

A027687


4perfect (quadruplyperfect or soustriple) numbers: sum of divisors of n is 4n.


25



30240, 32760, 2178540, 23569920, 45532800, 142990848, 1379454720, 43861478400, 66433720320, 153003540480, 403031236608, 704575228896, 181742883469056, 6088728021160320, 14942123276641920, 20158185857531904, 275502900594021408
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OFFSET

1,1


COMMENTS

It is conjectured that there are only finitely many terms.  N. J. A. Sloane, Jul 22 2012
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.
Conjecture: A010888(a(n)) divides a(n). Tested for n up to 36 incl.  Ivan N. Ianakiev, Oct 31 2013
From Farideh Firoozbakht, Dec 26 2014: (Start)
Theorem: If k>1 and p=a(n)/2^(k2)+1 is prime then for each positive integer m, 2^(k1)*p^m is a solution to the equation sigma(phi(x))=2*x2^k, which implies the equation has infinitely many solutions.
Proof: sigma(phi(2^(k1)*p^m)) = sigma(2^(k2)*(p1)*p^(m1)) = sigma(2^(k2)*(p1))*sigma(p^(m1)) = sigma(a(n))*(p^m1)/(p1) = 4*a(n)*(p^m1)/(p1) = 2^k*(p^m1) = 2*(2^(k1)*p^m)2^k.
It seems that for all such equations there exist such an infinite set of solutions. So I conjecture that the sequence is infinite! (End)
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


REFERENCES

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


LINKS

T. D. Noe, Table of n, a(n) for n=1..36 (complete sequence from Flammenkamp)
K. A. Broughan, Qizhi Zhou, Divisibility by 3 of even multiperfect numbers of abundancy 3 and 4, JIS 13 (2010) 10.1.5
Walter Nissen, Abundancy : Some Resources
Achim Flammenkamp, The Multiply Perfect Numbers Page
Fred Helenius, Link to Glossary and Lists
Eric Weisstein's World of Mathematics, Multiperfect Number.
Eric Weisstein's World of Mathematics, SousTriple.
Wikipedia, Multiply perfect number


EXAMPLE

From Daniel Forgues, May 09 2010: (Start)
30240 = 2^5*3^3*5*7
sigma(30240) = (2^61)/1*(3^41)/2*(5^21)/4*(7^21)/6
= (63)*(40)*(6)*(8)
= (7*3^2)*(2^3*5)*(2*3)*(2^3)
= 2^7*3^3*5*7
= (2^2) * (2^5*3^3*5*7)
= 4 * 30240 (End)


MATHEMATICA

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 *)


CROSSREFS

Cf. A000396, A005820, A007539, A046060, A046061.
Sequence in context: A254841 A247855 A113286 * A190475 A218029 A109485
Adjacent sequences: A027684 A027685 A027686 * A027688 A027689 A027690


KEYWORD

nonn


AUTHOR

JeanYves Perrier (nperrj(AT)ascom.ch)


EXTENSIONS

4 more terms from Labos Elemer


STATUS

approved



