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A027687 4-perfect (quadruply-perfect or sous-triple) numbers: sum of divisors of n is 4n. 24
30240, 32760, 2178540, 23569920, 45532800, 142990848, 1379454720, 43861478400, 66433720320, 153003540480, 403031236608, 704575228896, 181742883469056, 6088728021160320, 14942123276641920, 20158185857531904, 275502900594021408 (list; graph; refs; listen; history; text; internal format)



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^(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.

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.

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


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


T. D. Noe, Table of n, a(n) for n=1..36 (complete sequence from Flammenkamp)

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, Sous-Triple.

Wikipedia, Multiply perfect number


From Daniel Forgues, May 09 2010: (Start)

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

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


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 *)


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

Sequence in context: A254841 A247855 A113286 * A190475 A218029 A109485

Adjacent sequences:  A027684 A027685 A027686 * A027688 A027689 A027690




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


4 more terms from Labos Elemer



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Last modified December 1 15:05 EST 2015. Contains 264695 sequences.