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

10, 65, 20737

Offset: 1

Author

Mohamed Bouhamida, Jul 22 2008, Jul 27 2008

Keywords

• nonn
• bref

Comments

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?

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

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

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

Examples

			10 divides (1^2 + 2^2 + 5^2) giving 3 - the number of proper divisors of semiprime 10.
65 divides (1^2 + 5^2 + 13^2) giving 3 - the number of proper divisors of semiprime 65.
20737 divides (1^2 + 89^2 + 233^2) giving 3 - the number of proper divisors of semiprime 20737.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..5
• T. Cai, D. Chen, and Y. Zhang, Perfect numbers and Fibonacci primes, arXiv:1310.0898 [math.NT], 2013-2014.
• T. Cai, D. Chen, and Y. Zhang, Perfect numbers and Fibonacci primes (II), arXiv:1406.5684 [math.NT], 2014 (see case m=1 in Table 1).

Crossrefs

Cf. A000045, A001605, A046762.

Programs

PARI

isok(n) = sigma(n, 2) - n^2 == 3*n; \\ Michel Marcus, Jun 24 2014