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Numbers k such that k^2 - 1 and k^2 + 1 have 6 divisors each.
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%I #18 Jun 22 2024 00:24:35

%S 168,1368,97968,10374840,16104168,44049768,68674368,100741368,

%T 281803368,486775968,1177381968,1262878368,1336852968,2321986968,

%U 2404627368,3476635368,4374102768,5102102040,5142754368,5182128168,5385651768,6035269968,9218496168,10657878168

%N Numbers k such that k^2 - 1 and k^2 + 1 have 6 divisors each.

%C Each term is a number of the form k = sqrt(p^2 * q + 1) such that q = p^2 - 2 and k^2 + 1 = r^2 * s, where p, q, r, and s are distinct primes.

%F { k : tau(k^2 - 1) = tau(k^2 + 1) = 6}, where tau() is the number of divisors function, A000005.

%e 168 is a term: both 168^2 - 1 = 28223 = 13^2 * 167 and 168^2 + 1 = 28225 = 5^2 * 1129 have 6 divisors.

%Y Cf. A000005, A002522, A005563, A069062, A108278, A193432, A347191, A373209.

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jun 21 2024