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4-perfect (quadruply-perfect or sous-triple) numbers: sum of divisors of n is 4n.
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%I #63 Jul 01 2021 23:43:19

%S 30240,32760,2178540,23569920,45532800,142990848,1379454720,

%T 43861478400,66433720320,153003540480,403031236608,704575228896,

%U 181742883469056,6088728021160320,14942123276641920,20158185857531904,275502900594021408

%N 4-perfect (quadruply-perfect or sous-triple) numbers: sum of divisors of n is 4n.

%C It is conjectured that there are only finitely many terms. - _N. J. A. Sloane_, Jul 22 2012

%C Odd perfect number (unlikely to exist) and infinitely many Mersenne primes will make the sequence infinite - take the product of the OPN and coprime EPNs.

%C Conjecture: A010888(a(n)) divides a(n). Tested for n up to 36 incl. - _Ivan N. Ianakiev_, Oct 31 2013

%C From _Farideh Firoozbakht_, Dec 26 2014: (Start)

%C Theorem: If k>1 and p=a(n)/2^(k-2)+1 is prime then for each positive integer m, 2^(k-1)*p^m is a solution to the equation sigma(phi(x))=2*x-2^k, which implies the equation has infinitely many solutions.

%C Proof: sigma(phi(2^(k-1)*p^m)) = sigma(2^(k-2)*(p-1)*p^(m-1)) = sigma(2^(k-2)*(p-1))*sigma(p^(m-1)) = sigma(a(n))*(p^m-1)/(p-1) = 4*a(n)*(p^m-1)/(p-1) = 2^k*(p^m-1) = 2*(2^(k-1)*p^m)-2^k.

%C It seems that for all such equations there exist such an infinite set of solutions. So I conjecture that the sequence is infinite! (End)

%C If 3 were prepended to this sequence, then it would be the sequence of integers k such that numerator(sigma(k)/k)=4. - _Michel Marcus_, Nov 22 2015

%D R. K. Guy, Unsolved Problems in Number Theory, B2.

%H T. D. Noe, <a href="/A027687/b027687.txt">Table of n, a(n) for n=1..36</a> (complete sequence from Flammenkamp)

%H Abiodun E. Adeyemi, <a href="https://arxiv.org/abs/1906.05798">A Study of @-numbers</a>, arXiv:1906.05798 [math.NT], 2019.

%H K. A. Broughan and Qizhi Zhou, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Broughan/broughan17d.html">Divisibility by 3 of even multiperfect numbers of abundancy 3 and 4</a>, JIS 13 (2010) 10.1.5

%H S. Colbert-Pollack, J. Holdener, E. Rachfal, and Y. Xu <a href="https://doi.org/10.1080/10724117.2020.1849911">A DIY Project: Construct Your Own Multiply Perfect Number!</a>, Math Horizons, Vol. 28, pp. 20-23, February 2021.

%H F. Firoozbakht and M. F. Hasler, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Hasler/hasler2.html">Variations on Euclid's formula for Perfect Numbers</a>, JIS 13 (2010) #10.3.1.

%H Achim Flammenkamp, <a href="http://wwwhomes.uni-bielefeld.de/achim/mpn.html">The Multiply Perfect Numbers Page</a>

%H Fred Helenius, <a href="http://pw1.netcom.com/~fredh/index.html">Link to Glossary and Lists</a>

%H Walter Nissen, <a href="http://upforthecount.com/math/abundance.html">Abundancy : Some Resources </a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MultiperfectNumber.html">Multiperfect Number.</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Sous-Triple.html">Sous-Triple.</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiply_perfect_number">Multiply perfect number</a>

%e From _Daniel Forgues_, May 09 2010: (Start)

%e 30240 = 2^5*3^3*5*7

%e sigma(30240) = (2^6-1)/1*(3^4-1)/2*(5^2-1)/4*(7^2-1)/6

%e = (63)*(40)*(6)*(8)

%e = (7*3^2)*(2^3*5)*(2*3)*(2^3)

%e = 2^7*3^3*5*7

%e = (2^2) * (2^5*3^3*5*7)

%e = 4 * 30240 (End)

%t AbundantQ[n_]:=DivisorSigma[1, n]==4*n;a={};Do[If[AbundantQ[n], AppendTo[a, n]], {n, 10^6}];a (* _Vladimir Joseph Stephan Orlovsky_, Aug 16 2008 *)

%Y Cf. A000396, A005820, A007539, A046060, A046061.

%K nonn

%O 1,1

%A Jean-Yves Perrier (nperrj(AT)ascom.ch)

%E 4 more terms from _Labos Elemer_