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

1,2

Comments

Conjecture: the sequence is infinite.

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, ...

Links

Table of n, a(n) for n=1..16.

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:

Crossrefs

Cf. A002496, A005574.

Sequence in context: A113526 A113612 A180654 * A156358 A066266 A092845

Adjacent sequences: A260573 A260574 A260575 * A260577 A260578 A260579

Keyword

nonn

Author

Michel Lagneau, Jul 29 2015

Extensions

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

Status

approved