login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A174633 Let H(p) = p*tau(p)/sigma(p). Numbers 2^(H(p) - 1)*(2^H(p) - 1) where H(p) is integer (i.e., p in A001599). 1
1, 6, 28, 496, 2016, 496, 32640, 130816, 2096128, 523776, 8128, 536854528, 536854528, 134209536, 8589869056, 140737479966720, 140737479966720, 2199022206976, 33550336, 137438691328, 9007199187632128, 562949936644096 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
We prove that perfect numbers A000396 are included in this sequence. A number p is perfect if sigma(p)=2p where sigma(p) is the sum of the divisors of p (A000203). So, p is perfect if and only if H(p) = p*tau(p)/sigma(p) = p*tau(p)/2p = tau(p)/2. Because the numbers of the form 2^(q-1)*(2^q - 1) are perfect, where q is a prime such that 2^q - 1 is also prime, tau(p) = (q-1+1)*2 = 2q, and then H(p) = q. When p is perfect, we have p = 2^(H(p) - 1)*(2^H(p) - 1). Now, we prove that n = 2^(H(p) - 1)*(2^H(p) - 1) => n is an even perfect number. We have H(p) = p*tau(p)/sigma(p) and H is multiplicative. Because gcd(2^(H(p) - 1), 2^H(p) - 1) = 1, we obtain sigma(2^H(p) - 1)/tau(2^H(p) - 1) = 2^H(p) - 1 = 2^H(p)/2. Now, the equation sigma(m)/tau(m) = (m+1)/2 with m odd is possible only if m is prime. Thus, 2^H(p) - 1 is prime.
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, p. 4.
S. Bezuszka, Perfect Numbers, (Booklet 3, Motivated Math. Project Activities) Boston College Press, Chestnut Hill MA 1980.
J.M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres, Ellipses 2004, p. 73.
LINKS
C. K. Caldwell, Perfect number
J. J. O'Connor and E. F. Robertson, Perfect Numbers
EXAMPLE
For p = 1, H(1) = 1 and n=1; for p=2, H(2) = 2*tau(2)/sigma(2) = 2*2/3 = 4/3 (not integer). For p = 6, H(6) = 6*tau(6)/sigma(6) = 6*4/12 = 2, n = 2^(2-1)*(2^2 - 1) = 2*3 = 6 (first perfect number). Other perfect numbers: 28 (for p=28), 496 (for p=140), 8128 (for p = 8128).
MAPLE
for p from 1 to 10000000 do
H := p*numtheory[tau](p)/numtheory[sigma](p) ;
if type(H, 'integer') then
(2^(H-1))*(2^H-1) ;
printf("%d, ", %) ;
end if;
end do:
MATHEMATICA
h[p_] := p*DivisorSigma[0, p]/DivisorSigma[1, p]; hp=Select[Table[h[p], {p, 1, 10^6}], IntegerQ]; (2^(hp-1))*(2^hp-1) (* Jean-François Alcover, Sep 13 2011 *)
PROG
(GAP) H:=[];; for p in [1..240000] do if IsInt(p*Tau(p)/Sigma(p)) then Add(H, p*Tau(p)/Sigma(p)); fi; od; a:=List(H, i->(2^(i-1))*(2^i-1)); # Muniru A Asiru, Nov 28 2018
CROSSREFS
Cf. A000396.
Sequence in context: A325023 A138876 A326145 * A345004 A370789 A351440
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 24 2010
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified March 28 07:18 EDT 2024. Contains 371235 sequences. (Running on oeis4.)