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The perfect number $6$ is in the middle of the primes $5$ and $7$. It is the only perfect number with this property because odd numbers are not in the middle of two twin primes and even perfect numbers have the form $2^{n-1}(2^n-1)\ ,\ n\ge 2$ with a prime $2^n-1$, so they are greater than $4$, but not divisible by $3$ for $n>2$, hence not the middle of two twin-primes.

The least multi-perfect number $N$, for which $N-1$ is prime (except $6$), is $$N=13794\ 54720=2^8\times3\times5\times7\times19\times37\times73$$

, which is $4$-perfect, that means $\frac{\sigma(n)}{n}=4$ , where $\sigma(n)$ is the sum of the divisors of $n$.

  • Is there a multi-perfect-number $N$ greater than $6$, such that $N-1$ and $N+1$ are both prime ?
  • If yes, what is the minimal $N$ ?
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Looking at the list http://oeis.org/A007691/b007691.txt of the first 1600 multiply perfect numbers, you find that the smallest number you are looking for seems to be $$ \begin{align} n&=1928622300236318049928258133164032 \\ &=2^{33}\cdot3^4\cdot7\cdot11^3\cdot31\cdot61\cdot83\cdot331\cdot43691\cdot131071 \end{align} $$ which is $4$-perfect as well, with $n-1$ and $n+1$ being prime.

(The next $n$ in that sequence appears to be 20736673935772776371907300845135221460949851029611211752485268322919135405936727005685049658102947046034932245258365859207952308374437205033138073857921732017237200100173505791123852481233523248079738393931546624000000000 - don't ask me for $\sigma(n)/n$ for that one, though!)

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  • $\begingroup$ [2, 94; 3, 32; 5, 9; 7, 11; 11, 2; 13, 8; 17, 1; 19, 5; 23, 1; 29, 2; 31, 2; 37, 1; 43, 1; 53, 1; 59, 1; 61, 2; 67, 1; 71, 1; 73, 1; 83, 3; 97, 1; 107, 1; 127, 1; 181, 1; 191, 2; 331, 1; 353, 1; 487, 1; 521, 1; 607, 1; 1493, 1; 2609, 1; 385 1, 1; 5347, 1; 160967, 1; 524287, 1; 547889, 1; 1609669, 1; 2908363, 1; 6491257, 1; 420778751, 1; 947723521, 1; 2413941289, 1; 30327152671, 1] is the factorization of the large number, found by PARI/GP $\endgroup$ – Peter May 7 '15 at 22:15
  • $\begingroup$ The large number is $8$-perfect. $\endgroup$ – Peter May 7 '15 at 22:18
  • $\begingroup$ Great answer! Thanks a lot. $\endgroup$ – Peter May 7 '15 at 22:18

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