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

1, 3, 13, 327, 36673, 950117, 801495893, 5896798453, 760999599793, 3828797295053127, 520910599208391893, 2418812764637100821917, 793123421312468129647727, 6936392582189824489589830053, 31170731920863007986026123435697, 5284787778858696936313058199017107

Offset: 1

Author

Michel Lagneau, Jul 29 2015

Keywords

• nonn

Comments

Conjecture: the sequence is infinite.

From Robert Israel, Jun 23 2025: (Start)

Consider any finite set of primes p(i) = m(i)^2 + 1, i = 1 .. n.

Then k^2 + 1 == 0 (mod p(i)) if k == m(i) (mod p(i)).

By the Chinese Remainder Theorem, there exists k such that k == m(i) (mod p(i))

for i = 1 .. n. Thus the conjecture is true, and all terms a(n) exist.

(End)

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

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

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, ...

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

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

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

Crossrefs

Cf. A002496, A005574.

Programs

Maple

with(numtheory):lst:={2}:nn:=100:
for i from 1 to nn do:
   p:=i^2+1:
   if isprime(p)
   then
   lst:=lst union {p}:
   else fi:
od:
  pr:=1:
  for n from 1 to 7 do:
  pr:=pr*lst[n]:ii:=0:
   for j from 1 to 10^9 while(ii=0) do:
   if irem(j^2+1,pr)=0
   then
   ii:=1:
   printf("%d %d \n",n,j):
   fi:
   od:
  od:

Extensions

a(8)-a(17) from Hiroaki Yamanouchi, Aug 15 2015