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A348594
Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.
0
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).
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
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jan 26 2022
STATUS
approved