Triangular Numbers are numbers going from $1,3,6,10,15,21,28$ and so on . These numbers can be formed like a triangle , so called triangular numbers.
Perfect numbers are numbers going in the pattern $1,6,28$ and so on . Perfect numbers are numbers whose sum of the factors equals twice the number .
Cyclops numbers is a number in binary notation which has $1$'s only and a $0$ exactly in the middle . For example -
$0$ is a cyclops number , in base $10$ it is $0$ only.
$101$ is a cyclops number , in base $10$ it is $5$ .
$11011$ is a cyclops number , in base $10$ it is $27$ .
Now there is an interesting relation between these $3$ types of numbers . I notice that every perfect number is some triangular number , as well every perfect number is exactly $1$ less than their corresponding cyclops numbers when changed to base $10$ .
Can anyone tell my how this relation actually comes?
We know that every even perfect number is of the form $2^{p-1}(2^p-1)$. It is not hard to see that a cyclops number with $2n+1$ digits in its binary form is: $$1111 \ldots 1 (2n+1 \text{ ones})-1000\ldots0(n \text{ zeroes})=(2^{2n+1}-1)-2^n=2^{2n+1}-2^n-1$$
Moreover, triangular numbers are of the form: $$1+2+3+\ldots+k=\frac{k(k+1)}{2}$$
Now, notice that for even perfect numbers: $$2^{p-1}(2^p-1)=\frac{2^p(2^p-1)}{2}=\frac{k(k+1)}{2} \quad (k=2^p)$$ $$2^{p-1}(2^p-1)-1 = 2^{2p-1}-2^{p-1}-1=2^{2n+1}-2^n-1 \quad(n=p-1)$$
Thus, we have proven your observation for even perfect numbers. However, the existence of odd perfect numbers greater than $1$ is unknown.