Is $p(n) = 2^{n²+n-1} - n² - n + 1$ abundant for all $n >1$?

99 Views Asked by At

Let $p(n)=2^{n²+n-1}-n²-n+1 $, and let $\delta(n)$ be sum of proper divisors of $n\in\mathbb{N}$.

After some verifications according to the values of $n>1$ I noticed:

$$\delta(p(n))> p(n)$$

Is there some way to prove that for all $n>1$ : $\delta(p(n))> p(n)$ or is there a counterexample for it?

Note: For some verification look here.

1

There are 1 best solutions below

1
On

If referring to the sum of all divisors (which includes the number itself), trivially $\sigma_1(n) > n$ for all $n>1$ since $\sigma_1(n) = 1 + n + (\text{other stuff if n isn't prime or 0 if it is prime}) > n$

This trivially implies that $\sigma_1(p(n))>p(n)$ for all $n$ such that $p(n)>1$.

If your conjecture is instead about the sum of proper divisors, i.e., $s(n)$, then the conjecture fails for several small values of $n$.

$s(p(2)) = s(27) = 1+3+9=13 < 27 = p(2)$

$s(p(3)) = s(2037) = 1+3+7+21+97+291+679=1099 < 2037 = p(3)$

It is doubtful that many (if any) values hold true. This should reiterate the importance of understanding what definitions you and/or your calculator are using. In this case, keeping track of the difference between "the sum of divisors" and "the sum of proper divisors."