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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 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 (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, page 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 J. J. O'Connor & E. F. Robertson, Perfect Numbers C. K. Caldwell, Perfect number 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: A083387 A104511 A138876 * A201186 A060286 A000396 Adjacent sequences:  A174630 A174631 A174632 * A174634 A174635 A174636 KEYWORD nonn AUTHOR Michel Lagneau, Mar 24 2010 STATUS approved

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Last modified January 18 23:05 EST 2019. Contains 319282 sequences. (Running on oeis4.)