## n^2+1 prime k-tuplets

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• Mike,you are right. I had very few time and miscalculated. g(10)=30 because we always can find the g(12)=38 ---------- 66,.....,104 (nod 170) incuding the
Message 1 of 1 , Nov 2, 2004
Mike,you are right.
I had very few time and miscalculated.
g(10)=30
because we always can find the g(12)=38 ---------->66,.....,104 (nod 170)
incuding the 0¨s,of course. And they fit with 106,......,144 (mod 130)
or 116,......,154 (mod 130)

So g(10)=30
g(11)=34
g(12)=38
g(13)=42
g(14)=46
g(15)=50
g(16)=54

Sometimes,like Fermat,I make silly errors (almost all the Fermat primes are composite............)
Hope this is not again the case.

There is a typo in my previous post :must be--->

29 divides n^2+1 for n=46,104,186,244 (mod 290)
37 ---------------------------n=6,154,364 (mod 370)

"If the world was only represented in black and white ,it would seem grey
for us.So we´ll build a grey world" (Bill Gates,again......)

Happy binary day.

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