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A260576 Least k such that the product of the first n primes of the form m^2+1 (A002496) divides k^2+1.

%I A260576 #37 Jun 24 2025 00:52:49
%S A260576 1,3,13,327,36673,950117,801495893,5896798453,760999599793,
%T A260576 3828797295053127,520910599208391893,2418812764637100821917,
%U A260576 793123421312468129647727,6936392582189824489589830053,31170731920863007986026123435697,5284787778858696936313058199017107
%N A260576 Least k such that the product of the first n primes of the form m^2+1 (A002496) divides k^2+1.
%C A260576 Conjecture: the sequence is infinite.
%C A260576 From _Robert Israel_, Jun 23 2025: (Start)
%C A260576 Consider any finite set of primes p(i) = m(i)^2 + 1, i = 1 .. n.
%C A260576 Then k^2 + 1 == 0 (mod p(i)) if k == m(i) (mod p(i)).
%C A260576 By the Chinese Remainder Theorem, there exists k such that k == m(i) (mod p(i))
%C A260576 for i = 1 .. n.  Thus the conjecture is true, and all terms a(n) exist.
%C A260576 (End)
%C A260576 Let b(n) = Product_{k=1..n} A002496(k): 2, 10, 170, 6290, 635290, ...
%C A260576 b(1) divides k^2+1 for k = 1, 3, 5, ...
%C A260576 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 A260576 b(3) divides k^2+1 for k = 13, 47, 123, 157, 183, 217, 293, 327, 353, 387, 463, 497, 523, ...
%C A260576 b(4) divides k^2+1 for k = 327, 1067, 2707, 2843, 3447, 3583, 5223, 5963, 6617, 7357, 8997, 9133, 9737, 9873, ...
%C A260576 b(5) divides k^2+1 for k = 36673, 38067, 66347, 141087, 217443, 240087, 292183, 314827, 320463, ...
%p A260576 with(numtheory):lst:={2}:nn:=100:
%p A260576 for i from 1 to nn do:
%p A260576    p:=i^2+1:
%p A260576    if isprime(p)
%p A260576    then
%p A260576    lst:=lst union {p}:
%p A260576    else fi:
%p A260576 od:
%p A260576   pr:=1:
%p A260576   for n from 1 to 7 do:
%p A260576   pr:=pr*lst[n]:ii:=0:
%p A260576    for j from 1 to 10^9 while(ii=0) do:
%p A260576    if irem(j^2+1,pr)=0
%p A260576    then
%p A260576    ii:=1:
%p A260576    printf("%d %d \n",n,j):
%p A260576    fi:
%p A260576    od:
%p A260576   od:
%Y A260576 Cf. A002496, A005574.
%K A260576 nonn
%O A260576 1,2
%A A260576 _Michel Lagneau_, Jul 29 2015
%E A260576 a(8)-a(17) from _Hiroaki Yamanouchi_, Aug 15 2015