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A348594
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Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.
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0
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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50 = 2*5^2 is in the sequence because 50^2 + 1 = 41*61 with 50 = (41 + 61)/2 - 1.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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