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^(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
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)
Abiodun E. Adeyemi, A Study of @-numbers, arXiv:1906.05798 [math.NT], 2019.
K. A. Broughan and Qizhi Zhou, Divisibility by 3 of even multiperfect numbers of abundancy 3 and 4, JIS 13 (2010) 10.1.5
S. Colbert-Pollack, J. Holdener, E. Rachfal, and Y. Xu A DIY Project: Construct Your Own Multiply Perfect Number!, Math Horizons, Vol. 28, pp. 20-23, February 2021.
F. Firoozbakht and M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1.
Achim Flammenkamp, The Multiply Perfect Numbers Page
Fred Helenius, Link to Glossary and Lists
Walter Nissen, Abundancy : Some Resources
Eric Weisstein's World of Mathematics, Multiperfect Number.
Eric Weisstein's World of Mathematics, Sous-Triple.
Wikipedia, Multiply perfect number
EXAMPLE
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)
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
KEYWORD
nonn
AUTHOR
Jean-Yves Perrier (nperrj(AT)ascom.ch)
EXTENSIONS
4 more terms from Labos Elemer
STATUS
approved