- --- In primenumbers@yahoogroups.com, <bobb777@...> wrote:

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

Obviously.

> The second factor should be 5^n PLUS 2*5 etc.

> Another schoolboy howler.

No. Just someome typing good maths even faster

than his agile brain was working.

David ---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)`