the
following list of numbers, most of them of the form
2*prime. Beginning with n=446=2*223=2*p,
2^n+(n/2)^n has more than 1000 decimal digits,
exceeding my available prime testing tools'
capabilities.
446,454,458,466,478,481,502,508,514,524,526,538,
542,548,554,556,562,566,568,579,584,586,596,603,
604,614,618,622,626,628,632,634

Anyone out there to check primality of sum p_i^n
for this list of numbers?

Hugo Pfoertner

```