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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).

%I A174633 #36 Jan 09 2025 13:20:15
%S A174633 1,6,28,496,2016,496,32640,130816,2096128,523776,8128,536854528,
%T A174633 536854528,134209536,8589869056,140737479966720,140737479966720,
%U A174633 2199022206976,33550336,137438691328,9007199187632128,562949936644096
%N 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).
%C A174633 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.
%D A174633 T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, p. 4.
%D A174633 S. Bezuszka, Perfect Numbers, (Booklet 3, Motivated Math. Project Activities) Boston College Press, Chestnut Hill MA 1980.
%D A174633 J.M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres, Ellipses 2004, p. 73.
%H A174633 Muniru A Asiru, <a href="/A174633/b174633.txt">Table of n, a(n) for n = 1..725</a>
%H A174633 C. K. Caldwell, <a href="https://t5k.org/glossary/page.php?sort=PerfectNumber">Perfect number</a>
%H A174633 J. J. O'Connor and E. F. Robertson, <a href="https://mathshistory.st-andrews.ac.uk/HistTopics/Perfect_numbers/">Perfect Numbers</a>
%e A174633 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).
%p A174633 for p from 1 to 10000000 do
%p A174633         H := p*numtheory[tau](p)/numtheory[sigma](p) ;
%p A174633         if type(H,'integer') then
%p A174633                 (2^(H-1))*(2^H-1) ;
%p A174633                 printf("%d,",%) ;
%p A174633         end if;
%p A174633 end do:
%t A174633 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 *)
%o A174633 (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
%Y A174633 Cf. A000396.
%K A174633 nonn
%O A174633 1,2
%A A174633 _Michel Lagneau_, Mar 24 2010