A174633 Let H(p) = p*tau(p)/sigma(p). Numbers 2^(H(p) - 1)*(2^H(p) - 1) where H(p) is an integer (i.e., p in A001599).

1, 6, 28, 496, 2016, 496, 32640, 130816, 2096128, 523776, 8128, 536854528, 536854528, 134209536, 8589869056, 140737479966720, 140737479966720, 2199022206976, 33550336, 137438691328, 9007199187632128, 562949936644096

Offset: 1

Author

Michel Lagneau, Mar 24 2010

Keywords

• nonn

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.

Examples

			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).
		

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

Crossrefs

Cf. A000396.

Programs

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

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 *)