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Numbers k such that the sum of the prime distinct divisors of k^2+1 equals 2 times the largest prime divisor of k^2+1.
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%I #20 May 14 2023 09:40:28

%S 1123,1143,6235,8457,11565,21917,22857,33285,41319,58195,119571,

%T 124723,128363,173922,178703,188115,243939,280158,308859,309709,

%U 409485,430581,565571,703845,961237,1153362,1170291,1327998,1409794,1536651,1586195,1649395,1665868

%N Numbers k such that the sum of the prime distinct divisors of k^2+1 equals 2 times the largest prime divisor of k^2+1.

%H Amiram Eldar, <a href="/A200072/b200072.txt">Table of n, a(n) for n = 1..500</a>

%e 1123 is in the sequence because the distinct prime divisors of 1123^2 + 1 are 2, 5, 13, 89, 109 and the sum 2 + 5 + 13 + 89 + 109 = 218 = 2*109.

%t Select[Range[1700000],Plus@@(pl=First/@FactorInteger[#^2+1])==2*pl[[-1]]&]

%Y Cf. A002522, A014442, A193462, A200071.

%K nonn

%O 1,1

%A _Michel Lagneau_, Nov 13 2011