Re: How to do quickest

*To*: mathgroup at smc.vnet.net*Subject*: [mg73045] Re: [mg73035] How to do quickest*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Tue, 30 Jan 2007 23:52:38 -0500 (EST)*References*: <epf93g$buf$1@smc.vnet.net> <200701290939.EAA12706@smc.vnet.net> <200701301200.HAA16288@smc.vnet.net>

Artur wrote: > Probably these procedure isn't possible to do much quicker because limit > is time of single PrimeQ checking but mayby somebody is able do some > quickest (these procedure looking for prime numbers of the form > 1+4+4^2+4^3+...4^x (mayby number 5 is only one such number ???): > > a = {}; k = 0; Timing[Do[k = k + 4^x; If[ > PrimeQ[k], Print[k]; AppendTo[a, k]], {x, 0, 10000}]; a] > > PrimeQ is relatively quick but working only up to upper limit, later is > necessary uses NumberTheory packagae and much more slowest procedures. > > BEST WISHES > ARTUR First, you don't need NumberTheory packages, except possibly for minor post-processing. It is not known whether PrimeQ will ever give a false positive, but it cannot give a false negative. So if it claims your number is composite then it is composite. Worst case, you find contenders other than 5 and you have to prove primality. Worry about it when you get one. Second, there are various classes of values that can be skipped. We work with (changing your notation ever so slightly) 1+4+4^2+...+4^n, or Sum[4^j, {j,0,n}]. If n is odd then the total is divisible by 5 (either note the terms pair off as +1 and -1 mod 5, or else observe 1+x+...+x^n is divisible by 1+x for n odd). If n is a multiple of 3 then 3 divides the sum, because each summand is equal to 1 mod 3. Also easy to show 7 divides everyone in this class. Perhaps there is an obvious way to rule out all other contenders. Some seem to factor as (a+1)*(3*a+1). Offhand I don't know if there is an underlying pattern to this. Daniel Lichtblau Wolfram Research

**References**:**Re: NDSolve -- initial conditions***From:*Jens-Peer Kuska <kuska@informatik.uni-leipzig.de>

**How to do quickest***From:*Artur <grafix@csl.pl>