A140362 - OEIS

%I #33 Dec 30 2025 14:00:12

%S 10,65,20737

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

%C 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 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 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 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

%D Shen, Z., & Zhao, J. (2025). Unitary Perfect Numbers and Fibonacci Primes. The Fibonacci Quarterly, 63(3), 588-594. https://doi-org.proxy.libraries.rutgers.edu/10.1080/00150517.2025.2461642. Mentions this sequence.

%H T. D. Noe, <a href="/A140362/b140362.txt">Table of n, a(n) for n=1..5</a>

%H T. Cai, D. Chen, and Y. Zhang, <a href="https://arxiv.org/abs/1310.0898">Perfect numbers and Fibonacci primes</a>, arXiv:1310.0898 [math.NT], 2013-2014.

%H T. Cai, D. Chen, and Y. Zhang, <a href="https://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 10 divides (1^2 + 2^2 + 5^2) giving 3 - the number of proper divisors of semiprime 10.

%e 65 divides (1^2 + 5^2 + 13^2) giving 3 - the number of proper divisors of semiprime 65.

%e 20737 divides (1^2 + 89^2 + 233^2) giving 3 - the number of proper divisors of semiprime 20737.

%o (PARI) isok(n) = sigma(n, 2) - n^2 == 3*n; \\ _Michel Marcus_, Jun 24 2014

%Y Cf. A000045, A001605, A046762.

%K nonn,bref

%O 1,1

%A _Mohamed Bouhamida_, Jul 22 2008, Jul 27 2008