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A002144(n+1)^2+1 mod A002144(n), where A002144 are the Pythagorean primes (p=4k+1).
0

%I #22 Sep 03 2024 15:02:57

%S 0,0,0,28,17,39,4,72,79,65,17,65,17,29,145,65,84,65,145,17,109,17,65,

%T 0,145,65,17,145,88,17,64,145,17,28,257,65,17,65,145,145,257,65,17,

%U 269,145,401,257,145,65,257,65,145,17,577,145,65,145,17,577,65,577

%N A002144(n+1)^2+1 mod A002144(n), where A002144 are the Pythagorean primes (p=4k+1).

%C Motivated by the fact that the first terms are zero (which is of course a coincidence). Other values (17, 65, 145, 257...) occur much more frequently.

%C Conjecture: a(n) = A082073(n)^2 + 1 for all n > 159. - _Charles R Greathouse IV_, May 13 2012

%H K. Rose, <a href="http://groups.yahoo.com/group/primenumbers/message/24241">Law of small numbers</a>, primenumbers group, May 2012.

%e 5^2+1 = 2*13, 13^2+1 = 10*17, 17^2=10*29; therefore a(1)=a(2)=a(3)=0.

%e 29^2+1 = 22*37+28, therefore a(4)=28.

%e Kermit Rose's post in the primenumbers Yahoo group:

%e >>> (5**2+1)%13

%e 0

%e >>> (13**2+1)%17

%e 0

%e >>> (17**2+1)%29

%e 0

%e Looks remarkable.

%e >>> (29**2+1)%37

%e 28.

%e Oops: Break in the pattern. Another illustration of the law of small numbers. :)

%o (PARI) o=5;forprime(p=o+1,900,p%4==1||next;print1((o^2+1)%o=p","))

%K nonn

%O 1,4

%A _M. F. Hasler_, May 13 2012

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Last modified September 21 20:27 EDT 2024. Contains 376089 sequences. (Running on oeis4.)