## (k*M[n])^2+1

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• Some results from my sojourn into the world of primes of the form P=(k.M[n])^2+1, where M[n]=2*p[1]*p[2]...p[n], where p[] are the pythagorean primes (primes
Message 1 of 1 , May 18, 2006
Some results from my sojourn into the world of primes of the form
P=(k.M[n])^2+1, where M[n]=2*p[1]*p[2]...p[n], where p[] are the
pythagorean primes (primes which are 1mod4); and k is an integer

for k =1, P is prime at M[0], that is P=2^2+1=5
M[5],i.e. P= (1*2*5)^2+1 =101 is prime
M[13], i.e. (1*2*5*13)^2+1 is prime
M[17], M[101],M[181],M[277],M[593],M[641],M[1733] are prime.. all M
checked to M[10009].

I have also checked the first instance primes, in (k.M[n])^2+1 by
increasing the value of k, for all M[n] up to and including n=4993.

As expected, there is no formula, the pattern is a generally
increasing value of k.

Champions, where k is the highest value seen to date are at:
M[n] k
29 6
37 8
41 18
73 47
389 50
397 97
433 103
541 113
617 141
937 154
941 170
1249 200
1297 272
1381 670
2473 875
3457 877
4481 1091

Regards

Robert Smith
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