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
Muniru A Asiru, Table of n, a(n) for n = 1..725
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
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 24 2010
STATUS
approved