## pseudo-free numbers

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• classes of pseudo-free numbers... Let R = k*b^n + (b-1) be just such a number; the multiplier k is an odd natural number; and base b is a prime number,
Message 1 of 5 , Jul 20, 2009
classes of pseudo-free numbers...

Let R = k*b^n + (b-1) be just such a number;
the multiplier 'k' is an odd natural number; and
base 'b' is a prime number, preferably small; and
exponent 'n' is a prime number, preferably large;

if and only if... b^(R-1) == 1 (mod R), then 'R' is
a prime number, and the primality test with respect
to base 'b' will be pseudo-free.

does anyone agree with me or can someone find an ex-
plicit counter-example to refute the idea ???

the first instance is a Proth number, base 'b' = 2

Bill
• no takers ???  the formula s a great find... I invite you to search for a counter-example! ... From: leavemsg1 Subject: [PrimeNumbers]
Message 2 of 5 , Jul 22, 2009
no takers ???  the formula's a great find... I invite you to search for a counter-example!

--- On Mon, 7/20/09, leavemsg1 <leavemsg1@...> wrote:

From: leavemsg1 <leavemsg1@...>
Date: Monday, July 20, 2009, 11:22 AM

classes of pseudo-free numbers...

Let R = k*b^n + (b-1) be just such a number;
the multiplier 'k' is an odd natural number; and
base 'b' is a prime number, preferably small; and
exponent 'n' is a prime number, preferably large;

if and only if... b^(R-1) == 1 (mod R), then 'R' is
a prime number, and the primality test with respect
to base 'b' will be pseudo-free.

does anyone agree with me or can someone find an ex-
plicit counter-example to refute the idea ???

the first instance is a Proth number, base 'b' = 2

Bill

[Non-text portions of this message have been removed]
• ... Unproved assertions of the lack of pseudoprimes for numbers of a special form that get big fast are not particularly interesting. ... Counterexample with
Message 3 of 5 , Jul 22, 2009
> no takers ???  the formula's a great find... I invite you to search for a counter-example!

Unproved assertions of the lack of pseudoprimes for numbers of a special form that get big fast are not particularly interesting.

> the first instance is a Proth number, base 'b' = 2

Counterexample with largest prime exponent and smallish multiplier I've found

801*2^17+(2-1)=104988673=73.673.2137 is a 2-psp

also 262145*2^19+1 is a 2-psp

What's the reason for restricting to prime exponents?

Richard Heylen
• sorry the conjecture was missing one component... let R = k * b^n +(b-1) where k
Message 4 of 5 , Jul 23, 2009
sorry the conjecture was missing one component...

let R = k * b^n +(b-1) where k <= b^n +(b-1), k odd; b is a preferably
small prime and n is prime, but larger than 3; now, the restriction on
k is compatible with the Proth definition.
are there still counter-examples ??? I don't have the computer power
to verify it; Bill

--- On Thu, 7/23/09, Maximilian Hasler <maximilian.hasler@...> wrote:

> From: Maximilian Hasler <maximilian.hasler@...>
> Subject: Re: [PrimeNumbers] Re: pseudo-free numbers
> Date: Thursday, July 23, 2009, 4:38 AM
> Counter-examples [k,n, b^(R-1) mod
> R]
> for b=2 :
> ? bbcheck(2)
> [85, 2, Mod(1, 341)]
> [161, 2, Mod(1, 645)]
> [675, 2, Mod(1, 2701)]
> [705, 2, Mod(1, 2821)]
> [819, 2, Mod(1, 3277)]
> [585, 3, Mod(1, 4681)]
> [825, 3, Mod(1, 6601)]
> [77, 5, Mod(1, 2465)]
> [265, 5, Mod(1, 8481)]
> [495, 5, Mod(1, 15841)]
> [585, 5, Mod(1, 18721)]
> [805, 5, Mod(1, 25761)]
> [65, 7, Mod(1, 8321)]
> [259, 7, Mod(1, 33153)]
> [273, 7, Mod(1, 34945)]
> [801, 17, Mod(1, 104988673)]
>
> ? bbcheck(3) \\ counter-examples for b=3
> [171, 2, Mod(1, 1541)]
> [551, 2, Mod(1, 4961)]
> [57, 3, Mod(1, 1541)]
> [563,  Mod(1, 15203)]
>
> (18:00) gp > bbcheck(5)\\ counter-examples for b=5
> [69, 2, Mod(1, 1729)]
> [321, 2, Mod(1, 8029)]
> [93, 5, Mod(1, 290629)]
>
> (18:01) gp > bbcheck(7)\\ counter-examples for b=7
> [235, 2, Mod(1, 11521)]
> [527, 2, Mod(1, 25829)]
> [521, 3, Mod(1, 178709)]
>
> (18:02) gp > bbcheck(11)\\ counter-examples for b=11
> [463, 2, Mod(1, 56033)]
>
> Maximilian
>
> wrote:
> >
> >  > no takers ???  the formula's a great
> find... I invite you to search for a counter-example!
> >
> >
> > Unproved assertions of the lack of pseudoprimes for
> numbers of a special form that get big fast are not
> particularly interesting.
> >
> >
> >  > the first instance is a Proth number, base
> 'b' = 2
> >
> >
> > Counterexample with largest prime exponent and
> smallish multiplier I've found
> >
> >  801*2^17+(2-1)=104988673=73.673.2137 is a 2-psp
> >
> >  also 262145*2^19+1 is a 2-psp
> >
> >  What's the reason for restricting to prime
> exponents?
> >
> >  Richard Heylen
> >
> >
> >
> >
> >  ------------------------------------
> >
> >  Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> >  The Prime Pages : http://www.primepages.org/
> >
> >
> >
> >
> >
> >
> >
> >
> >
>
• I believe that the following class of numbers is pseudo-free... Let R = k * b^n +(b-1) where k is odd positive, b is a small prime base, n is a larger prime
Message 5 of 5 , Aug 1 7:29 AM
I believe that the following class of numbers is pseudo-free...

Let R = k * b^n +(b-1) where k is odd positive, b is a small prime base,
n is a larger prime number and k is restricted to be <= b^n +(b-1).

when tested against the base 'b', iff b^(R-1) ==1 (mod R), then 'R' is
prime, never a pseudo-prime number.

Bill

can anyone refute this ???
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