- --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

Aurelius [or somebody] he say:

> --- In primenumbers@yahoogroups.com, "Kevin Acres" <research@> wrote:

>

> > > Definition: Let F(n) = ((5^n-9)/4)^2-5 for integer n > 0.

> > > Exercise 2: For odd n > 1, prove that F(n)/4 is composite.

> >

> > I did work out an algorithm to derive one divisor of F(n)/4 for odd n, but

> > that's probably a long way from proving it composite.

>

> What is needed is a formula: F(n)=L(n)*M(n). Then it suffices

> to show that each of L(n) and M(n) has an odd prime divisor

> for odd n > 1. Here is a clue:

>

> 4*F(11) = 24414063^2 - 15625^2

>

16*F(n) = 5^(2*n)-18*5^n+1=

(5^n-2*5^[(n+1)/2]+1) * (5^n-2*5^[(n+1)/2]+1)

Mike ---In primenumbers@yahoogroups.com, <thefatphil@...> wrote:

> You only chose that target after

> the arrow had landed, I'm sure.It happened thus:

1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for

n <= 300, completely. As later shown in "factordb", I succeeded.2) Meanwhile I ran OpenPFGW on n in [301,600], hoping for a

quick outlier and found none.3) I estimated the probability of an easily discoverable

completely factorization for n>600 and found it to be small.4) Recalling how I had once been caught out before by

a "probably no more" heuristic, I set a lone process running on

n in [601, 10000] so as not to be caught out again by Jens.5) When I later looked and pfgw.log, it had found a hit at

n=608.So yes, Phil, you are quite correct that the puzzle was set

after this finding. However the heuristic that I gave was

made prior to my discovery, else I would not have said that

I was surprised.The point that you are making (I think) is that I do such

expsriments often and only notice when the result is unexpected.

I don't tell folk about all the boring times when a negative

heuristic is borne out by a null result. That is the selectioneffect.

David

`(guilty of not boring folk with what is routine)`