Re: Carmichael question
- I've found another Carmichael of the form n^2-n-1
It is n=151068195602 so n^2-n-1=22821599722293063946801
It has the 8 factor factorisation
The lcm of all of these factors minus 1 is
2^3.3^3.5^2.7.13 and this of course divides n^2-n-2
Now this should shed some light on the why the Carmichael tends to
end in 1. It's because the prime factors of the Carmichael tend to be
a smooth number plus one. 5 is a very useful small prime for
rendering numbers smooth. From Richard Pinch's website it appears
that only about a tenth of Carmichael numbers in the 10^15 range that
don't have 5 as a factor don't end in a 1.
- Apologies for the aborted attempt at a post. 'Tab' followed by either
'space' or 'return' (too quick to know what happened) seems to be a
lethal key combination.
From: "mcnamara_gio" <mcnamara_gio@...>
>2, 3, and 5 can never be factors. This boosts density by a factor of
> What do you think about this sequence a(n)=n^2+7n-1. Its terms are
> usually prime. I have calculated that 72% of a(n) so that n<500 is
> prime. 81% of a(n) is prime when n<5000. 85.6% of a(n) is prime when
> n<50000 and 88.5% of a(n) is prime when n<500000. I am going to find
> more prime terms in this sequence. What do you think about it?
(2/1)*(3/2)*(5/4) over arbitrary ranges. However, 7,11, 13 and 17 both
divide 2 of the p possible residues. This decreases density by a factor
of (5/6)*(9/10)*(11/12)*(15/16) over arbitrary ranges.
Looking at primes up to 10000, the density boost is almost exactly 2.75.
This is pretty feeble compared with Euler's famous trinomials.
Run this script in Pari/GP:
"rthis" "rnorm" "roots))
Research the Euler trinomials, and try the above script on them too, to see why
I say 2.75 is pretty feeble.
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