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%I #8 Sep 08 2022 08:46:08
%S 6,11,45,108
%N Numbers n such that the largest prime factor of n^2 - 2 is 17.
%C Irregular triangle of numbers k such that A038873(n) is the largest prime factor of k^2 - 2:
%C n\k| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
%C ------------------------------------------------------------------------
%C 1 | 2 | | | | | | | | | |
%C 2 | 3 | 4 | 10 | | | | | | | |
%C 3 | 6 | 11 | 45 | 108 | | | | | | |
%C 4 | 5 | 18 | 28 | 74 | 156 | 235 | | | | |
%C 5 | 8 | 23 | 39 | 116 | 1201 | | | | | |
%C 6 | 17 | 24 | 58 | 147 | 304 | 550 | 2272 | | | |
%C 7 | 7 | 40 | 54 | 87 | 101 | 181 | 557 | 1558 | | |
%C 8 | 12 | 59 | 130 | 225 | 414 | 1077 | 1124 | 2686 | 3420 | 4035 |
%C 9 | 32 | 41 | 178 | 333 } 698 | 844 | 1638 | 4567 | | |
%C 10 | 9 | 70 | 88 | 228 | 386 | 465 | 623 | 878 | 1431 | 7654 | 9313
%C ..., where A038873(n) = primes p such that x^2 = 2 has a solution mod p.
%C a(5) > 10^7. - _Tom Edgar_, May 16 2014
%e 6 is in this sequence because (6^2 - 2)/17 = 2 and 2 < 17;
%e 11 is in this sequence because (11^2 - 2)/17 = 7 and 7 < 17;
%e 45 is in this sequence because (45^2 - 2)/17^2 = 7 and 7 < 17;
%e 108 is in this sequence because (108^2 - 2)/17 = 686 = 2*7^3 and 7 < 17.
%o (Magma) [n: n in [2..120] | k eq 17 where k is D[#D] where D is PrimeDivisors(n^2-2)];
%Y Cf. A223701.
%K nonn
%O 1,1
%A _Juri-Stepan Gerasimov_, May 16 2014
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