A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

8, 50, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000, 2576450, 2761250, 2832200

Offset

1,1

Comments

Subsequence of A085722.

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), ...

Property:

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

From the relations:

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

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

We obtain:

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

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

with m = 2*k^2 we obtain:

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

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

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

Links

Table of n, a(n) for n=1..34.

Example

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

Maple

with(numtheory):nn:=250:printf(`%d, `, 8):
for k from 0 to nn do:
n:=50*k^2:d:=factorset(n^2+1):
  if bigomega(n^2+1)=2 and (d[1]+d[2])/2 - 1 = n
   then
    printf(`%d, `, n):
    else
  fi:
od:

Mathematica

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A195231 A162236 A215874 * A078304 A000851 A054620

Adjacent sequences: A348591 A348592 A348593 * A348595 A348596 A348597

Keyword

nonn

Author

Michel Lagneau, Jan 26 2022

Status

approved