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Least k such that the product of the first n primes of the form m^2+1 (A002496) divides k^2+1.
0

%I #34 Feb 28 2022 04:14:07

%S 1,3,13,327,36673,950117,801495893,5896798453,760999599793,

%T 3828797295053127,520910599208391893,2418812764637100821917,

%U 793123421312468129647727,6936392582189824489589830053,31170731920863007986026123435697,5284787778858696936313058199017107

%N Least k such that the product of the first n primes of the form m^2+1 (A002496) divides k^2+1.

%C Conjecture: the sequence is infinite.

%C Let b(n) = Product_{k=1..n} A002496(k): 2, 10, 170, 6290, 635290, ...

%C b(1) divides k^2+1 for k = 1, 3, 5, ...

%C b(2) divides k^2+1 for k = 3, 7, 13, 17, 23, 27, 33, 37, 43, 47, 53, 57, 63, 67, 73, 77, 83, ...

%C b(3) divides k^2+1 for k = 13, 47, 123, 157, 183, 217, 293, 327, 353, 387, 463, 497, 523, ...

%C b(4) divides k^2+1 for k = 327, 1067, 2707, 2843, 3447, 3583, 5223, 5963, 6617, 7357, 8997, 9133, 9737, 9873, ...

%C b(5) divides k^2+1 for k = 36673, 38067, 66347, 141087, 217443, 240087, 292183, 314827, 320463, ...

%p with(numtheory):lst:={2}:nn:=100:

%p for i from 1 to nn do:

%p p:=i^2+1:

%p if isprime(p)

%p then

%p lst:=lst union {p}:

%p else fi:

%p od:

%p pr:=1:

%p for n from 1 to 7 do:

%p pr:=pr*lst[n]:ii:=0:

%p for j from 1 to 10^9 while(ii=0) do:

%p if irem(j^2+1,pr)=0

%p then

%p ii:=1:

%p printf("%d %d \n",n,j):

%p fi:

%p od:

%p od:

%Y Cf. A002496, A005574.

%K nonn

%O 1,2

%A _Michel Lagneau_, Jul 29 2015

%E a(8)-a(17) from _Hiroaki Yamanouchi_, Aug 15 2015