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A174633
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Let H(p) = p*tau(p)/sigma(p). Numbers (2^(H(p)-1))*(2^H(p) -1) when H(p) is integer (ie., p in A001599).
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0
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1, 6, 28, 496, 2016, 496, 32640, 130816, 2096128, 523776, 8128, 536854528, 536854528, 134209536, 8589869056, 140737479966720, 140737479966720, 2199022206976, 33550336, 137438691328, 9007199187632128, 562949936644096
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OFFSET
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1,2
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COMMENTS
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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, so 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 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 m is prime. Then, 2^H(p)-1 is prime.
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REFERENCES
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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 problemes en theorie classique des nombres, Ellipses 2004, p.73
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LINKS
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Table of n, a(n) for n=1..22.
J. J. O'Connor & E. F. Robertson, Perfect Numbers
C. K. Caldwell, Perfect number
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EXAMPLE
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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)
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MAPLE
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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:
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MATHEMATICA
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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) (* From Jean-François Alcover, Sep 13 2011 *)
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CROSSREFS
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Cf. A000396.
Sequence in context: A083387 A104511 A138876 * A201186 A060286 A000396
Adjacent sequences: A174630 A174631 A174632 * A174634 A174635 A174636
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KEYWORD
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nonn
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AUTHOR
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Michel Lagneau, Mar 24 2010
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STATUS
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approved
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