%I #15 Jan 26 2022 09:00:25

%S 8,50,1250,1800,2450,9800,14450,20000,24200,101250,105800,135200,

%T 162450,168200,204800,304200,336200,451250,480200,490050,530450,

%U 696200,924800,966050,1008200,1125000,1155200,1428050,1805000,2332800,2420000,2576450,2761250,2832200

%N Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

%C Subsequence of A085722.

%C The corresponding pairs (p, q) of the sequence are (5, 13), (41, 61), (1201, 1301), (1741, 1861), (2381, 2521), (9661, 9941), (14281, 14621), (19801, 20201), (23981, 24421), (100801, 101701), ...

%C Property:

%C a(n) = 2* A109306(n)^2 and a(n) == 0 (mod 50) for n > 1. Proof:

%C From the relations:

%C (1) m^2 + 1 = p*q

%C (2) (p + q)/2 = m + 1

%C We obtain:

%C (3) p = m + 1 - sqrt(8*m)/2

%C (4) q = m + 1 + sqrt(8*m)/2

%C with m = 2*k^2 we obtain:

%C (5) p = k^2 + (k-1)^2

%C (6) q = k^2 + (k+1)^2

%C For n > 1, A109306(n) == 0 (mod 5) => 2*A109306(n)^2 == 0 (mod 50).

%e 50 = 2*5^2 is in the sequence because 50^2 + 1 = 41*61 with 50 = (41 + 61)/2 - 1.

%p with(numtheory):nn:=250:printf(`%d, `,8):

%p for k from 0 to nn do:

%p n:=50*k^2:d:=factorset(n^2+1):

%p if bigomega(n^2+1)=2 and (d[1]+d[2])/2 - 1 = n

%p then

%p printf(`%d, `,n):

%p else

%p fi:

%p od:

%t q[n_] := Module[{f = FactorInteger[n^2 + 1]}, f[[;; , 2]] == {1, 1} && f[[1, 1]] + f[[2, 1]] == 2*n + 2]; Select[Range[3*10^5], q] (* _Amiram Eldar_, Jan 26 2022 *)

%o (PARI) isok(m) = my(x); if (bigomega(x=m^2+1)==2, my(f=factor(x)); (f[1,1]+f[2,1] == 2*(m+1))); \\ _Michel Marcus_, Jan 26 2022

%Y Cf. A014442, A085722, A089120, A109306, A144255.

%K nonn

%O 1,1

%A _Michel Lagneau_, Jan 26 2022