cp's OEIS Frontend

A140362 Semiprimes pq that divide the sum of the squares of their divisors, 1+p^2+q^2+(pq)^2.

%I A140362 #30 Dec 30 2024 17:20:53
%S A140362 10,65,20737
%N A140362 Semiprimes pq that divide the sum of the squares of their divisors, 1+p^2+q^2+(pq)^2.
%C A140362 6 is the smallest integer n which is the product of two distinct primes and which divides the sum of the cubes of the divisors of n. Are there other numbers with this property?
%C A140362 Using Pell equations and a Fibonacci identity, Max Alekseyev and I have shown that all terms are the product of prime Fibonacci numbers whose indices are twin primes. The first three terms are Fib(3)*Fib(5), Fib(5)*Fib(7) and Fib(11)*Fib(13). The other two known terms are Fib(431)*Fib(433) and Fib(569)*Fib(571), huge numbers that are in the b-file. The sequence probably has no additional terms. - _T. D. Noe_, Jul 27 2008
%C A140362 Let a, b, c and d be consecutive odd-indexed Fibonacci numbers. Then it can be proved that 1 + b^2 + c^2 + (bc)^2 = abcd, which shows that bc divides 1 + b^2 + c^2 + (bc)^2. Hence if b and c are prime, then bc is in this sequence. - _T. D. Noe_, Jul 27 2008
%C A140362 Empirical search suggests that A067558(a(n))/A032741(a(n)) = a(n). A032741(a(n)) = 3 for all n by definition of semiprime. A067558(a(n)) must also then be divisible by 3. a(n) can be called the n-th "perfect mean square aliquot number". - _William Krier_, Dec 16 2024
%H A140362 T. D. Noe, <a href="/A140362/b140362.txt">Table of n, a(n) for n=1..5</a>
%H A140362 T. Cai, D. Chen, and Y. Zhang, <a href="http://arxiv.org/abs/1310.0898">Perfect numbers and Fibonacci primes</a>, arXiv:1310.0898 [math.NT], 2013-2014.
%H A140362 T. Cai, D. Chen, and Y. Zhang, <a href="http://arxiv.org/abs/1406.5684">Perfect numbers and Fibonacci primes (II)</a>, arXiv:1406.5684 [math.NT], 2014 (see case m=1 in Table 1).
%e A140362 10 divides (1^2 + 2^2 + 5^2) giving 3 - the number of proper divisors of semiprime 10.
%e A140362 65 divides (1^2 + 5^2 + 13^2) giving 3 - the number of proper divisors of semiprime 65.
%e A140362 20737 divides (1^2 + 89^2 + 233^2) giving 3 - the number of proper divisors of semiprime 20737.
%o A140362 (PARI) isok(n) = sigma(n, 2) - n^2 == 3*n; \\ _Michel Marcus_, Jun 24 2014
%Y A140362 Cf. A000045, A001605, A046762.
%K A140362 nonn,bref
%O A140362 1,1
%A A140362 _Mohamed Bouhamida_, Jul 22 2008, Jul 27 2008