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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: A325023 A138876 A326145 * A325637 A336702 A325654

Adjacent sequences:  A174630 A174631 A174632 * A174634 A174635 A174636

KEYWORD

nonn

AUTHOR

Michel Lagneau, Mar 24 2010

STATUS

approved

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Last modified August 12 17:17 EDT 2020. Contains 336439 sequences. (Running on oeis4.)