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%I #9 May 03 2022 03:20:18
%S 2,4,8,10,16,18,20,24,28,30,32,36,40,42,48,54,56,60,64,70,72,80,84,88,
%T 90,96,100,104,108,112,120,126,128,132,136,140,144,150,152,156,160,
%U 162,168,176,180,182,184,192,196,198,200,204,208,210,216,220,224,228
%N Numbers that are the number of divisors of p^2 - 1 for some prime p.
%C For all primes p > 73, tau(p^2 - 1) >= A309906(2) = 32.
%e 184 is a term: p = 111149057 is a prime, and p^2 - 1 = (p-1)*(p+1) = 2^22 * 3 * 53 * 18524843, which has 23*2*2*2 = 184 divisors.
%e 190 is not a term: 190 = 2 * 5 * 19, so a number with 190 divisors must be of the form q^189, q^94 * r, q^37 * r^4, q^18 * r^9, or q^18 * r^4 * s, and for every prime p > 3, p^2 - 1 is a multiple of 24 = 2^3 * 8, so all the forms with 190 divisors are easily ruled out except for q^18 * r^4 * s. If p^2 - 1 = q^18 * r^4 * s, then it's one of the products 2^18 * 3^4 * s, 2^18 * r^4 * 3, 3^18 * 2^4 * s, or q^18 * 2^4 * 3. Each of these can be shown to be impossible by examining all possible ways of factoring the product into two even factors (p-1 and p+1) that differ by exactly two.
%Y Cf. A000005, A309906, A341660.
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, May 02 2022
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