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For integers $m\geq 1$, we denote with $\sigma(m)=\sum_{d\mid m}d$ the sum of divisors function. If $m$ is even and satisfies $\sigma(m)=2m$, then we say that $m$ is an even perfect number. See this MathWorld.

Let $$N_1<N_2<\ldots<N_n$$ a sequence of different even perfect numbers $N_k=2^{p_k-1}(2^{p_k}-1)$, $1\leq k\leq n$, and thus $2^{p_k}-1$ is the corresponding Mersenne prime. And let $\mathcal{N}$ also an even perfect number greater than previous, that is $N_n<\mathcal{N}$.

Thus there exist a positive integer $\gamma$ such that the Mersenne prime associted with $\mathcal{N}$ has the form $$2^{\left(\sum_{k=1}^n p_k\right)+\gamma}-1.$$

Claim. Under previous assumptions one has that $$\sigma\left(2^{\gamma-1}\sigma\left(2^{\gamma-1}\prod_{k=1}^n\sigma(N_k)\right)\right)=2\mathcal{N}.$$

As remark also we can write the factor $2^{\gamma-1}\cdot\prod_{k=1}^n\sigma(N_k)$ as $2^{n+\gamma-1}\cdot\prod_{k=1}^n N_k$.

Question. Let $m$ and $\delta$ positive integers and $1\leq M_1<M_2<\ldots<M_m<\mathcal{M}$, also positive integers, satisfying $$\sigma\left(2^{\delta-1}\sigma\left(2^{\delta-1}\prod_{k=1}^m\sigma(M_k)\right)\right)=2\mathcal{M}.$$ Is it possible to prove that each $M_k$ is an even perfect number $2^{q_k-1}(2^{q_k}-1)$, and the Mersenne prime associated with the even perfect nubmer $\mathcal{M}$ is $2^{\left(\sum_{k=1}^m q_k\right)+\delta}-1$? Many thanks.

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  • $\begingroup$ That is I am asking if it is possible to get a characterization of a finite subset of even perfect numbers with previous identity, or well you can show a counterexample. In previous calculations and reasoning we presume that our set of different even perfect numbers there exist. Many thanks all users. $\endgroup$ – user243301 Jul 16 '17 at 1:27
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Thus there exist a positive integer $\gamma$ such that the Mersenne prime associted with $\mathcal{N}$ has the form $$2^{\left(\sum_{k=1}^n p_k\right)+\gamma}-1.$$

False, at least if you meant the next perfect even number. Consider $n = 3$, then we have $p_1 = 2, p_2 = 3, p_3 = 5$, but the next prime is $7 < 2 + 3 + 5$, meaning $\gamma$ would have to be negative. We can't even assume we can choose $\mathcal{N}$ large enough, because we don't know if there are an infinite number of Mersenne primes. But let's assume there are an infinite number of Mersenne primes and choose $\mathcal{N}$ large enough.

Then there is your claimed equation.

$$\sigma\left(2^{\gamma-1}\sigma\left(2^{\gamma-1}\prod_{k=1}^n\sigma(N_k)\right)\right)=2\mathcal{N}$$

$$\sigma\left(2^{\gamma-1}\sigma\left(2^{\gamma-1} \cdot \prod_{k=1}^n2 \cdot2^{p_k-1}(2^{p_k}-1)\right)\right)=2\mathcal{N}$$

$$\sigma\left(2^{\gamma-1}\sigma\left(2^{\gamma-1+\sum_{k=1}^n p_k} \cdot \prod_{k=1}^n(2^{p_k}-1)\right)\right)=2\mathcal{N}$$

Since the power of two and the product of primes is coprime we have:

$$\sigma\left(2^{\gamma-1} \cdot \sigma\left(2^{\gamma-1+\sum_{k=1}^n p_k}\right) \cdot \prod_{k=1}^n \sigma (2^{p_k}-1)\right)=2\mathcal{N}$$

$$\sigma\left(2^{\gamma-1}\cdot\left(2^{\gamma+\sum_{k=1}^n p_k} - 1\right) \cdot \prod_{k=1}^n 2^{p_k}\right)=2\mathcal{N}$$

$$\sigma\left(2^{\gamma-1 + \sum_{k=1}^n p_k}\left(2^{\gamma+\sum_{k=1}^n p_k} - 1\right) \right)=2\mathcal{N}$$

$$\sigma\left(\mathcal{N} \right)=2\mathcal{N}$$

So if we assume that $N_k$ are all even perfect numbers and we choose $\mathcal{N}, \gamma$ such that $\mathcal{N}$ is also a perfect even number, the equation holds.

Now to answer your question, absolutely not. Choose $\delta = m = 1$ and (almost) arbitrary $M_m$. Then the equation is:

$$\sigma(\sigma(\sigma(M_m))) = 2\mathcal{M}$$

A simple counterexample would be $M_m = 8$, $\mathcal{M} = 30$.

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  • $\begingroup$ Many thanks @orlp now read your answer. $\endgroup$ – user243301 Jul 16 '17 at 12:30
  • $\begingroup$ Many thanks for your nice answer. Iam going to study it in next hours. $\endgroup$ – user243301 Jul 16 '17 at 12:40

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