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• Thanks to Jens for showing that t
Message 1 of 4 , Nov 1, 2004
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Thanks to Jens for showing that t <=4 for n=m+10*t+ 0 / 2 (t=0,1,2,3,4)
such that n^2+1 is prime for all t.
And to Mike for finding tha smallest n^2+1 prime k-tuplets.

29 divides n^2+1 for n=45,104,186,244 (mod 290)
37 " n=6,154,564 (mod 370)

And greater 4k+1 primes allow wider consecutive t values.

5 pairs of consecutive t¨s must be 114 or 624 or 1544 (mod 2210) for the first number .(there is another residue) .
Could Mike try to find the least such 5 pair? (I only have a 220 Mhz Cyrix)
If the first number is m=624 (mod 2210) a 16- n^2+1 tuplet could be found:
m-4 +0,4,6,10,14,16,20,24,26,34,36,40,44,46,50,54
If m=1544 (mod 2210) another 16-tuplet
m-8 +0,4,8,10,14,18,20,28,30,34,38,40,44,48,50,54
So it seems that a 10-tuplet is s,....,s+32

Since there are only 9 n=4 or 6 (mod 130) values such that n^2+1 is not divisible by 13 there is at most a k=18 set of closest n^2+1 twin primes.
Combining with 17 divisibility it seems that the largest is k=16 -------->

(m=114 (mod 2210) +0/2 ) +0,10,20,30,40,70,80,90

Check that,please.I hope i didn¨t mess somewhere.

Robin Garcia

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